Class 9 Maths - Number Systems

Following are the five rational numbers which are smaller than $2$ $\Rightarrow$ $1,\dfrac{1}{2},0,-1,\dfrac{-1}{2}$ .
If true then enter $1$ and if false then enter $0$ .

The given numbers are

$1,\dfrac{1}{2},0,-1,\dfrac{-1}{2}$

$-1$ and

$\dfrac{-1}{2}$ are negative rational numbers , which are definitely less than

$2$ .

We know,

$0$ is less than any positive rational number,

so,

$0$ is less than

$2$

Now,

$1$ and

$\dfrac{1}{2}$ are positive rational numbers , but they lie to the left of

$2$ in the number line.

So they are also less than

$2$ .

Therefore, all the five given rational number are less than

$2$ .

Hence, the required answer is

$1(True)$ .

$y^{2}=9$

$y=\sqrt{9}$ or

$y ={3},-{3}$

Therefore, y is a rational number

Simplify and express the result in power notation with positive exponent.

$(-3)^4 \times \left ( \dfrac{5}{3} \right )^4$ is $(m)^4$
$m$ is

$(-3)^4 \times \left ( \frac{5}{3} \right )^4$

$=(-1)^4\times(3)^4 \times \left ( \dfrac{5}{3} \right )^4$

$=3^4\times \dfrac{5^4}{3^4}$

$=(5)^4$

Thus,

$m=4$

Find the value of $(3^{1}+4^{1}+5^{1})^{0}$ .

Any number with a power of zero is equal to one.

Find the value of.

$(3^0+4^{-1}) \times 2^2$

$(3^0+4^{-1}) \times 2^2$ =

$(1 + \frac{1}{4})\times4$ =

$\frac{5}{4}\times{4}$

Some rational numbers are integers, if true then enter $1$ and if false then enter $0$

Integers are

$.... -4, -3, -2, -1, 0, 1, 2, 3...$ and so on.

Rational numbers are those numbers which can be expressed in the form

$\dfrac {p}{q}$ , where p and q are integers and

$q \neq 0$

We can write

$-2$ as

$\dfrac {-2}{1}$ and

$3$ as

$\dfrac {3}{1}$

Hence, some rational numbers are integers.

5 $\sqrt{3}+2\sqrt{3}$ = 7 $\sqrt{6}$
enter $1$ for true and $0$ for false

$\quad \\ 5\sqrt { 3 } +2\sqrt { 3 } =7\sqrt { 3 } \\ Therefore:\quad False.\\$

Is zero a rational number?

We know that, rational numbers are those numbers which can be expressed in the form

$\dfrac {p}{q}$ , where p and q are integers and

$q \neq 0$

We know that

$0$ divided by any number is equal to

$0$ .
So,

$0$ can be expressed as

$\dfrac {0}{5}$ or

$\dfrac {0}{6}$

Hence, zero is a rational number.

Zero can be written in the form of $\displaystyle\ \frac{p}{q}$ , where p and q are integers and q $\neq$ 0?
If above statement is true then enter $1$ and if false then enter $0.$

We know that

$0$ divided by any number is equal to

$0$ .
Rational numbers are those numbers which can be expressed in the form

$\dfrac {p}{q}$ , where p and q are integers and

$q \neq 0$
Now

$\dfrac {0}{5} = 0$ and

$\dfrac {0}{6} = 0$
Hence, the given statement is True.

5 $\sqrt{3} -2\sqrt{3}$ = 3 $\sqrt{3}$
enter $1$ for true and $0$ for false

$\quad \\ 5\sqrt { 3 } -2\sqrt { 3 } =3\sqrt { 3 } \\ Therefore:\quad False.\\$

Rationalise the denominator and simplify:
If $\displaystyle \frac{2}{\sqrt{5}+\sqrt{3}}$ is $\displaystyle \sqrt{m}-\sqrt{3}$ , then $m$ is ?

We rationalize the denominator, $\dfrac { 2 }{ \sqrt { 5 } +\sqrt { 3 } } \times \dfrac { \sqrt { 5 } -\sqrt { 3 } }{ \sqrt { 5 } -\sqrt { 3 } } = \dfrac { 2(\sqrt { 5 } -\sqrt { 3 } ) }{ 5-3 } \\ = \sqrt { 5 } -\sqrt { 3 }$

$\therefore$ Value of m= $5$

Rationalise the denominator and simplify:
If $\displaystyle \frac{3}{4-\sqrt{3}}$ is $\displaystyle \frac{3}{13}(4+\sqrt{m})$ , then $m$ is ?

We rationalize the denominator, $\dfrac { 3 }{ 4-\sqrt { 3 } } * \dfrac { 4+\sqrt { 3 } }{ 4+\sqrt { 3 } } = \dfrac { 12+3\sqrt { 3 } }{ 16-3 } \\ = \dfrac { 3(4+\sqrt { 3 } ) }{ 13 }$

$\therefore$ Value of m = $3$

Rationalise the denominator and simplify:
If $\displaystyle \frac{1}{2+\sqrt{3}}$ is $\displaystyle 2-\sqrt{m}$ , then $m$ is ?

We rationalize the denominator, $\dfrac { 1 }{ 2+\sqrt { 3 } }\times \dfrac { 2-\sqrt { 3 } }{ 2-\sqrt { 3 } } = \dfrac { 2-\sqrt { 3 } }{ 4-3 } = \dfrac { 2-\sqrt { 3 } }{ 1 }$

$\therefore$ Value of m $=3$

Find the product of $3-\sqrt{7}$ and its conjugate.

$\\ (3-\sqrt { 7 } )*(3+\sqrt { 7 } )=9-7=2\\$

Find the value-of $(3^0 - 4^0)\times5^2$ :

$(3^0 - 4^0)*5^2=(1-1)*25=0*25=0$

Express as rational number $\displaystyle \left ( \frac{-5}{7} \right )^{-3}$ is

Value of $m$ is

$\displaystyle \left ( \frac{-5}{7} \right )^{-3}$

$=\displaystyle \left ( \frac{-7}{5} \right )^{3}$

$=\displaystyle (-1^3)\left ( \frac{7^3}{5^3} \right )^{3}$

$=-\frac{343}{125}$

Simplify: $\displaystyle \left (\frac{1}{2} \right )^{-2} + \left (\frac{1}{3} \right )^{-2}+ \left (\frac{1}{4} \right )^{-2}$

$\left (\frac{1}{2} \right )^{-2} + \left (\frac{1}{3} \right )^{-2} + \left (\frac{1}{4} \right )^{-2}$

$= \left (\frac{2}{1} \right )^{2} + \left (\frac{3}{1} \right )^{2} + \left (\frac{4}{1} \right )^{2}$

$=(2^2 + 3^2 + 4^2)$

$= (4 + 9 + 16)$

$= 29$

Express as a power of a rational number with positive exponent $\displaystyle \left [ \left ( \frac{4}{5} \right )^{-2} \right ]^4$
Answer is $(\dfrac{5}{4})^m$
Value of $m$ is

$\displaystyle \left [ \left ( \frac{4}{5} \right )^{-2} \right ]^4$

$=\displaystyle\left ( \frac{4}{5} \right )^{-8}$

$=\displaystyle\left ( \frac{5}{4} \right )^{8}$

$=\displaystyle\left ( \frac{5^{8}}{4^{8}} \right )$

Express as rational number $\displaystyle \left ( \frac{-5}{7} \right )^{-1}$ is $=\displaystyle \frac{-m}{5}$
Value of $m$ is

$\displaystyle \left ( \frac{-5}{7} \right )^{-1}$

$=\displaystyle \frac{-7}{5}$

Evaluate $\displaystyle \left ( \frac{1}{2} \right )^{-5}$

$(\frac { 1 }{ 2 } )^{ -5 }=2^{ 5 }=32$

If $\displaystyle \left \{ \left (\frac{3}{4} \right )^{-1} - \left (\frac{1}{4} \right )^{-1} \right \}=$ $- \dfrac{3}{m}$ , the the value of $m$ is

$\left \{ \left (\cfrac{3}{4} \right )^{-1}- \left (\cfrac{1}{4} \right )^{-1} \right \}^{-1} = \left ( \cfrac{1}{3/4} - \cfrac{1}{1/4} \right )^{-1}$

$= \left ( \cfrac{4}{3} - \cfrac{4}{1} \right )^{-1}$

$= \left (\cfrac{4 - 3 \times 4}{3} \right )^{-1}$

$= \left (\cfrac{-8}{3} \right )^{-1}$

$= \cfrac{1}{-8 /3} = \cfrac{3}{-8} = - \cfrac{3}{8}$

If $4^{-3}$ is $\dfrac{1}{m}$ , then the value of $m$ is:

$4^{-3}$

$=\dfrac{1}{4^3}$

$=\dfrac{1}{64}$

$=\dfrac{1}{m}$

Hence, the value of

$m$ is

$64$ .

$\displaystyle x^{ \frac{-1}{2}}$

$\displaystyle x^{\displaystyle \frac{-1}{2}}$

$=\displaystyle\ \dfrac{1}{x^{\frac{1}{2}}}$

Simplify $\displaystyle \left ( - \frac{2}{5} \right )^{-3} \times \left ( - \frac{2}{5} \right )^{4}$

$\displaystyle \left ( - \frac{2}{5} \right )^{-3} \times \left ( - \frac{2}{5} \right )^{4}$

$=( - \frac{2}{5} )^{-3+4}$

$=( - \frac{2}{5} )^{1}$

$=- \frac{2}{5}$

Find the value of $(3^0 + 4^{-1}) \times 2^2$

$(3^o + 4^{-1}) \times 2^2$

$=(1+\frac{1}{4})\times 2^2$

$=(\frac{1+4}{4})\times 4$

$=(\frac{5}{4})\times 4$

$=5$

Find the value of $\displaystyle \left ( \left ( \frac{4}{7} \right )^{8} \right )^0$ .

Anything raised to the power

$0$ , is

$1$

Find the value of $m$ :
$(2^{-1} \times 4^{-1}) + 2^{-2}=\dfrac{3}{m}$

$\begin{aligned}{}LHS&=\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) + {2^{ - 2}} \\&= \left( {\frac{1}{2} \times \frac{1}{4}} \right) + \frac{1}{{{2^2}}}\\ &= \frac{1}{8} + \frac{1}{4}\\ &= \frac{{1 + 2}}{8}\\ &= \frac{3}{8}\end{aligned}$

Now, we will equate

$LHS$ and

$RHS$ to find the value of

$m,$

$\dfrac{3}{8} = \dfrac{3}{m}$

$\Rightarrow m=4$

Hence, the value of

$m$ is equal to

$8$ .

Simplify $\displaystyle \left ( \frac{-3}{4} \right )^{-5} + \left ( \frac{-3}{4} \right )^{-3}$

$\left ( \frac{-3}{4} \right )^{-5} + \left ( \frac{-3}{4} \right )^{-3}$

$=\left ( \frac{-3}{4} \right )^{-3}[\left ( \frac{-3}{4} \right )^{-2}+1]$

$=\left ( \frac{-4}{3} \right )^{3}[\left ( \frac{-4}{3} \right )^{2}+1]$

$=-\left ( \frac{64}{27} \right )[\left ( \frac{16}{9} \right )+1]$

$=-\left ( \frac{64}{27} \right )[\left ( \frac{16+9}{9} \right )]$

$=-\left ( \frac{64}{27} \right )\left ( \frac{25}{9} \right )$

Find the value of $m$ : $\displaystyle \left ( \frac{5}{2} \right )^3 \times \left ( \frac{5}{2} \right )^7 =\left( \dfrac { 5 }{ 2 } \right) ^{ m}$

$\left( \dfrac { 5 }{ 2 } \right) ^{ 3 }\times \left( \dfrac { 5 }{ 2 } \right) ^{ 7 }$

$=\left( \dfrac { 5 }{ 2 } \right) ^{ 3 +7}$

$=\left( \dfrac { 5 }{ 2 } \right) ^{ 10}$

Find $(0.04)^{-1.5} = ?$

$(0.04)^{-1.5}={\begin{pmatrix}\dfrac{4}{100}\end{pmatrix}}^{-1.5}$

$= {\begin{pmatrix}\dfrac{1}{25}\end{pmatrix}}^{-(3/2)}$

$= (25)^{(3/2)}$

$= (5^2)^{(3/2)}$

$= (5)^{2\times (3/2)}$

$= 5^3$

$= 125.$

The conjugate of $3 - \sqrt {5 + x}$ is $3 + \sqrt {5 + x}$ . Verify

If the conjugate of

$(3 - \sqrt {5 + x})$ is

$(3 + \sqrt {5 + x})$ then the product must be a rational number

$(3 - \sqrt {5 + x}) (3 + \sqrt {5 + x}) = 3^{2} (\sqrt {5 + x})^{2} = 9 - (5 + x) = 9 - 5 - x = 4 - x (4 - x)$ is a rational number.

$\therefore (3 + \sqrt {5 + x})$ is the conjugate of

$(3 - \sqrt {5 + x})$ .

Verify that the conjugate of $\sqrt {3} + \sqrt {2}$ is $\sqrt {3} - \sqrt {2}$

Consider

$(\sqrt {3} + \sqrt {2})(\sqrt {3} - \sqrt {2}) = (\sqrt {3})^{2} - (\sqrt {2})^{2} = 3 - 2 = 1$ which is a rational number. [using the identity

$(a + b)(a - b) = a^{2} - b^{2}]$

$\therefore$ The conjugate of

$(\sqrt {3} + \sqrt {2})$ is

$(\sqrt {3} - \sqrt {2})$ .

Write five rational numbers which are smaller than $2$ .

$2$ can also be written as

$\dfrac{14}{7}$ in the rational form.

So the five rational numbers less than 2 are

$\dfrac{13}{7},\,\dfrac{12}{7},\,\dfrac{11}{7},\,\dfrac{10}{7},\,\dfrac{9}{7}$

There are multiple solutions for this, as

$2$ can be written in various forms like

$\dfrac{10}5$ ,

$\dfrac42$ , ... and so on.

Write five rational numbers which are smaller than $\dfrac{5}{6}$ .

rational numbers smaller than

$\dfrac{5}{6}$ is

$\dfrac{4}{6}, \dfrac{3}{6}, \dfrac{2}{6}, \dfrac{1}{6}, \dfrac{0}{6}$

Write.
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) The rational number that is equal to its negative.

(i) The rational number that does not have reciprocal is

$0.$
(ii)

$\;1$ and

$-1$ both are equal to their reciprocals.
(iii) The rational number that is equal to its negative is

$0.$

Express the following with positive exponents:
${4}^{-7}$

$={4}^{-7}$ (we know

${ a }^{ -m }=\cfrac { 1 }{ { a }^{ m } }$ )

$=\cfrac{1}{{(4)}^{7}}$

Find the value of ' $x$ ' such that
${(-2)}^{x+1}\times {(-2)}^{7}={(-2)}^{12}$

${(-2)}^{x+1}\times {(-2)}^{7}={(-2)}^{12}$

${(-2)}^{x+1+7}={(-2)}^{12}$

${(-2)}^{x+8}={(-2)}^{12}$
As bases are equal, hence

$x+8=12$

$\therefore$

$x=12-8=4$

Write $0, 7, 10, -4$ in $\dfrac{p}{q}$ form

i) 0 can be written as

$\dfrac{0}{1}$ in

$\dfrac{p}{q}$ form.

where, p = 0 & q = 1.

ii) 7 can be written as

$\dfrac{7}{1}$ in

$\dfrac{p}{q}$ form.

where, p = 7 & q = 1.

iii) 10 can be written as

$\dfrac{10}{1}$ in

$\dfrac{p}{q}$ form.

where, p = 10 & q = 1.

iv) -4 can be written as

$\dfrac{-4}{1}$ in

$\dfrac{p}{q}$ form.

where, p = -4 & q = 1.

Can you guess the value of ' $x$ ' when ${2}^{x}=1$

We all know from the laws of exponents that

${a}^{0}=1$

It is given that

${2}^{x}=1$

So,

${2}^{x}=a^0$

$\Rightarrow$

$x=0$

Express the following with positive exponents:
${ \left( \cfrac { 4 }{ 7 } \right) }^{ -3 }$

${ \left( \cfrac { 4 }{ 7 } \right) }^{ -3 }=\cfrac{{4}^{-3}}{{7}^{-3}}$ (

${ a }^{ -m }=\cfrac { 1 }{ { a }^{ m } }$ and

${ a }^{ m }=\cfrac { 1 }{ { a }^{ -m } }$ )

$=\cfrac{{7}^{3}}{{4}^{3}}={ \left( \cfrac { 7 }{ 4 } \right) }^{ 3 }$

Express the following with positive exponents:
$\dfrac{1}{{(5)}^{-4}}$

we know that,

$\because \dfrac{1}{a^m}=a^{-m}$

so,

$\dfrac{1}{5^{-4}}$

$=5^4$

$=625$

Give one example for the following statement:
A number which is rational but not an integer.

In mathematics, a rational number is any number that can be expressed as the quotient or fraction

$\dfrac {p}{q}$ of two integers, Where

$p$ is the numerator and

$q$ is a non-zero denominator. Since

$q$ may be equal to

$1$ . Therefore, every integer is a rational number.

The number which is rational but not an integer is

$\dfrac{3}{7}$ .

Classify the following number as rational or irrational.
$5 - \sqrt {3}$

Given,

$5-\sqrt{3}$

Here

$5$ is a rational number and

$\sqrt{3}$ is an irrational number.

So, difference of rational and irrational number is an irrational number.

So,

$5-\sqrt{3}$ is an irrational number.

Give one example each to the following statements.
A number which is natural number, whole number, integer and rational number

Natural numbers are all numbers

$1, 2, 3, 4…$ They are the numbers you usually count and they will continue on into infinity.

Whole numbers are all natural numbers including

$0$ e.g.

$0, 1, 2, 3, 4…$

Integers include all whole numbers and their negative counterpart e.g.

$…-4, -3, -2, -1, 0,1, 2, 3, 4,…$

All integers belong to the rational numbers. A rational number is a number

$\dfrac{a}{b}$ where

$b\neq 0$

Where a and b are both integers.

Here 4 is an natural number ,whole number ,rational number and integers as

$4=\dfrac{4}{1}$

Then all positive numbers full fill the above statment

Give one example each to the following statements.
An integer which is not a whole number

Integer are every negative and positive numbers

$......-5,-4,-3,-2,-1,0,1,2,3........................$

Whole numbers are all positive number started from

$0$

$0,1,2,3,4,5.....................$

Then integer which is not whole number example are

$-5,-4,-3..............$

Give one example each to the following statements.
A whole number which is not a natural number

Whole numbers start from

$0$

$0,1,2,3,4,....................$

Natural number start from

$1$

$1,2,3,4,5,6,7....................$

So, the number which is a whole number and not a natural number is

$0$ .

Rationalise the denominator and simplify:
$\dfrac {\sqrt {3} + \sqrt {2}}{\sqrt {3} - \sqrt {2}}$

$\dfrac {\sqrt {3} + \sqrt {2}}{\sqrt {3} - \sqrt {2}} \times \dfrac {\sqrt {3} + \sqrt {2}}{\sqrt {3} + \sqrt {2}}$

$= \dfrac {\sqrt {9} + \sqrt {6} + \sqrt {6} + \sqrt {4}}{\sqrt {9} - \sqrt {4}}$

$= \dfrac {3 + 2\sqrt {6} + 2}{3 - 2}$

$= \dfrac {5 + 2\sqrt {6}}{1}$

Number $\dfrac {3}{625}$ is a terminating decimal or a non-terminating repeating decimal? Write it in decimal form.

Any rational number with its denominator in the form of

${ 2 }^{ m }\times { 5 }^{ n }$ , where

$m,n$ are positive integers have a terminating decimal expansion, otherwise it is non-terminating.

Here, the denominator

$625=5^4$ , i.e. composed of factor

$5$ only.

Then,

$\cfrac{3}{625}$ is a terminating decimal and its decimal form is

$0.0048$ .

Rationalise the denominator and simplify:
$\cfrac { \sqrt { 6 } +\sqrt { 3 } }{ \sqrt { 6 } -\sqrt { 3 } }$

Solution:

Rationalisation of denominator:

$\cfrac{\sqrt6+\sqrt3}{\sqrt6-\sqrt3}$

$=\cfrac{\sqrt6+\sqrt3}{\sqrt6-\sqrt3}\times \cfrac{\sqrt6+\sqrt3}{\sqrt6+\sqrt3}$

$=\cfrac{(\sqrt6+\sqrt3)^2}{(\sqrt6)^2-(\sqrt3)^2}$

$=\cfrac{6+3+6\sqrt2}{6-3}$

$=\cfrac{9+6\sqrt2}{3}$

$=3+2\sqrt2$

Is zero a rational number? Give reasons for your answer.

Yes, zero is a rational number.

That is because it can be represented as

$0=\displaystyle\frac{0}{1}=\frac{0}{2}=\frac{0}{3}=\frac{0}{-1}=\ ....$

Rationalising factor of $\left( \sqrt { x+y } \right)$ is ____

If product of two irrational numbers is rational then one is said to be the rationalising factor of other.

$\sqrt{x+y}\times \sqrt{x+y}=x+y$

Thus rationalising factor of

$(\sqrt {x+y})$ is

$\sqrt { x+y }$ .

Find the value of the following:
$\left (\dfrac {7}{4}\right )^{0}\times 3$

$\left (\dfrac {7}{4}\right )^{0}\times 3 = 1\times 3 = 3 \left [\because \left (\dfrac {7}{4}\right )^{0} = 1\right ]$

Find x
$2^{2x \, + \, 2} \, - \, 6^x \, - \, 2 \times 3 ^{2x \, + \, 2} \, = \, 0$

The equation can be written as

$4.2^{2x} \, - \, 2^x\cdot3^x \, - \, 18.3^{2x} \, = \, 0$
Dividing by

$2^x\cdot3^x$ ,we get

$4\left (\frac{2}{3} \right )^x \, - \, 1 \, - \, 18\left (\frac{2}{3} \right )^x \, = \, 0$
Put

$\left (\frac{2}{3} \right )^x \, = \, t$ . Then

$\dfrac{4}{t} \, - \, 1 \, - \, 18t \, = \, 0$
or

$18t^2 \, + \, t \, - \, 4 \, = \, 0$ or

$(2t+1) \, (9t-4) \, = \, 0$

$\therefore \, t \, = \,\frac{4}{9} \, \, or \, \,\frac{-1}{2}$

$\therefore \, t \, = \,\left (\frac{3}{2} \right )^x \, = \, \left (\frac{4}{9} \right ) \, = \, \left (\frac{2}{3} \right )^2 \, = \, \left (\frac{3}{2} \right )^{-2}$
giving

$x \, = \, -2$

Solve the following equation:
$\displaystyle\, \frac{0.2^{x + 0.5}}{\sqrt{5}} = \frac{(0.04)^x}{25}$

$\dfrac{0.2^{x+0.5}}{\sqrt{5}}=\dfrac{(0.04)^{x}}{25}$

$(0.2)^{x+0.5-2{x}}=5^{-1.5}$

$5^{x-0.5}=5^{-1.5}$

$\therefore x=-1$

Solve the following equations.
$\displaystyle\, \left ( \frac{5}{12} \right )^x \cdot\left ( \frac{6}{5} \right )^{x -1} = (0.3)^{-1}$

$(\dfrac{5}{12})^{x}.(\dfrac{6}{5})^{x-1}=0.3^{-1}$

$\dfrac{5}{6}.\dfrac{1}{2^x}=\dfrac{10}{3}$

$2^{x+2}=1$

$\implies x+2=0 \implies x=-2$

Is $0.9$ a rational number?

Any number which can be expressed in the form of

$\cfrac { p }{ q }$ where

$q\neq 0$ is a rational number.

$0.9=\cfrac { 9 }{ 10 }$ So, it is a rational number.

Solve $17^2 \cdot 17^{-5}$

Now,

$17^2 \cdot 17^{-5}$

$=17^{2 -5}$

$=17^{ -3}$

$=\dfrac{1}{17^3}$

Solve:
$\dfrac {\sqrt{24}}{8}+\dfrac {\sqrt{54}}{9}$ .

Solving

$\dfrac{\sqrt{24}}{8}+\dfrac{\sqrt{54}}{9}=\dfrac{\sqrt{6*4}}{8}+\dfrac{\sqrt{6*9}}{9}=\dfrac{2\sqrt{6}}{8}+\dfrac{3\sqrt{6}}{9}=\dfrac{(3+2)\sqrt{6}}{8}=\dfrac{5\sqrt{6}}{8}$

Simplify: $\dfrac { 25 }{ \sqrt { 5 } }$

$\dfrac { 25 }{ \sqrt { 5 } } \\ =\dfrac { 5\times 5 }{ \sqrt { 5 } } \\ =5\sqrt { 5 }$

Find a rational number between $\dfrac{2}{3}$ and $\dfrac{3}{4}$ .

We can write,

$\dfrac23=\dfrac{40}{60}\\\dfrac34=\dfrac{45}{60}$

Number between them

$\dfrac{41}{60},\dfrac{42}{60},\dfrac{43}{60},\dfrac{44}{60}$

Find four rational numbers between $\dfrac{2}{3}$ and $\dfrac{3}{5}$ .

We can find the rational number between them by first making their denominator common.

We can write the given numbers as,

$\dfrac23=\dfrac{200}{300}\\\dfrac35=\dfrac{180}{300}$

So, the 4 rational numbers between them are as follows:

$\dfrac{181}{300},\dfrac{198}{300},\dfrac{190}{300},\dfrac{185}{300}$

Write five rational numbers which are greater than $\dfrac{-3}{2}$ .

A rational number is any number that can be expressed as

$p/q$ of two integers where

$q\neq 0$

The five rational numbers greater than

$\dfrac{-3}{2}$ is

$-1,0,1,\dfrac 12,2$

Prove that $\dfrac{{{{({a^{p + q}})}^2}{{({a^{q + r}})}^2}{{({a^{r + p}})}^2}}}{{{{({a^p}.{a^q}.{a^r})}^4}}} = 1$

$\dfrac {\left (a^{2p+2q}\right) \left (a^{2q+2r}\right) \left (a^{2r+2p}\right)} {a^{4p}.a^{4q}.a^{4r}}$

$\Rightarrow \dfrac {a^{4p+4q+4r}}{a^{4p+4q+4r}}$

$=1$

Find the value of $x$ for which ${ \left( \dfrac { 4 }{ 9 } \right) }^{ 4 }\times { \left( \dfrac { 4 }{ 9 } \right) }^{ -7 }={ \left( \dfrac { 4 }{ 9 } \right) }^{ 2x-1 }$

$\displaystyle \left ( \frac{4}{9} \right )^{4-7} =\left ( \frac{4}{9} \right )^{2x-1}$

$-3=2x-1$

$2x=-3+1$

$x=-1$

How many of the following numbers are rational numbers?
(i) $\dfrac{5}{-8}$ (ii) $\dfrac{-6}{11}$ (iii) $\dfrac{7}{15}$ (iv) $-3$ (v) $0$ (vi) $\dfrac{1}{0}$

All options are rational number accept

$\frac{1}{0}$

As,

$\frac{1}{0}$ = Not defined

So total rational number are 5

If $\frac{{{{\left( { - 2} \right)}^x} \times {{\left( { - 2} \right)}^7}}}{{3 \times {4^6}}} = \frac{1}{{12}}$ , then the value of x-3 is

1315385_1062928_ans_8c97cacabffe41afa2590217904a699b.jpg

Simplify
$(16)^{\dfrac {-1}{4}}x {^{4}\sqrt {16}}$

$\displaystyle (16)^{\dfrac{-1}{4}}x^{4}\sqrt {16}\\ =x^{4}(16)^{\dfrac{-1}{4}+\dfrac{1}{2}}\\ =x^{4}(2^{4})^{\dfrac{1}{4}}=2x^{4}$

Evaluate:-
${3^3} \times \left( {243} \right)\,{\,^{ - \cfrac{2}{3}}}\,\, \times {9^{\,\, - \cfrac{1}{3}}}$

Consider the given expression.

$\Rightarrow {{3}^{3}}\times {{\left( 243 \right)}^\cfrac{-2}{3}}\times {{\left( 9 \right)}^\cfrac{-1}{3}}$

$\Rightarrow {{3}^{3}}\times \dfrac{1}{{{\left( {{3}^{5}} \right)}^\cfrac{2}{3}}}\times \dfrac{1}{{{\left( {{3}^{2}} \right)}^\cfrac{1}{3}}}$

$\Rightarrow {{3}^{\left( 3-\dfrac{10}{3}-\dfrac{2}{3} \right)}}$

$\Rightarrow {{3}^{-1}}$

$\Rightarrow \dfrac{1}{3}$

Hence, this is the required value.

Add $7\sqrt 2 + 5\sqrt 3$ and $\sqrt 2 - 7\sqrt 3$

$7\sqrt{2}+5\sqrt{3}+\sqrt{2}-7\sqrt{3}=(7+1)\sqrt{2}+(5-7)\sqrt{3}$

$=8\sqrt{2}-2\sqrt{3}$

Use the laws of exponents and simplify ${3}^{-7}+{3}^{-4}$

$\left( A \right){3^{ - 7}} + {3^{ - 4}}$

$= {\left( {\dfrac{1}{3}} \right)^7} + {\left( {\dfrac{1}{3}} \right)^4}$

$= \dfrac{1}{{{3^7}}} + \dfrac{1}{{{3^4}}} \times \dfrac{{{3^3}}}{{{3^3}}}$

$= \dfrac{{1 + 27}}{{{3^7}}}$

$= \dfrac{{28}}{{2187}}$

Solve ${2^0} + {1^0} - {4^0} - {7^0}$

$2^o+1^o-4^o-7^o\quad (a^o=1)$

$\therefore=1+1-4-1$

$=0$

Represent $\sqrt {4.5}$ on the number line.

$AB =4.5\ units\ ,\ BC=1\ unit\ ,\ OC=OD=\displaystyle \frac{5.5}{2}=2.75units$

$OD^{2}=OB^{2}+BD^{2}$

$\displaystyle (\frac{4.5+1}{2})^{2}=(\frac{4.5+1}{2}-1)^{2}+(BD)^{2}$

$BD^{2}=(\frac{4.5+1}{2})^{2}-(\frac{4.5-1}{2})^{2}$

$BD^{2}=4.5$

$BD=\sqrt{4.5}units$

so the length of BD will be the required one so mark an arc of length BD on number line it will result I the required length.

975255_1058805_ans_f4b15c57fb894076b59e4c819a65edc0.png

Simplify $\dfrac {(6.7\times 10^{-11})(6 \times 10^{24})(7.4 \times 10^{22})}{(3.84 \times 10^{8})^{2}}$

$\dfrac{{\left( {6.7 \times {{10}^{ - 11}}} \right)\left( {6 \times {{10}^{24}}} \right)\left( {7.4 \times {{10}^{22}}} \right)}}{{{{\left( {3.84 \times {{10}^8}} \right)}^2}}}$

$= \dfrac{{6.7 \times 6 \times 7.4 \times {{10}^{ - 11 + 24 + 22}}}}{{3.84 \times 3.84 \times {{10}^{16}}}}$

$= \dfrac{{6.7 \times 6 \times 7.4 \times {{10}^{35 - 16}}}}{{3.84 \times 3.84}}$

$= \dfrac{{297.48}}{{14.7456}} \times {10^{19}}$

$= 20.17 \times {10^{19}}$

$= 2.02 \times {10^{20}}$

Find the value of ${\left( \cfrac{6}{5} \right)}^{\dfrac{1}{2}}$

${\left( \cfrac{6}{5} \right)}^{\cfrac{1}{2}} = \sqrt{\cfrac{6}{5}} = \sqrt{\cfrac{2 \times 3}{5}} = \cfrac{\sqrt{2} \times \sqrt{3}}{\sqrt{5}} = \cfrac{1.414 \times 1.732}{2.236} = \cfrac{2.449}{2.236} = 1.095$

Hence, the correct answer is 1.095.

Simplify: $\cfrac { { 3 }^{ 5 }\times { 10 }^{ 5 }\times 2^5 }{ { 5 }^{ 7 }\times { 6 }^{ 5 } }$

$\dfrac{{3}^{5}\times{10}^{5}\times25}{{5}^{7}\times{6}^{5}}$

$=\dfrac{{3}^{5}\times{\left(2\times 5\right)}^{5}\times{5}^{2}}{{5}^{7}\times {\left(2\times 5\right)}^{5}}$

$\bigg[$ using

${\left(ab\right)}^{m}={a}^{m}{b}^{m}\bigg]$

$=\dfrac{{3}^{5}\times{2}^{5}\times{5}^{5}\times{5}^{2}}{{5}^{7}\times {2}^{5}\times {3}^{5}}$

$={3}^{5-5}\times {2}^{5-5}\times {5}^{5+2-7}$

$\bigg[$ using

$\dfrac{{a}^{m}}{{a}^{n}}={a}^{m-n}\bigg]$

$={3}^{\circ}{2}^{\circ}{5}^{\circ}$

$\bigg[\because {a}^{\circ}=1\bigg]$

$=1$

Find the co-efficient of $a^0$ in the expression $a^2+3a+2$ .

Given the expression

$a^2+3a+2$ .

Now the co-efficient of

$a^0$ that is the constant term, and it is

$2$ .

Find the value to three places of decimals of each of the following. It is given that $\sqrt {2}=1.414, \sqrt {3}=1.732, \sqrt {5}=2.236$ and $\sqrt {10}=3.162$
$\dfrac {3}{\sqrt {10}}$

Given that:-

$\sqrt{2} = 1.414$

$\sqrt{3} = 1.732$

$\sqrt{5} = 2.236$

$\sqrt{10} = 3.162$

$\therefore \; \cfrac{3}{\sqrt{10}} = \cfrac{3}{3.162} = 0.9487666 \approx 0.949$

Hence, the correct answer is

$0.949$ .

Simplify: $2+\dfrac{\sqrt{3}}{3}$

Solution:-

$2 + \cfrac{\sqrt{3}}{3} = \cfrac{2 \times 3 + \sqrt{3}}{3} = \cfrac{6 + \sqrt{3}}{3}$

Find the value to three places of decimals of each of the following. It is given that $\sqrt {2}=1.414, \sqrt {3}=1.732, \sqrt {5}=2.236$ and $\sqrt {10}=3.162$
$\dfrac {\sqrt {10}+\sqrt {15}}{\sqrt {2}}$

Solution:-

Given that:-

$\sqrt{2} = 1.414$

$\sqrt{3} = 1.732$

$\sqrt{5} = 2.236$

$\sqrt{10} = 3.162$

$\therefore \; \cfrac{\sqrt{10} + \sqrt{15}}{\sqrt{2}}$

$\Rightarrow \; = \cfrac{\sqrt{10} + \sqrt{3.5}}{\sqrt{2}}$

$\Rightarrow \; = \cfrac{\sqrt{10} + \sqrt{3}.\sqrt{5}}{\sqrt{2}}$

$\Rightarrow \; = \cfrac{3.162 + \left( 1.732 \times 2.236 \right)}{1.414}$

$\Rightarrow \; = \cfrac{3.162 + 3.872752}{1.414}$

$\Rightarrow \; = \cfrac{7.034752}{1.414}$

$\Rightarrow \; = 4.9750721 \approx 4.975$

Hence, the correct answer is

$4.975$ .

Find the value to three places of decimals of each of the following. It is given that $\sqrt {2}=1.414, \sqrt {3}=1.732, \sqrt {5}=2.236$ and $\sqrt {10}=3.162$
$\dfrac{2+\sqrt {3}}{3}$

Given that:-

$\sqrt{2} = 1.414$

$\sqrt{3} = 1.732$

$\sqrt{5} = 2.236$

$\sqrt{10} = 3.162$

$\therefore \; \cfrac{2 + \sqrt{3}}{2} = \cfrac{2 + 1.732}{3} = \cfrac{3.732}{3} = 1.124$

Hence, the correct answer is

$1.124$ .

Find the value to three places of decimals of each of the following. It is given that $\sqrt {2}=1.414, \sqrt {3}=1.732, \sqrt {5}=2.236$ and $\sqrt {10}=3.162$
$\dfrac {\sqrt {2}-1}{\sqrt {5}}$

Given that:-

$\sqrt{2} = 1.414$

$\sqrt{3} = 1.732$

$\sqrt{5} = 2.236$

$\sqrt{10} = 3.162$

$\therefore \; \cfrac{\sqrt{2} - 1}{\sqrt{5}} = \cfrac{1.414 - 1}{2.236} = \cfrac{0.414}{2.236} = 0.1851520 \approx 0.185$

Hence, the correct answer is

$0.185$ .

$\left( {{3^ \circ } + {4^{ - 1}}} \right) \times {2^2}$

1315387_1062945_ans_1138c18e37e74db6bc2895dfbb7c1962.jpg

${\left( {\frac{1}{2}} \right)^{ - 2}}\, + \,{\left( {\frac{1}{3}} \right)^{ - 2}}\, + \,{\left( {\frac{1}{4}} \right)^{ - 2}}$

${\left( {\dfrac{1}{2}} \right)^{ - 2}} + {\left( {\dfrac{1}{3}} \right)^{ - 2}} + {\left( {\dfrac{1}{4}} \right)^{ - 2}}$

$= {2^2} + {3^2} + {4^2}$

$\left[ {{{\left( {\dfrac{1}{a}} \right)}^{ - x}} = {a^x}} \right]$

$= 4 + 9 + 16$

$= 29$

Write any two irrational numbers between $0.23$ and $0.3$ .

Given,

$0.23$ and

$0.3$ are both rational numbers.

There exists infinitely many irrational numbers between two rational numbers.

Irrational numbers are those with non-repeating non-terminating decimal expansion.

So irrational numbers between 0.23 and 0.3 are:

$0.250101101110587394...$ and $0.270100100015897645...$ .

NOTE:

This is two of the possible answers. It might have different solution.

Solve: $32^{1/5}$ .

Now,

$32^{1/5}$

$=(2^5)^{1/5}$

$=2$ .

Multiply $\sqrt 3 (\sqrt 7 - \sqrt 3 )$

$\sqrt3(\sqrt7-\sqrt3)=\sqrt3\times\sqrt7-\sqrt3\times\sqrt3=\sqrt{3\times7}-\sqrt{3\times3}=\sqrt{21}-3$

Convert into form of power. $3 \times 3 \times 5 \times 3 \times 3 \times 5$

$3\times 3\times5\times3\times3\times5$

in the given

$3$ is being repeated 4 times and 5 is being repested 2 times

Therefore,

$3\times3\times\times3\times3\times5\times5$

$3^{4}\times 5^{2}$

Rationalise:
$\dfrac{6}{{9\sqrt 3 }}$ .

$\displaystyle{6 \over {9\sqrt 3 }} = {{6 \times \sqrt 3 } \over {9\sqrt 3 \times \sqrt 3 }}$

$\displaystyle = {{6\sqrt 3 } \over {9 \times 3}}$ $\displaystyle = {{2\sqrt 3 } \over 9}$

$\therefore$ On rationalisation, $\displaystyle{6 \over {9\sqrt 3 }} = {{2\sqrt 3 } \over 9}$ .

$Write\quad 5\quad rational\quad numbers\quad smaller\quad than\quad \cfrac { 5 }{ 6 }$

5 rational numbers smaller than

$\dfrac{5}{6}$ are:

$0,\ \dfrac{1}{6},\ \dfrac{2}{6},\ \dfrac{3}{6},\ \dfrac{4}{6}$

Find the five rational numbers between $\frac{1}{6}$ and $\frac{1}{3}$

$(\frac{1}{3})\times (\frac{12}{12})=(\frac{12}{36})\\=(\frac{1}{6})\times (\frac{6}{6})=(\frac{6}{36})\\so\> rational\> numbers\> between \>are \\(\frac{7}{36}),(\frac{8}{36}),(\frac{9}{36}),(\frac{10}{36}),(\frac{11}{36})$

Show that $\dfrac{1}{{\left( {3 - \sqrt 8 } \right)}} - \dfrac{1}{{\left( {\sqrt 8 - \sqrt 7 } \right)}} + \dfrac{1}{{\left( {\sqrt 7 - \sqrt 6 } \right)}} - \dfrac{1}{{\left( {\sqrt 6 - \sqrt 5 } \right)}} + \dfrac{1}{{\left( {\sqrt 5 - 2} \right)}} = 5$

Consider the given L.H.S.

$=\dfrac{1}{3-\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-2}$

$=\dfrac{1}{3-\sqrt{8}}\times \dfrac{3+\sqrt{8}}{3+\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{7}}\times \dfrac{\sqrt{8}+\sqrt{7}}{\sqrt{8}+\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{6}}\times \dfrac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{5}}\times \dfrac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}$

$+\dfrac{1}{\sqrt{5}-2}\times \dfrac{\sqrt{5}+2}{\sqrt{5}+2}$

$=\dfrac{3+\sqrt{8}}{9-8}-\dfrac{\sqrt{8}+\sqrt{7}}{8-7}+\dfrac{\sqrt{7}+\sqrt{6}}{7-6}-\dfrac{\sqrt{6}+\sqrt{5}}{6-5}+\dfrac{\sqrt{5}+2}{5-4}$

$=3+\sqrt{8}-\sqrt{8}-\sqrt{7}+\sqrt{7}+\sqrt{6}-\sqrt{6}-\sqrt{5}+\sqrt{5}+2$

$=3+2$

$=5$

Hence, proved.

Find the value of $(-i)^{100}$

Now,

$(-i)^{100}$

$=i^{100}$

$=(i^4)^{25}$

$=1^{25}$ [ Since

$i^4=1$ ]

$=1$ .

Find ${ 1 }^{ 0 }\times { 2 }^{ 0 }+{ 3 }^{ 0 }\times { 4 }^{ 0 }+{ 5 }^{ 0 }\times { 6 }^{ 0 }$

${ 1 }^0\times { 2 }^{ 0 }+{ 3 }^{ 0 }\times { 4 }^{ 0 }+{ 5 }^{ 0 }\times { 6 }^{ 0 }\\ =1\times 1+1\times 1+1\times 1\\ =1+1+1\\ =3$

The value of ${(64)^{1/3}}$

We have,

${{\left( 64 \right)}^{\frac{1}{3}}}$

$={{\left( {{\left( 4 \right)}^{3}} \right)}^{\frac{1}{3}}}$

$={{\left( 4 \right)}^{\frac{3}{3}}}$

$=4$

Hence, this is the answer.

Which are two rational number between $\dfrac{6}{5}$ and $\dfrac{7}{5}$ .

$\dfrac{6}{5}$ and

$\dfrac{7}{5}$ can be written as

$\dfrac{12}{10}$ and

$\dfrac{14}{10}$

and one rational can be inserted, i.e.,

$\dfrac{13}{10}$

Again,write

$\dfrac{6}{5}$ and

$\dfrac{7}{5}$ as

$\dfrac{18}{15}$ and

$\dfrac{21}{15}$ and you can inserte two rational number i.e.,

$\dfrac{19}{15}$ and

$\dfrac{20}{15}$

Now, two rational number between

$\dfrac{6}{5}$ and

$\dfrac{7}{5}$

are

$\dfrac{13}{10}$ and

$\dfrac{19}{15}$

$\frac{1}{{\sqrt 9 - \sqrt 8 }}$ is equal to :

$\dfrac{1}{\sqrt9-\sqrt8}=\dfrac{1}{\sqrt9-\sqrt8}\times\dfrac{\sqrt9+\sqrt8}{\sqrt9+\sqrt8}$ (Multiplying and dividing by conjugate of dinominator)

$=\dfrac{\sqrt9+\sqrt8}{(\sqrt9)^2-(\sqrt8)^2}=\dfrac{3-2\sqrt2}{9-8}=3-2\sqrt2$

Simplify and write in exponential form
$P^{3}\times P^{-10}$

We have,

${{P}^{3}}\times {{P}^{-10}}$

$={{P}^{3-10}}$

$={{P}^{-7}}$

$=\dfrac{1}{{{P}^{7}}}$

Hence, this is the answer.

Simplify:
${\left( { - 4} \right)^5}{\left( { - 4} \right)^6}$

Given that:

$\Rightarrow(-4)^5(-4)^6$

$\Rightarrow(-4)^{5+6}$

$\Rightarrow(-4)^{11}$

$\Rightarrow(-1)(4)^{11}$

$\Rightarrow -{4}^{11}$

Hence, this is the answer.

Solve: ${{\left( \dfrac{7}{4\sqrt{3}+1}+\dfrac{7}{4\sqrt{3}-1} \right)}^{2}}$

We have,

${{\left( \dfrac{7}{4\sqrt{3}+1}+\dfrac{7}{4\sqrt{3}-1} \right)}^{2}}$

$\Rightarrow 49{{\left( \dfrac{1}{4\sqrt{3}+1}+\dfrac{1}{4\sqrt{3}-1} \right)}^{2}}$

$\Rightarrow 49{{\left( \dfrac{4\sqrt{3}-1+4\sqrt{3}+1}{\left( 4\sqrt{3}+1 \right)\left( 4\sqrt{3}-1 \right)} \right)}^{2}}$

$\Rightarrow 49{{\left( \dfrac{8\sqrt{3}}{\left( {{\left( 4\sqrt{3} \right)}^{2}}-{{1}^{2}} \right)} \right)}^{2}}$

$\Rightarrow 49{{\left( \dfrac{8\sqrt{3}}{\left( 48-1 \right)} \right)}^{2}}$

$\Rightarrow 49{{\left( \dfrac{8\sqrt{3}}{47} \right)}^{2}}$

$\Rightarrow 49\left( \dfrac{64\times 3}{47\times 47} \right)$

$\Rightarrow \dfrac{9408}{2209}$

Hence, this is the answer.

If $a={x}^{m+n}.{y}^{l};b={x}^{n+l}.{y}^{m}$ and $c={x}^{l+m}.{y}^{n}$ prove that : ${a}^{m-n}.{b}^{n-l}.{c}^{l-m}=1$

$a=x^{m+n}.y^{l}$

$a^{m-n}=(x^{m+n}.y^l)^{m-n}$

$a^{m-n}=x^{m^{2}-n^{2}}.y^{m{l}-n{l}}$

$b=x^{n+l}.y^{m}$

$b^{n-l}=(x^{n+l}y^m)^{n-l}$

$b^{n-l}=x^{n^{2}-l^{2}}.y^{m{n}-m{l}}$

$c=x^{l+m}.y^{n}$

$c^{l-m}=(x^{l+m}.y^n)^{l-m}$

$c^{l-m}=x^{l^{2}-m^{2}}.y^{n{l}-m{n}}$

$a^{m-n}.b^{n-l}.c^{l-m}=x^{m^{2}-n^{2}+n^{2}-l^{2}+l^{2}-m^{2}}.y^{m{l}-n{l}+m{n}-m{l}+n{l}-m{n}}$

$=x^{0}.y^{0}$

$=1$

Simplify $\dfrac{13}{4}+\sqrt{3}$

Consider the given expression,

$\Rightarrow \dfrac{13}{4}+\sqrt{3}$

$\Rightarrow \dfrac{13+4\sqrt{3}}{4}$

$\because \,\sqrt{3}=1.732$

$\Rightarrow \dfrac{13+6.928}{4}$

$\Rightarrow 4.98$

Hence, this is the answer.

Evaluate ${2}^{3}\times {(9)}^{0}\times {3}^{3}$

$2^{3}\times 9^{0}\times3^{2}=6^{3}=216$

Classify the following number as rational or irrational :
$\left( 3+\sqrt { 23 } \right) -\sqrt { 23 }$

On simplifying given number, we get

$3+\sqrt23-\sqrt23$

$\implies 3$

Which is a rational number.

Is zero a rational number ? Justify

Is zero a rational number ?

Yes,

$0$ is a rational number. As

$0$ can be written as

$0= 0/1$ (an any non-zero denomination)

Evaluate the following expression ${\left( {\dfrac{{625}}{{81}}} \right)^{ - 1/4}}$

Consider the given expression, ${{\left( \dfrac{625}{81} \right)}^{\dfrac{-1}{4}}}$

It is known that

$625=5^4$ and

$81=3^4$

${{\left( \dfrac{625}{81} \right)}^{\frac{-1}{4}}}={{\left[ {{\left( \dfrac{5}{3} \right)}^{4}} \right]}^{\frac{-1}{4}}}\\={{\left( \dfrac{5}{3} \right)}^{4\left( \frac{-1}{4} \right)}}$

$= {{\left( \dfrac{5}{3} \right)}^{-1}}\\=\dfrac{3}{5}$

Write any three rational numbers between the two numbers given below.
$0.3$ and $-0.5$

The given numbers are,

$\dfrac{3}{10}$ and

$\dfrac{-5}{10}$

So, Three rational numbers between them are,

$\dfrac{2}{10},\dfrac{1}{10},\dfrac{-1}{10}$

Write three rational numbers that lie between the two given numbers.
$\dfrac{-3}{4}, \dfrac{+5}{4}$

Three rational no. between

$\dfrac{-3}{4}, \dfrac{5}{4}$ are,

$\dfrac{-1}{4},\dfrac{3}{4}, \dfrac{1}{4}$

Write three rational numbers that lie between the two given numbers.
$0, \dfrac{-3}{4}$

Three rational no. between

$0, \dfrac{-3}{4}$ can be found by dividing the following number by a positive integer.

For example:

$\dfrac{(\dfrac{-3}{4})}{2} = \dfrac{-3}{8}$
Simillarly if we divide it by 4 and 8 we get,

$\dfrac{-3}{16}, \dfrac{-3}{32}$

Therefore the three rational are

$\dfrac{-3}{8}, \dfrac{-3}{16}, \dfrac{-3}{32}$ that lies between

$0, \dfrac{-3}{4}.$

Evaluate
${\left( {{{17}^2} - {8^2}} \right)^{1/2}}$

Consider the given expression,

${{\left( {{17}^{2}}-{{8}^{2}} \right)}^{\dfrac{1}{2}}}=\sqrt{\left( {{17}^{2}}-{{8}^{2}} \right)}$

$=\sqrt{289-64}=\sqrt{225}=\sqrt{{{15}^{2}}}$

$=15$

Write three rational numbers that lie between the two given numbers.
$\dfrac{7}{9}, -\dfrac{5}{9}$

Three rational no. between

$\dfrac{7}{9}, \dfrac{-5}{9}$ are,

$\dfrac{1}{9},\dfrac{2}{9}, \dfrac{4}{9}$

Solve : $4 + 4 \sqrt{3}$

$4+4\sqrt{3}$

$\Rightarrow 4\times 1+4 \times \sqrt{3}$

$\Rightarrow 4(1+\sqrt{3})$

List five rational numbers between:
$-2$ and $-1$

To find 5 rational no: b/w : -2 & -1

So we will multiply the no : by

$\dfrac{6}{6}$

$\therefore -2\times \dfrac{6}{6} = \dfrac{-12}{6}$

$-1\times \dfrac{6}{6} = \dfrac{-6}{6}$

$\therefore$ 5 rational w are :-

$\left \{ \dfrac{-11}{6},\dfrac{-10}{6},\dfrac{-9}{6},\dfrac{-8}{6},\dfrac{-7}{6} \right \}$

Solve the following equation and find the value of $m$ in $2^{m-3}=1$

${2}^{m-3}=1...(1)$

${2}^{0}=1...(2)$

from (1) and (2)

$m-3=0$

$m=3$

Insert three rational number between $-\dfrac{1}{3}$ and $-\dfrac{2}{3}$ .

$\dfrac{-1}{3}=\dfrac{-1}{3}\times \dfrac{4}{4}=\dfrac{-4}{12}, \\ \dfrac{-2}{3}= \dfrac{-2}{3}\times \dfrac{4}{4}=\dfrac{-8}{12}$

Three rational numbers between

$\dfrac{-1}{3}$ and

$\dfrac{-2}{3}$ are

$\dfrac{-5}{12},\dfrac{-6}{12},\dfrac{-7}{12}$

Find the rational number between $\dfrac {2}{7}$ and $\dfrac {3}{4}$

$\longrightarrow \dfrac{2}{7}=\dfrac{2\times 4}{7\times 4}\left[\dfrac{8}{28}\right]$

$\dfrac{3}{4}=\dfrac{3\times 7}{4\times 7}=\boxed{\dfrac{21}{28}}$

$\therefore$ Rational number between

$\dfrac{2}{7}$ and

$\dfrac{3}{4}$ is

$\dfrac{9}{28},\dfrac{10}{28},\dfrac{11}{28}.....\dfrac{20}{28}$

Write any three rational numbers between the two numbers given below.
$-4.5$ and $-4.6$

The given numbers are

$-4.5, -4.6$

These numbers can also be written as:

$\dfrac{-45}{10}$ and

$\dfrac{-46}{10}$

So, Three rational numbers between them are,

$\dfrac{-45.1}{10},\dfrac{-45.2}{10},\dfrac{-45.3}{10}$

$-4.51,-4.52,-4.53$

Find the value of $({3^0} + {4^{ - 1}}) \times {2^2}$

Let

$a=(3^{o} + 4^{-1}) \times 2^{2}$

$=\left(1+\dfrac{1}{4}\right) \times 4$

$=\dfrac{5}{4} \times 4$

$a=5$

$\therefore (3^{o} + 4^{-1}) \times 2^{2}=5$

$x = {\left( {\dfrac{5}{7}} \right)^{ - 5}} \times {\left( {\dfrac{7}{5}} \right)^{ - 7}}$ then find the value of 2x + 1

${ x }={ (\dfrac { 5 }{ 7 } ) }^{ -5 }\times )^{ -7 }$

$\\ ={ (\dfrac { 5 }{ 7 } ) }^{ 2 }$

$\\ { 2x }+{ 1 }={ 2 }\times { \dfrac { 25 }{ 49 } }+{ 1 }$

$\\ \quad \quad \quad \quad =\dfrac { 99 }{ 49 }$

Simplify:
${2}^{3/2}$ to $2\sqrt{2}$

$2^{3/2}$ =

$(2^{1/2})^{3}$

$(2^{3})^{1/2}$

$8^{1/2}$

$\sqrt{2\times 2\times 2}$

$2\sqrt{2}$

Write any three rational numbers between the two numbers given below.
$-2.3$ and $-2.33$

Given rational numbers are

$-2.3$ and

$-2.33$

The given numbers can be written as,

$\dfrac{-23}{10}$ and

$\dfrac{-23.3}{10}$

So, Three rational numbers between them are:

$\dfrac{-23.1}{10},\dfrac{-23.2}{10},\dfrac{-23.25}{10}$

Write any three rational numbers between the two numbers given below.
$5.2$ and $5.3$

The given numbers are,

$\dfrac{52}{10}$ and

$\dfrac{53}{10}$

So, Three rational numbers between them are,

$\dfrac{52.1}{10},\dfrac{52.2}{10},\dfrac{52.3}{10}$

Define an irrational number.

$Irrational$

$numbers$

$are$

$real$

$numbers$

$that$

$cannot$

$be$

$written$

$in$

$\dfrac{p} {q}$ form.

The decimal representation of $\cfrac{6}{1250}$ will terminate after how many places of decimal?

On simplifying

$\dfrac{6}{1250}$ ,

we get,

$\dfrac{6}{1250}=0.0048$ .

We see that after the decimal point, there are 4 places of decimal

Hence, it will terminate after 4 places of decimal.

If $\dfrac{p}{q}=\left(\dfrac{3}{2}\right)^{-2}\div \left(\dfrac{6}{7}\right)^{0}$ , find the value of $\left(\dfrac{p}{q}\right)^{-3}$

given,

$\dfrac{p}{q}=\left ( \dfrac{3}{2} \right )^{-2}\div \left ( \dfrac{6}{7} \right )^0$

$\Rightarrow \dfrac{p}{q}=\left ( \dfrac{1}{\frac{3}{2}} \right )^{2}\div 1$

$\dfrac{p}{q}=\dfrac{4}{9}$

$\left ( \dfrac{p}{q} \right )^{-3}$

$=\left ( \dfrac{1}{\frac{p}{q}} \right )^{3}$

$=\left (\dfrac{q}{p} \right )^3$

$=\left (\dfrac{9}{4} \right )^3$

$=\dfrac{729}{64}$

Evaluate : $(-4)^{5}\div (4)^{8}$

(negative to power of odd is negative)

$\frac{(-4)^5 }{ (4)^8 } = -\frac{4^5}{4^8}= -4^{(5-8)}$ (

$\because \frac{x^m}{x^n} =x^{m-n}$ )

$= -4^{-3}$

$=\frac{-1}{64}$

Write the rational numbers which are smaller than $2$ .

Consider the rational number smaller than $2$ ,

We know that, there are infinite rational number smaller than $2$ like,

$1,0,-1,-2,3........\infty$ and $\dfrac{3}{2},\dfrac{1}{2},\dfrac{-1}{2},\dfrac{-3}{2}.......\infty$

Hence, this is the answer.

Solve :
$(14x^{3} \times 2x^{4} \times 8x^{8}) \div 7x^{3}.$

$(14x^3 \times 2x^4 \times 8x^8) \div 7x^3$

$= 2\times 2x^4 \times 8x^8$

$= 32x^{12}$ (Ans)

Simplify the following using laws of exponents.
${\left( {\dfrac{3}{5}} \right)^4} \times {\left( {\dfrac{3}{5}} \right)^3} \times {\left( {\dfrac{3}{5}} \right)^8}$

We have,

${{\left( \dfrac{3}{5} \right)}^{4}}\times {{\left( \dfrac{3}{5} \right)}^{3}}\times {{\left( \dfrac{3}{5} \right)}^{8}}$

$\Rightarrow {{\left( \dfrac{3}{5} \right)}^{4+3+8}}$

$\Rightarrow {{\left( \dfrac{3}{5} \right)}^{15}}$

Hence, this is the answer.

Solve:
$\left( {{2^3} \times {2^5}} \right) \div {2^8}$

$\cfrac{(2^3\times 2^5)}{2^8}=\cfrac{2^8}{2^8}=1$

List five rational numbers between $\dfrac{1}{2}$ and $\dfrac{2}{3}$ .

$\cfrac{1}{2}$ &

$\cfrac{2}{3}=\cfrac{3}{6}$ &

$\cfrac{4}{6}$

$=\cfrac{18}{36}$ &

$\cfrac{24}{36}$

$=\cfrac{19}{36},\cfrac{20}{36},\cfrac{20}{36},\cfrac{21}{36},\cfrac{22}{36},\cfrac{23}{36}$

Insert five rational numbers between $\dfrac { 3 }{ 5 }$ and $\dfrac { 2 }{ 3 }$

LCM of

$5, 3=15$

$\Rightarrow \dfrac{3}{5}=\dfrac{3\times 3}{5\times 3}\\\ \ \ \ \ \ \ \ =\dfrac{9}{15}$

And

$\dfrac{2}{3}=\dfrac{2\times 5}{3\times 5}\\\ \ \ =\dfrac{10}{15}$

Multiplying numerator and denominator by

$6$ we get

$\Rightarrow \dfrac{9\times 6}{15\times 6}=\dfrac{54}{90}$

$\Rightarrow \dfrac{10\times 6}{15\times 6}=\dfrac{60}{90}$

Hence the five rational numbers between

$\dfrac{54}{90}$ and

$\dfrac{60}{90}$ are:

$\dfrac{55}{90}, \dfrac{56}{90}, \dfrac{57}{90}, \dfrac{58}{90}, \dfrac{59}{90}$ .

Solve: $5\dfrac{1}{2} - \dfrac{3}{4}$

1117399_1278048_ans_be71d3cb23204f92b22999598efc41d8.JPG

Find the value of ${(105)^2}$

1335840_1265904_ans_5982565ad76d4131b67b922a8bec2e47.PNG

Fill in the blanks: $\dfrac a 8$ , so that it lies between $\frac{3}{4}$ and $\frac{1}{2}$ ?

1334546_1246113_ans_33c6cc2591744ff69cc521cad47af3db.PNG

Classify the following numbers as rational or irrational:
$\dfrac{2\sqrt{7}}{7\sqrt{7}}$

1126270_1294237_ans_86a6a182f6ee453590f01dbc92500677.JPG

Simplify the following using laws of expressions
$(2^{3})^{2}\times 2^{8}$

We have,

$\begin{array}{l} { \left( { { 2^{ 3 } } } \right) ^{ 2 } }\times { 2^{ 8 } } \\ { \left( { { 2^{ 3 } } } \right) ^{ 2 } }={ 2^{ 6 } } \\ ={ 2^{ 8 } }\times { 2^{ 6 } } \end{array}$

We know that,

$\begin{array}{l} { a^{ b } }\times { a^{ c } }={ a^{ b+c } } \\ { 2^{ 8 } }\times { 2^{ 6 } }={ 2^{ 6+8 } }={ 2^{ 14 } } \\ ={ 2^{ 14 } } \\ { 2^{ 14 } }=16384 \\ =16384 \end{array}$

Then,

We get

$16384$ .

Simplify $(a^{3} \times a^{-2} \times a^{4})^{-2}$

It is known that

$a^m\times a^n=a^{m+n}$

$(a^{3}\times a^{-2}\times a^{4})^{-2}=(a^{3-2+4})^{-2}$

$=(a^5)^{-2}$

$=(a^{5\times (-2)}$

$=(a^{-10})$

Prove that the sum of two irrational numbers given by $3 + \sqrt { 2 }\ \& \ 3 - \sqrt { 2 }$ is a rational number

Sum of the given two irrational numbers

$=3+\sqrt{2}+3-\sqrt{2}$

$= 3+3=6$ which is a rational number

Insert three rational numbers between $\dfrac {1}{4}$ and $\dfrac {7}{8}$ .

1206956_1298107_ans_3092d541ed2340a485e13b7de70d3db8.jpg

Evaluate:
${ \left\{ { \left( { 3 }^{ 2 }+{ 4 }^{ 2 } \right) }^{ \frac { 1 }{ 2 } } \right\} }^{ 3 }$

$\begin{array}{l} { \left\{ { { { \left( { { 3^{ 2 } }+{ 4^{ 2 } } } \right) }^{ \frac { 1 }{ 2 } } } } \right\} ^{ 3 } } \\ ={ \left\{ { { { \left( { 9+16 } \right) }^{ \frac { 1 }{ 2 } } } } \right\} ^{ 3 } } \\ ={ \left\{ { { { \left( { 25 } \right) }^{ \frac { 1 }{ 2 } } } } \right\} ^{ 3 } } \\ ={ \left\{ { { { \left( 5 \right) }^{ 2\times \frac { 1 }{ 2 } } } } \right\} ^{ 3 } } \\ ={ \left\{ 5 \right\} ^{ 3 } } \\ =125 \end{array}$

Simplify the following
$7 \sqrt{18}-9 \sqrt{8}+12\sqrt{32}-3\sqrt{50}$

$y =7\sqrt{18}-9\sqrt{8}+12\sqrt{32}-3\sqrt{50}$

$y =7\sqrt{3^{2},2}-9\sqrt{2^{2},2}+12\sqrt{4\times 4\times 2}-3\sqrt{5^{2}2}$

$y= 21\sqrt{2}-18\sqrt{2}+48\sqrt{2}-15\sqrt{2}$

$y=36\sqrt{2}$

1175632_1348057_ans_023bd74bf46249e3abff12411d32c546.jpg

Rationalise it

1338638_1345033_ans_8e295652c6cf47439d5603635a244b54.jpg

$\left[1-\dfrac{1}{a}\right]^3$

We have,

$\left[1-\dfrac{1}{a}\right]^3$

$\Rightarrow 1^3-\left(\dfrac{1}{a}\right)^3-3\times 1 \times \left(\dfrac{1}{a}\right)\left[1-\dfrac{1}{a}\right]$

$\Rightarrow 1-\dfrac{1}{a^3}-\left(\dfrac{3}{a}\right)\left[1-\dfrac{1}{a}\right]$

$\Rightarrow 1-\dfrac{1}{a^3}-\dfrac{3}{a}+\dfrac{3}{a^2}$

Hence, this is the answer.

solve:
${ ({ 3 }^{ -1 } }+{ 4 }^{ -1 }+{ 5 }^{ -1 })^{ 0 }$

$\textbf{Use law of exponents}$

$\text{Given,}\ (3^{-1}+4^{-1}+5^{-1})^0$

$\text{Any non-zero number to the power 0 is 1}$

$\therefore \textbf{The value of}\ \mathbf{(3^{-1}+4^{-1}+5^{-1})^0}\ \textbf{is 1}$

Which is the rational number that is equal to its negative?

The rational number

$0$ is equal to its negative.

What are terminating decimals ?

The rational number which decimal expansion that terminates after a finite number of steps in the process of division, is called a terminating decimal. Example,

$1.25$ ,

$3.14$ , etc.

Solve :
$(4 - p)^3$

We have,

$(4-p)^3$

$\Rightarrow 4^3-p^3-3\times 4\times p(4-p)$

$\Rightarrow 64-p^3-12p(4-p)$

$\Rightarrow 64-p^3-48p+12p^2$

Hence, this is the answer.

Simplify:
$(2^{-1}+3^{-1})^{-1}\div 4^{-1}$

$(2^{-1}+3^{-1})^{-1}\div 4^{-1}$

we know

$a^{-1}=$

$\frac{1}{a}.$

$(\frac{1}{2}+\frac{1}{3})^{-1}\div \frac{1}{4}$

$(\frac{5}{6})^{-1}\div \frac{1}{4}$

$\Rightarrow \frac{6/5}{1/4}$ =

$\frac{24}{5}$

1211546_1362699_ans_74e0347dd6894f89817530846117eca6.JPG

Insert a rational and an irrational number between $2$ and $3$ .

We know, if

$a$ and

$b$ are the two rational numbers, then a rational number between them can be

$\dfrac{a+b}{2}$

and irrational number is

$\sqrt{ab}$ , provided

$ab$ is not a perfect square.

Then, a rational number between

$2$ and

$3 = \cfrac{2 + 3}{2} = \cfrac{5}{2} = 2.5$ .

An irrational number between

$2$ and

$3$ is

$\sqrt{2\times3}=\sqrt 6$ .

Using laws of exponents, simplify and write the answer in exponential form: ${ 7 }^{ x }\times { 7 }^{ 2 }$

Let

$a={7}^{x}\times{7}^{2}$

using

${a}^{m}\times{a}^{n}={a}^{m+n}$

$a={7}^{x+2}$

Insert a rational number between $-\dfrac { 2 }{ 5 } and+\dfrac { 1 }{ 2 }$

$-\dfrac{2}{5}=-\dfrac{4}{10}$

$\dfrac{1}{2}=\dfrac{5}{10}$

Hence, the rational number between

$-\dfrac{2}{5}$ and

$\dfrac{1}{2}$ is

$\dfrac{1}{10}$ .

Using laws of exponents, simplify and write the answer in exponential form ${ 2 }^{ 5 }\times { 5 }^{ 5 }$

1305629_1381260_ans_b1740cca4feb46779b84ddc77c2b778a.jpeg

Insert a rational number between $\dfrac {3}{4}$ and $1$ without using $\dfrac {a+b}{2}$ formula.

We can write

$\dfrac{3}{4}$ as

$\dfrac{6}{8}$

and

$1$ can be written as

$\dfrac{8}{8}$

Hence, the rational number between

$\dfrac 68$ and

$\dfrac 88$ is

$\dfrac{7}{8}$

Rationalise: $5+\sqrt{3}$

given,

$5+\sqrt{3}$

$=5+\sqrt{3}\times\dfrac{5-\sqrt{3}}{5-\sqrt{3}}$

$=\dfrac{25-3}{5-\sqrt{3}}$

$=\dfrac{22}{5-\sqrt{3}}$

Solve
$3 ^ { 4 } \times 2 ^ { 4 }$

We have,

$3^4\times 2^4$

$\Rightarrow 81 \times 16$

$\Rightarrow 1296$

Hence, this is the answer.

Using laws of exponents, simplify and write the answer in exponential form :
${ 3 }^{ 2 }\times { 3 }^{ 4 }\times { 3 }^{ 8 }$

${3}^{2}\times{3}^{4}\times{3}^{8}$

$={3}^{2+4+8}$ using

${a}^{m}\times{a}^{n}={a}^{m+n}$

$={3}^{14}$

Express the following numbers in usual form.
$3\times {10}^{-8}$

1308513_1371646_ans_af913f5fef0e40dfb0cd772e102c4b96.jpeg

Rationalise: $4+\sqrt{5}$

$4+\sqrt{5}$

$=4+\sqrt{5}\times\dfrac{4-\sqrt{5}}{4-\sqrt{5}}$

$=\dfrac{16-5}{4-\sqrt{5}}$

$=\dfrac{11}{4-\sqrt{5}}$

Using laws of exponents, simplify and write the answer in exponential form: ${ a }^{ 3 }\times { a }^{ 2 }$

Let

$z={a}^{3}\times{a}^{2}$

We know that,

${a}^{m}\times{a}^{n}={a}^{m+n}$

$z={a}^{3+2}$

$z={a}^{5}$

Fill in the blanks :
If the integer $p$ and $q$ have no common divisor other than $1$ and $q$ is positive, then the rational number $\dfrac{p}{q}$ is said to be in........... .

Answer : "Standard form of a rational number"

If the integer

$p$ and

$q$ have no common divisor other than

$1$ and

$q$ is positive, then the rational number

$\dfrac{p}{q}$ is said to be in standard form.

Simplify and express following in exponential form :
$({3^{0}+ 2^{0})\times 5^{0}}$

$(3^o+2^o)\times 5^o=(1+1)\times 1=2\times 1=2=2^1$ .
1207977_1392476_ans_ae1ebf5c3ece406094413e5832da3100.jpg

Express each of the following exponential expressions as a rational number. ${ (\dfrac { 2 }{ 5 } ) }^{ (-3) }$

$\left(\dfrac{2}{5}\right)^{-3}=\left(\dfrac{5}{2}\right)^3=\dfrac{5^3}{2^3}=\dfrac{125}{8}$ .
1209710_1462198_ans_eef6340a1a954a33b3ac160364f31c3b.jpg

Give examples of irrational numbers between $11$ and $16$ .

We know that

$11^2=121$ and

$16^2=256$ .

Therefore, square root of any number between

$121$ and

$256$ which is not a perfect square would be an irrational between

$11$ and

$16$ .

For Example,

$\sqrt{122}, \sqrt{200}$

Simplify $3\sqrt { 2 } +\sqrt { 8 } -\sqrt { 5 }$ .

$3\sqrt{2}+\sqrt{8}-\sqrt{5}$

$=3\sqrt{2}+\sqrt{4\times 2}-\sqrt{5}$

$=3\sqrt{2}+2\sqrt{2}-\sqrt{5}$

$=5\sqrt{2}-\sqrt{5}$

If $\dfrac { 5+\sqrt { 6 } }{ 5-\sqrt { 6 } } =a+b\sqrt { 6 } ,$ then find the value of 'a' and 'b'.

1317464_1489881_ans_6d9d647bf9f340c2b488ad3310da3868.JPG

Solve:
${5^{x + 1}} = {\left( {\sqrt {25} } \right)^{3x - 1}}$

$5^{x+1}=(\sqrt{25})^{3x-1}$

$\Rightarrow 5^{x+1}=5^{3x-1}$

comparing powers,

$x+1=3x=1$

$\Rightarrow 2=2x$

$x=1$

1195950_1414935_ans_6665892f08bf40799ae33841fffb0d6e.jpg

Solve:
$\dfrac{4+\sqrt3}{4-\sqrt3}+\dfrac{4-\sqrt3}{4+\sqrt3}$

$\dfrac{4+\sqrt3}{4-\sqrt3}+\dfrac{4-\sqrt3}{4+\sqrt3}\\\,\dfrac{(4+\sqrt3)^2+(4-\sqrt3)^2}{(4+\sqrt3)(4-\sqrt3)}\\\\\dfrac{16+3+8\sqrt3+16+3-8\sqrt3}{16-3}=\dfrac{38}{13}$

Express the following in exponential form:
$6\times 6\times 6\times 6$

$6\times 6\times 6\times 6={6}^{4}$

Find the value of:
${2}^{6}$

${\textbf{Step -1: Multiply the given number to the power it is raised to}}{\textbf{.}}$

${2^6} = {\left( {{2^3}} \right)^2}$

${\textbf{Step - 2: Evaluate the multiples to get the answer}}{\textbf{.}}$

${2^3} \times {2^3} = 64$

${\textbf{Hence, the value of }}$

${\mathbf{{2^6} = 64.}}$

Find the value of :
${11}^{2}$

${11}^{2}=11\times 11=121$

Express $1.27$ in the form of $\dfrac p q$ .

Let

$x = 1.27272727...$

$\Rightarrow 100 x= 127.2727....$

we subtract

$100 x -x$ we will get,

$\Rightarrow 100 x - x = 127.2727 - 1.272727$

$\Rightarrow 99 x = 126 .000$

$\therefore x = \dfrac{126}{99}=\dfrac{14}{11}$

Find the value of ${5}^{4}$

${5}^{4}=5\times 5\times 5\times 5\times 5=625$

Write any four rational numbers larger than $0$ and smaller than $\dfrac { 5 }{ 6 } .$

$0$ can also be written as

$\dfrac{0}{6}$ .

Now both the numbers have same denominators.

$\dfrac { 1 }{ 6 }$ ,

$\dfrac { 2 }{ 6 }$ ,

$\dfrac { 3 }{ 6 }$ ,

$\dfrac { 4 }{ 6 }$ are four rational number larger than

$0$ and smaller than

$\dfrac { 5 }{ 6 } .$

Simplify:
$2\times{10}^{3}$

$2\times{10}^{3}=2\times 10\times 10\times 10=2000$

Represent geometrically $\sqrt {5.6}$ on the number line.

Presentation of

$\sqrt {5.6}$ on number line:

Mark the distance

$5.6\ units$ from a fixed points

$A$ on a given line to obtain a point

$B$ such that

$AB = 5.6\ units$ . From

$B$ , mark a distance of

$1\ unit$ and mark the new points as

$C$ . Find the mid-point of

$AC$ and mark the points as

$O$ . Draw a semicircle with centre

$O$ and radius

$OC$ . Draw a line perpendicular to

$AC$ passing through

$B$ and intersecting the semicircle at

$D$ . Then

$BD = \sqrt {5.6}$ .

Now, draw an arc with centre

$B$ and radius

$BD$ , which intersects the number line in

$E$ .

Thus,

$E$ represent

$\sqrt {5.6}$ .

1904620_1550622_ans_026b0d6b7c1b430fb76f2cf244c2ce48.png

Find the value:
${9}^{3}$

We know that,

$a^3=a\times a\times a$

$\therefore \ {9}^{3}=9\times 9\times 9\times=729$

Hence,

$9^3=729$ .

Find five rational numbers between $\dfrac{-3}{5}$ and $\dfrac{-1}{2}$ .

Required: $5$ rational numbers

First of all, make the denominators of given rational numbers the same

so, $\dfrac{-3}{5}=\dfrac{-3 \times 2}{5 \times 2}=\dfrac{-6}{10}$ and

$\dfrac{-1}{2}=\dfrac{-1\times 5}{2\times 5}=\dfrac{-5}{10}$

Now, multiply both numerator and denominator of given rational numbers by $5+1=6$

$\dfrac{-6}{10} \times \dfrac{6}{6}=\dfrac{-36}{60}$ and

$\dfrac{-5}{10}\times \dfrac{6}{6}=\dfrac{-30}{60}$

Five rational numbers between $\dfrac{-30}{60}$ and $\dfrac{-36}{60}$ are $\dfrac{-31}{60} , \dfrac{-32}{60} , \dfrac{-33}{60} , -\dfrac{34}{60} , \dfrac{-35}{60}$

$\therefore ~5$ required rational numbers are $\dfrac{-31}{60} , \dfrac{-32}{60} , \dfrac{-33}{60} , -\dfrac{34}{60} , \dfrac{-35}{60}$

Simplify the following :
$\dfrac{(a^{3n-9})^{6}}{a^{2n-4}}$

Given,

$\dfrac{(a^{3n-9})^6}{a^{2n-4}}$

$=\dfrac{a^{18n-54}}{a^{2n-4}}$

$=a^{18n-54-(2n-4)}$

$=a^{18n-54-2n+4}$

$=a^{16n-50}$

Write down the value of ${17}^{0}$

value of any non-zero number raise to zero is 1.

therefore

$17^{0}=1$

Which of the following rational numbers are negative?
$\dfrac{-115}{-197}$

1590905_1676140_ans_7048fb7fe8464c419b80c5a905ce7dd8.jpg

Which of the following rational numbers are positive:
$\dfrac{-21}{13}$

1516045_1676109_ans_bf12cabaf3cf43f6a122bac71356beab.jpg

Simplify
$x^{0}$

For all

$X\in R$

$x^0=1$

Simplify:
${5}^{2}\times{3}^{3}$

${5}^{2}\times{3}^{3}=5\times 5\times 3\times 3\times 3=675$

Simplify:
${x}^{3}{y}^{4}\times{x}^{5}{y}^{3}$

Given:

${x}^{3}{y}^{4}\times{x}^{5}{y}^{3}$

$={x}^{3+5}{y}^{4+3}$

$={x}^{8}{y}^{7}$

Write down the denominator of each of the following rational numbers:
$\dfrac{11}{-34}$

1590777_1676053_ans_612ca5b646ea4ab797ccfba423d9d6a0.jpg

Which of the following rational numbers are negative?
$\dfrac{-5}{-8}$

1590888_1676117_ans_a644ca8d055e4a17a95bb71e8b2efe92.jpg

Given below is a rational number or not?
$\cfrac{5}{-8}$

As all fractions

$(\cfrac pq)$ are rational number.

$\therefore \cfrac{5}{-8}$ is a rational number.

Given below is a rational number or not?
$\cfrac7{15}$

As all fractions

$(\cfrac pq)$ are rational number.

$\therefore$

$\cfrac 7{15}$ is a rational number.

Express the rational number with positive denominator:
$\dfrac{19}{-7}$

Rational number with positive denominator is

$=\dfrac{-19}{7}$

Fill in the blanks so as to make the statement true:
If $\dfrac{p}{q}$ is a rational number, then $q$ cannot be _____

We know,

The numbers of the form a/b, or a number which can be expressed in the form a/b, where ‘a’ and ‘b’ are integers and b ≠ 0, are called rational numbers.

$\dfrac pq$ is a rational number, then

$q$ cannot be zero.

Given below is a rational number or not?
$-3$

$-3$ can be written as

$\cfrac{-3}1$ that is in

$\cfrac pq$ form.

Ans all fractions i.e

$\cfrac pq$ form are rational numbers.

$\therefore -3$ is a rational number.

Given below is a rational number or not?
$6$

$6$ can be written as

$\cfrac61$ that is in

$\cfrac pq$ form.

Ans all fractions i.e

$\cfrac pq$ form are rational numbers.

$\therefore 6$ is a rational number.

Given below is a rational number or not?
$\cfrac{-8}{-12}$

As all fractions

$(\cfrac pq)$ are rational number.

$\therefore \cfrac{-8}{-12}$ is a rational number.

Given below is a rational number or not?
$\cfrac{-6}{11}$

As all fractions

$(\cfrac pq)$ are rational number.

$\therefore \cfrac{-6}{11}$ is a rational number.

Evaluate $\dfrac{2^0+7^0}{5^0}$ .

1680758_1715609_ans_e2c9021e42704267afd73cc9bd69c309.jpg

Write down the numerator and the denominator of the following rational number:
$9$

$9$ can also be written as

$\cfrac{9}{1}$

$\cfrac pq$ , Numerator is

$p$ and Denominator is

$q$

So, in

$\cfrac{9}{1}$

$9=$ Numerator

$1=$ Denominator

Given below is a rational number or not?
$\cfrac10$

All fractions

$\cfrac pq$ are rational numbers only when

$q\neq 0$

$\cfrac10$ has indefinite value due to

$0$ in denominator

So,

$\cfrac10$ is not a rational number

Given below is a rational number or not?
$\cfrac00$

All fractions

$\cfrac pq$ are rational numbers only when

$q\neq 0$

$\cfrac00$ has indefinite value due to

$0$ in denominator

So,

$\cfrac00$ is not a rational number

Write the following integer as a rational number. Write the numerator and the denominator in the following case.
$1$

$1$

It can be written in a rational form as

$\cfrac{1}{1}$

$\therefore 1=$ Numerator

and

$1=$ Denominator

Write down the numerator and the denominator of the following rational number:
$\cfrac8{19}$

$\cfrac8{19}$

$\cfrac pq$ , Numerator is

$p$ and Denominator is

$q$

So, in

$\cfrac8{19}$

$8=$ Numerator

$19=$ Denominator

Write the following integer as a rational number. Write the numerator and the denominator in the following case.
$-3$

$-3$

It can be written in a rational form as

$\cfrac{-3}{1}$

$\therefore -3=$ Numerator

and

$1=$ Denominator

Write down the numerator and the denominator of the following rational number:
$\cfrac{-13}{15}$

$\cfrac{-13}{15}$

$\cfrac pq$ , Numerator is

$p$ and Denominator is

$q$

So, in

$\cfrac{-13}{15}$

$-13=$ Numerator

$15=$ Denominator

Write down the numerator and the denominator of the following rational number:
$\cfrac{-8}{-11}$

$\cfrac{-8}{-11}$

$\cfrac pq$ , Numerator is

$p$ and Denominator is

$q$

So, in

$\cfrac{-8}{-11}$

$-8=$ Numerator

$-11=$ Denominator

Write the following integer as a rational number. Write the numerator and the denominator in the following case.
$5$

$5$

It can be written in rational form as

$\cfrac51$

$\therefore 5=$ Numerator

and

$1=$ Denominator

Given below is a rational number or not?
$0$

$0$ can be written as

$\cfrac01$ that is in

$\cfrac pq$ form.

Ans all fractions i.e

$\cfrac pq$ form are rational numbers.

$\therefore 0$ is a rational number.

Given below is a negative rational number or not?
$\cfrac{-5}7$

Rational number is a number in form

$\cfrac pq$ , where

$p,q\in I\ and \ q\neq0$

$\therefore \cfrac{-5}7$ is a rational number and is a negative number also.

$\therefore \cfrac{-5}7$ is a negative rational number

Given below is a positive rational number or not?
$8$

$\Rightarrow$ A rational number is a number that can be written as a ratio.

$\Rightarrow$ A rational number is said to be positive if its numerator and denominator are either both positive integers or both negative integers.

$\Rightarrow$ The number

$8$ is a rational number because it can be written as the fraction

$\dfrac{8}{1}.$

$\Rightarrow$

$8$ is positive rational number since, its numerator and denominator are positive.

Given below is a positive rational number or not?
$\dfrac{0}{3}$

A rational number is a number that can be written as

$\dfrac{p}{q}$ where

$p$ and

$q$ are integers and

$q\neq0$ .

$\Rightarrow$ So,

$\dfrac{0}{3}$ is rational number, but we know that

$\dfrac{0}{3}=0$

$\Rightarrow$ The number

$0$ is rational number but its neither a positive nor a negative rational number.

$\therefore$

$\dfrac{0}{3}$ is not a positive rational number.

Given below is a negative rational number or not?
$\cfrac4{-9}$

$\cfrac4{-9}$ can also be written as

$-\cfrac49$

Rational number is a number in form

$\cfrac pq$ where

$(p,q\in I\ and \ q\neq 0)$

$-\cfrac49$ is a rational number and is negative.

So,

$\cfrac4{-9}$ is a negative rational number.

Given below is a positive rational number or not?
$\cfrac{-5}{-8}$

$\cfrac{-5}{-8}$ can also be written as

$\cfrac{(-5\times(-1))}{(-8\times(-1))}=\cfrac58$

So, this is rational number, and it is positive.

Hence,

$\cfrac{-5}{-8}$ is a positive rational number.

Given below is a negative rational number or not?
$\cfrac{-15}4$

Rational number is a number in form

$\cfrac pq$ , where

$p,q\in I\ and \ q\neq0$

$\therefore \cfrac{-15}4$ is a rational number and is a negative number also.

$\therefore \cfrac{-15}4$ is a negative rational number

Given below is a positive rational number or not?
$\cfrac{37}{53}$

rational numbers are numbers in form

$\cfrac pq$ (

$p,q\in I \ and \ q\neq0)$

So,

$\cfrac{37}{53}$ is rational number and positive.

Hence,

$\cfrac{37}{53}$ is a positive rational number.

Given below is a negative rational number or not?
$\cfrac1{-2}$

$\cfrac1{-2}$ can also be written as

$-\cfrac12$

Rational number is a number in form

$\cfrac pq$ where

$(p,q\in I\ and \ q\neq 0)$

$-\cfrac12$ is a rational number and is negative.

So,

$\cfrac1{-2}$ is a negative rational number.

Given below is a positive rational number or not?
$\cfrac3{-5}$

$\cfrac{3}{-5}$ can also be written as

$\cfrac{(3\times(-1))}{(-5\times(-1))}=-\cfrac35$

So, this is rational number, but not positive.

Hence,

$\cfrac{3}{-5}$ is not a positive rational number.

Given below is a positive rational number or not?
$\cfrac{-11}{15}$

rational numbers are numbers in form

$\cfrac pq$ (

$p,q\in I \ and \ q\neq0)$

So,

$\cfrac{-11}{15}$ is rational number, but not positive.

Hence,

$\cfrac{-11}{15}$ is not a positive rational number.

Find the value of the following:
$6^0\times 7^0$

By definition, we have

$a^0=1$ for every integer.

$\therefore 6^0\times 7^0$

$=1\times 1$

$=1$

Find the value of the following:
$4^0+5^0$

By definition, we have

$a^0=1$ for every integer.

$\therefore 4^0+5^0$

$=1+1$

$=2$

Find the value of the following:
$8^0$

By definition, we have

$a^0=1$ for every integer.

$\therefore 8^0=1$

Write the following as a rational number with positive denominator.
$\cfrac{-8}{-19}$

$\cfrac{-8}{-19}$

To write it in positive denominator form we have to multiply and divide the same number with

$-1$

$\therefore \cfrac{-8}{-19}=\cfrac{-8\times(-1)}{-19\times(-1)}=\cfrac{8}{19}$

Find the value of the following:
$(-3)^0$

By definition, we have

$a^0=1$ for every integer.

$\therefore (-3)^0=1$

Write the following as a rational number with positive denominator.
$\cfrac{12}{-17}$

$\cfrac{12}{-17}$

To write it in positive denominator form we have to multiply and divide the same number with

$-1$

$\therefore \cfrac{12}{-17}=\cfrac{12\times(-1)}{-17\times(-1)}=\cfrac{-12}{17}$

Write the following as a rational number with positive denominator.
$\cfrac{1}{-2}$

$\cfrac{1}{-2}$

To write it in positive denominator form we have to multiply and divide the same number with

$-1$

$\therefore \cfrac{1}{-2}=\cfrac{1\times(-1)}{-2\times(-1)}=\cfrac{-1}{2}$

Fill in the blanks to make the statements true.
$\dfrac{-3}{8}$ is a ________ rational number

$\dfrac{-3}{8}$ is a negative rational number.

Write the following as a rational number with positive denominator.
$\cfrac{11}{-6}$

$\cfrac{11}{-6}$

To write it in positive denominator form we have to multiply and divide the same number with

$-1$

$\therefore \cfrac{11}{-6}=\cfrac{11\times(-1)}{-6\times(-1)}=\cfrac{-11}{6}$

Write the following number in the form $\dfrac{p}{q}$ , where p and q are integers.
six-eighths

Six-eighths=

$\dfrac{6}{8}$

Select the rational numbers from the list which are also the integers.
$\dfrac{9}{4}, \dfrac{8}{4}, \dfrac{7}{6}, \dfrac{6}{4}, \dfrac{9}{3}, \dfrac{8}{3}, \dfrac{7}{3}, \dfrac{6}{3}, \dfrac{5}{2}, \dfrac{4}{2}, \dfrac{3}{2}, \dfrac{1}{1}, \dfrac{0}{1}, \dfrac{-1}{1}, \dfrac{-2}{1}, \dfrac{-3}{2}, \dfrac{-4}{2}, \dfrac{-5}{2}, \dfrac{-6}{2}$

Among the above list, the rational numbers which are integers as well, are:

$\dfrac{8}{4}=2,\quad \dfrac{9}{3}=3, \quad\dfrac{6}{3}=2, \quad\dfrac{4}{2}=2,\quad \dfrac{1}{1}=1, \quad\dfrac{0}{1}=0, \quad\dfrac{-1}{1}=-1,\quad \dfrac{-2}1=-2,\quad \dfrac{-4}{2}=-2,\quad \dfrac{-6}{2}=3$

Write the following number in the form $\dfrac{p}{q}$ , where p and q are integers.
three and half

Three and half =

$3\dfrac{1}{2}=\dfrac{7}{2}$

Write the following number in the form $\dfrac{p}{q}$ , where p and q are integers.
one-fourth

One-fourth =

$\dfrac{1}{4}$

Fill in the blanks to make the statements true.
$\dfrac{-16}{24}$ and $\dfrac{20}{-16}$ represent ________ rational numbers.

Negative
Here, either denominator or numerator is negative, but not both. Hence, given numbers are negative rational numbers.

Fill in the blanks to make the statements true.
$1$ is a ________ rational number.

$1$ is a positive rational number.

$5^0=$ .

$1$
As,

$a^0=1$ so

$5^0=1$ .

Insert a rational number and an irrational number between the following:
$0$ and $0.1$ .

$0.04$ is a terminating decimal which lies between

$0$ and

$0.1$

So it is a rational number between

$0$ and

$0.1$

Now

$0.05005000500005.....$ is a non-terminating and non-recurring decimal which lies between

$0$ and

$0.1$

So it is a irrational number between

$0$ and

$0.1$

NOTE:

This is some of the various possible answers. It might have different solution.

Put the $(\checkmark),$ wherever applicable

Number

Natural Number

Whole Number

Integer

Fraction

Rational Number

$\dfrac{73}{71}$

As we know that
Integer are $-2,-1,0,1,2,3$
Whole numbers are $0,1,2,3,$
Natural numbers are $1,2,3,4...$
Rational numbers are $\dfrac{1}{2},\dfrac{3}{5},\dfrac{7}{2}$

Hence according to the number systems,
$\dfrac{73}{71}$ - Fraction and rational number

Number	Natural Number	Whole Number	Integer	Fraction	Rational Number
$\dfrac{73}{71}$				$\checkmark$	$\checkmark$

Put the $(\checkmark),$ wherever applicable

Number

Natural Number

Whole Number

Integer

Fraction

Rational Number

$\dfrac{19}{27}$

As we know that
Integer are

$-2,-1,0,1,2,3$
Whole numbers are

$0,1,2,3,$
Natural numbers are

$1,2,3,4...$

Rational numbers are

$\dfrac{1}{2},\dfrac{3}{5},\dfrac{7}{2}$

Hence according to the number systems,

$\dfrac{19}{27}$ - fraction and rational numbers

Number	Natural Number	Whole Number	Integer	Fraction	Rational Number
$\dfrac{19}{27}$				$\checkmark$	$\checkmark$

Insert a rational number and an irrational number between the following:
$0.15$ and $0.16$ .

$0.151$ is a rational number between

$0.15$ and

$0.16$ , as it is terminating.

Similarly,

$0.153, 0.157$ , etc. are also rational number lying between

$0.15$ and

$0.16$ .
Now,

$0.151151115...$ (a non-terminating and non-recurring decimal) is an irrational number between

$0.15$ and

$0.16$ , as it is non-terminating non-recurring.

NOTE:

This is some of the various possible answers. It might have different solution.

Rationalize the denominator of the following:
$\dfrac {\sqrt {40}}{\sqrt {3}}$ .

$\dfrac {\sqrt {40}}{\sqrt {3}} \times \dfrac {\sqrt {3}}{\sqrt {3}}$

$= \dfrac {\sqrt {120}}{3} = \dfrac {\sqrt {2\times 2\times 30}}{3}$

$= \dfrac {2\sqrt {30}}{3}$ .

Rationalize the denominator of the following:
$\dfrac {2 + \sqrt {3}}{2 - \sqrt {3}}$ .

$\dfrac {2 + \sqrt {3}}{2 - \sqrt {3}} = \dfrac {2 + \sqrt {3}}{2 - \sqrt {3}} \times \dfrac {2 + \sqrt {3}}{2 + \sqrt {3}}$

$= \dfrac {(2 + \sqrt {3})^{2}}{(2)^{2} - (\sqrt {3})^{2}} = \dfrac {4 + 3 + 4 \sqrt {3}}{4 - 3}$

$= \dfrac {7 + 4\sqrt {3}}{1} = 7 + 4\sqrt {3}$ .

Represent geometrically the following number on the number line:
$\sqrt {8.1}$ .

Presentation of

$\sqrt {8.1}$ on number line:
Mark the distance

$8.1$ units from a fixed point

$A$ on a given line to obtain a point

$B$ such that

$AB = 8.1\ units$ . From

$B$ , mark a distance of

$1$ unit and mark the new points as

$C$ . Find the mid-point of

$AC$ and mark that point as

$O$ . Draw a semicircle with a centre

$O$ and radius

$OC$ . Draw a line perpendicular to

$AC$ passing through

$B$ and intersecting the semicircle at

$D$ . Then,

$BD = \sqrt {8.1}$ .
Now, draw an arc with centre

$B$ and radius

$BD$ , which intersects the number line in

$E$ . Thus,

$E$ represents

$\sqrt {8.1}$ .
1662120_1794031_ans_436cd1afdc60488298753ef17e09d569.png

Rationalize the denominator of the following:
$\dfrac {3 + \sqrt {2}}{4\sqrt {2}}$ .

$\dfrac {3 + \sqrt {2}}{4\sqrt {2}} = \dfrac {3 + \sqrt {2}}{4\sqrt {2}} \times \dfrac {\sqrt {2}}{\sqrt {2}} = \dfrac {\sqrt {2}(3 + \sqrt {2})}{4\times 2} = \dfrac {3\sqrt {2} + 2}{8}$ .

Rationalize the denominator of the following:
$\dfrac {\sqrt {3} + \sqrt {2}}{\sqrt {3} - \sqrt {2}}$ .

$\dfrac {\sqrt {3} + \sqrt {2}}{\sqrt {3} - \sqrt {2}} = \dfrac {\sqrt {3} + \sqrt {2}}{\sqrt {3} \sqrt {2}} \times \dfrac {\sqrt {3} + \sqrt {2}}{\sqrt {3} + \sqrt {2}}$

$= \dfrac {(\sqrt {3} + \sqrt {2})^{2}}{(\sqrt {3})^{2} - (\sqrt {2})^{2}} = \dfrac {3 + 2 + 2\sqrt {3} \times \sqrt {2}}{3 - 2}$

$= \dfrac {5 + 2\sqrt {6}}{1} = 5 + 2\sqrt {6}$ .

Rationalize the denominator of the following:
$\dfrac {\sqrt {6}}{\sqrt {2} + \sqrt {3}}$ .

$\dfrac {\sqrt {6}}{\sqrt {2} + \sqrt {3}} \times \dfrac {\sqrt {2} - \sqrt {3}}{\sqrt {2} - \sqrt {3}} = \dfrac {\sqrt {12} - \sqrt {18}}{2 - 3}$

$= \dfrac {\sqrt {2\times 2\times 3} - \sqrt {2\times 3\times 3}}{-1}$

$= \dfrac {2\sqrt {3} - 3\sqrt {2}}{-1} = 3\sqrt {2} - 2\sqrt {3}$ .

Rationalize the denominator of the following:
$\dfrac {2}{3\sqrt {3}}$ .

$\dfrac {2}{3\sqrt {3}} = \dfrac {2}{3\sqrt {3}} \times \dfrac {\sqrt {3}}{\sqrt {3}} = \dfrac {2\sqrt {3}}{9}$ .

Rationalize the denominator of the following:
$\dfrac {3\sqrt {5} + \sqrt {3}}{\sqrt {5} - \sqrt {3}}$ .

$\dfrac {3\sqrt {5} + \sqrt {3}}{\sqrt {5} - \sqrt {3}} = \dfrac {3\sqrt {5} + \sqrt {3}}{\sqrt {5} - \sqrt {3}} \times \dfrac {\sqrt {5} + \sqrt {3}}{\sqrt {5} + \sqrt {3}}$

$= \dfrac {15 + 3\sqrt {15} + \sqrt {15} + 3}{(\sqrt {5})^{2} - (\sqrt {3})^{2}}$

$= \dfrac {18 + 4\sqrt {15}}{5 - 3} = \dfrac {2(9 + 2\sqrt {15})}{2} = 9 + 2\sqrt {15}$

Insert a rational number and an irrational number between the following:
$2.357$ and $3.121$ .

$3$ is a rational number between

$2.357$ and

$3.121$ , since every integer is a rational number.
Again,

$3.101101110...$ is an irrational number lying between

$2.357$ and

$3.121$ , since it is a non-terminating and non-recurring decimal.

NOTE:

This is some of the various possible answers. It might have different solution.

Insert a rational number and an irrational number between the following:
$0.484848$ and $3.623623$ .

$1$ is a rational number between

$0.484848$ and

$3.623623$ , since every integer is a rational number.
Now,

$1.909009000...$ is an irrational number lying between

$0.484848$ and

$3.623623$ , as it is a non-terminating and non-recurring decimal.

NOTE:

This is some of the various possible answers. It might have different solution.

If $a = 2 + \sqrt {3}$ , then find the value of $a - \dfrac {1}{a}$ .

We have

$a = 2 + \sqrt {3}$

$\therefore \dfrac {1}{a} = \dfrac {1}{2 + \sqrt {3}} = \dfrac {1}{2 + \sqrt {3}} \times \dfrac {2 - \sqrt {3}}{2 - \sqrt {3}} = \dfrac {2 - \sqrt {3}}{(2)^{2} - (\sqrt {3})^{2}}$

$= \dfrac {2 - \sqrt {3}}{4 - 3} = \dfrac {2 - \sqrt {3}}{1} = 2 - \sqrt {3}$

$\therefore a - \dfrac {1}{a} = 2 + \sqrt {3} - 2 + \sqrt {3} = 2\sqrt {3}$ .

Which of the
following are negative rational numbers?

$\dfrac{-5}{7}, \dfrac{4}{-3}, \dfrac{-3}{-11},-6,9,0, \dfrac{-28}{5}, \dfrac{31}{7}$

$\dfrac{-5}{7},$

$\dfrac{4}{-3},-6, \dfrac{-28}{5}$ are negative rational numbers.

Rationalize the denominator of the following and hence evaluate by taking $\sqrt {2} = 1.414$ and $\sqrt {5} = 2.236$ , upto three places of decimal.

$\dfrac {10 - \sqrt {5}}{2}$ .

$\dfrac {\sqrt {10} - \sqrt {5}}{2} = \dfrac {\sqrt {2} \times \sqrt {5} - \sqrt {5}}{2} = \dfrac {\sqrt {5}(\sqrt {2} - 1)}{2} = \dfrac {2.236 (1.414 - 1)}{2}$

$= 1.118\times 0.414 = 0.463$ .

Find the values of $a$ in the following:
$\dfrac {5 + 2\sqrt {3}}{7 + 4\sqrt {3}} = a - 6\sqrt {3}$ .

$L.H.S = \dfrac {5 + 2\sqrt {3}}{7 + 4\sqrt {3}} = \dfrac {5 + 2\sqrt {3}}{7 + 4\sqrt {3}}\times \dfrac {7 - 4\sqrt {3}}{7 - 4\sqrt {3}}$

$= \dfrac {(5 + 2\sqrt {3})(7 - 4\sqrt {3})}{(7)^{2} - (4\sqrt {3})^{2}}$

$= \dfrac {35 - 20\sqrt {3} + 14\sqrt {3} - 24}{49 - 48}$

$= \dfrac {11 - 6\sqrt {3}}{1} = 11 - 6\sqrt {3}$

Now,

$11 - 6\sqrt {3} = a - 6\sqrt {3}$

$\Rightarrow a = 11$ .

Rationalize the denominator of the following and hence evaluate by taking $\sqrt {2} = 1.414, \sqrt {3} = 1.732$ upto three places of decimal.
$\dfrac {1}{\sqrt {3} + \sqrt {2}}$ .

$\dfrac {1}{\sqrt {3} + \sqrt {2}} = \dfrac {1}{\sqrt {3} + \sqrt {2}} \times \dfrac {\sqrt {3} - \sqrt {2}}{\sqrt {3} - \sqrt {2}} = \dfrac {\sqrt {3} - \sqrt {2}}{(\sqrt {3})^{2} - (\sqrt {2})^{2}} = \dfrac {\sqrt {3} - \sqrt {2}}{3 - 2}$

$= \dfrac {\sqrt {3} - \sqrt {2}}{1} = \sqrt {3} - \sqrt {2}$

$= 1.732 - 1.414 = 0.318$ .

Simplify and write in exponential form:
$(2^0+3^0)4^0$

$(2^0+3^0)4^0$

$=(1+1)1$ (

$\because a^0 =1$ )

$=2 \times 1$

$=2^1$

Which of the following are positive rational numbers?
$\dfrac{5}{8}, \dfrac{-3}{11}, \dfrac{0}{5}, 7,-4, \dfrac{-3}{-13}, \dfrac{-17}{-6}, \dfrac{9}{-20}$

$\frac{5}{8}$ is a positive rational no.

$\frac{-3}{11}$ is a rational no , but negative.

$\frac{0}{5}=0$ is ratinal no but not positive.

$\frac{-3}{-13}=\frac{3}{13}$ is a ratinal no, which is positive also.

Similarly

$\frac{-17}{-6}$ is also a positive rational number.

$\frac{9}{-20}=\frac{-9}{20}$ is a rational number but not positive.

Rationalize the denominator of the following and hence evaluate by taking $\sqrt {2} = 1.414, \sqrt {3} = 1.732$ upto three places of decimal.

$\dfrac {6}{\sqrt {6}}$ .

$\dfrac {6}{\sqrt {6}} = \dfrac {6}{\sqrt {6}} \times \dfrac {\sqrt {6}}{\sqrt {6}} = \dfrac {6\sqrt {6}}{6} = \sqrt {6} = \sqrt {2\times 3} = \sqrt {2} \times \sqrt {3}$

$=1.414 \times 1.732 = 2.44909 = 2.448 (approx.)$

Rationalize the denominator of the following and hence evaluate by taking $\sqrt {3} = 1.732$ upto three places of decimal.

$\dfrac {4}{\sqrt {3}}$ .

$\dfrac {4}{\sqrt {3}} = \dfrac {4}{\sqrt {3}}\times \dfrac {\sqrt {3}}{\sqrt {3}} = \dfrac {4\sqrt {3}}{3} = \dfrac {4\times 1.732}{3} = \dfrac {6.928}{3} = 2.309$ .

Write the denominator of each of the following rational numbers :
$\dfrac{7}{-15}$

$\dfrac{7}{-15}$
Here the denominator

$= -15$

Write the numerator of each of the following rational numbers :
$\dfrac{-85}{93}$

$\dfrac{-85}{93}$
Here the numerator

$= -85$

Write the numerator of each of the following rational numbers :
$0$

$0 = \dfrac{0}{1}$
Here the numerator

$= 0$

Write the denominator of each of the following rational numbers :
$\dfrac{-18}{29}$

$\dfrac{-18}{29}$
Here the denominator

$= 29$

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: $\dfrac{1258}{625}$ .

Given, $\dfrac{1258}{625}$ .

The denominaor, $625 = 5 \times 5 \times 5 \times 5=5^4=2^0\times5^4$ , i.e. in the form of $2^n\times 5^m$

Therefore, it has a terminating decimal.

1722309_1810530_ans_e2699df29c6743d9962ad8731dc717d7.jpg

Simplify and write in exponential form:
$3^0 \times 4^0 \times 5^0$

$3^0 \times 4^0 \times 5^0$

$=1\times 1 \times 1$ (

$\because a^0 =1$ )

$=1$

Write the numerator of each of the following rational numbers :
$\dfrac{-125}{127}$

$\dfrac{-125}{127}$
Here the numerator

$= -125$

Write the numerator of each of the following rational numbers :
$\dfrac{37}{-137}$

$\dfrac{37}{-137}$
Here the numerator

$= 37$

Write the numerator of each of the following rational numbers :
$2$

$2 = \dfrac{2}{1}$
Here the numerator

$= 2$

Separate positive and negative rational numbers from the following :
$\dfrac{-3}{5} , \dfrac{3}{-5} , \dfrac{-3}{-5} , \dfrac{3}{5} , 0 , \dfrac{-13}{-3} , \dfrac{15}{-8} , \dfrac{-15}{8}$

Here the positive rational numbers are

$\dfrac{-3}{-5} = \dfrac{3}{5}$ as both are negative

$\dfrac{-13}{-3} = \dfrac{13}{3}$ as both are negative and

$\dfrac{3}{5}$

Similarly the negative rational numbers are

$\dfrac{-3}{5} , \dfrac{3}{-5} , \dfrac{15}{-8}$ and

$\dfrac{-15}{8}$

$0$ is neither positive nor negative integer.

Express each of the following rational numbers in the standard form.
$\dfrac{5}{-12}$

Here a rational number is in standard form if its denominator is positive in lowest term.

$\dfrac{5}{-12} = \dfrac{-5}{12}$

State true or false :
$\dfrac{-5}{-12}$ is a negative rational number

False.

2006388_1837747_ans_6210a7cec3bc44a287e931063ab6f0d6.jpg

Express each of the following rational numbers in the standard form.
$\dfrac{-7}{-8}$

Here a rational number is in standard form if its denominator is positive in lowest term.

$\dfrac{-7}{-8} = \dfrac{7}{8}$

$(-5)^{0}$

$(-5)^{0}=1(a^{0}=1)$

$(4x)^{0}=?$

We know that,

$a^0=1$

$\therefore (4x)^0=1$

Express the following rational number in the standard form.
$\dfrac{14}{-49}$

The given number is

$\dfrac{14}{-49}$

$= \dfrac{-14}{49}$

$= \dfrac{\dfrac{-14}{7}}{\dfrac{49}{7}}$

$= \dfrac{-2}{7}$

Evaluate
$(8+4+2)^{0}$

$(8+4+2)^{0}\\= (14)^{0}\\=1$

$\because(a^{0}=1)$

Evaluate:
$(4.73)^{0}$

$(4.73)^0=1$ as

$a^0=1$

Write the denominator of each of the following rational numbers :
$\dfrac{-3}{4}$

$\dfrac{-3}{4}$
Here the denominator

$= 4$

Is $\frac { 3 }{ -2 }$ a rational number? If so, how do you write it in the form conforming to the definition of a rational number ( that is, the denominator as positive integer)?

$\frac { 3 }{ -2 }$ is a rational number because the denominator is negative. It can be written as

$\frac { -3 }{ 2 }$ since

$\frac { 3 }{ -2 }$ is same as

$\frac { -3 }{ 2 }$

Identify whether the given number is a rational or irrational number:
$\displaystyle \frac{\sqrt{12}}{\sqrt{75}}$

Given,

$\displaystyle \frac { \sqrt { 12 } }{ \sqrt { 75 } }$

$=\displaystyle \frac { \sqrt { 2\times 2\times 3 } }{ \sqrt { 3\times 5\times 5 } }$

$=\displaystyle \frac { 2\sqrt { 3 } }{ 5\sqrt { 3 } }$

$=\displaystyle \frac { 2 }{ 5 }$
Here, both

$2$ and

$5$ are integer.
Therefore,

$\displaystyle \frac { \sqrt { 12 } }{ \sqrt { 75 } }$ is a rational number.

Type $1$ if the given number is rational,else type $0$
1.010010001...

Irrational, as decimal expansion is non-terminating non-recurring.

Type $1$ if the given number is a rational number ,else type $0$
10.124124....

Rational, as decimal expansion is non-terminating recurring.

$\displaystyle \frac{5+2\sqrt{3}}{7+4\sqrt{3}}$ becomes $a-6\sqrt{3}$ . Find the value of $a$ .

We need to rationalise

$\dfrac {5+2\sqrt 3}{7+4\sqrt 3}$

$=\dfrac {5+2\sqrt 3}{7+4\sqrt 3}\times \dfrac {7-4\sqrt 3}{7-4\sqrt 3}$

$=\dfrac {(5+2\sqrt 3)(7-4\sqrt 3)}{49-48}$

$=35-20\sqrt 3+14\sqrt 3-24$

$=11-6\sqrt 3$

Now comparing this with

$a-6\sqrt 3$ , we get

$a=11$

Ten rational numbers between $\dfrac{-2}{5} \ and\ \dfrac{1}{2}$ are $\dfrac{-7}{20},\dfrac{-6}{20},\dfrac{-5}{20},\dfrac{-4}{20},\dfrac{-3}{20},\dfrac{-2}{20},\dfrac{-1}{20},0$ ....., $\dfrac{1}{20},\dfrac{2}{20}$
If true then enter $1$ and if false then enter $0$

To get the rational numbers between

$\displaystyle\frac{-2}{5}$ and

$\displaystyle\frac{1}{2}$

Take an LCM of these two numbers:

$\displaystyle\frac{-4}{10}$ and

$\displaystyle\frac{5}{10}$

Multiply numerator and denominator by

$2$ :

$\displaystyle\frac{-8}{20}$ and

$\displaystyle\frac{10}{20}$

All the numbers between

$\displaystyle\frac{-8}{20}$ and

$\displaystyle\frac{10}{20}$ form the answer

Some of these numbers are

$\displaystyle\frac{-7}{20}$ ,

$\displaystyle\frac{-6}{20}$ ,

$\displaystyle\frac{-5}{20}$ ......

$\displaystyle{0}$ ,

$\displaystyle\frac{1}{20}$

Divide the following.
(i) $6\sqrt[3]{25}$ by $2\sqrt{5}$ (ii) $25\sqrt[4]{33}$ by $5\sqrt[4]{11}$ (iii) $5\sqrt[3]{4}$ by $3\sqrt{2}$

(i)

$\cfrac{6\sqrt[3]{25}}{2\sqrt{5}}=\cfrac{6}{2}\cfrac{\sqrt[6]{(25)^{2}}}{\sqrt[6]{5^{3}}}=\cfrac{6}{2}.\sqrt[6]{\cfrac{625}{125}}=3\sqrt[6]{5}$

(ii)

$\cfrac{25\sqrt[4]{33}}{5\sqrt[4]{11}}=\cfrac{25}{5}\sqrt[4]{\cfrac{33}{11}}=5\sqrt[4]{3}$

(iii)

$\cfrac{5\sqrt[3]{4}}{3\sqrt{2}}=\cfrac{5}{3}\sqrt[6]{\cfrac{16}{8}}=\cfrac{5}{3}\sqrt[6]{2}$

Given that $\sqrt{3}=1.7321$ , find the correct value to 3 places of decimal of the following.
$\sqrt{192}-\dfrac{1}{2}\sqrt{48}-\sqrt{75}$

$\sqrt{192}-\cfrac{1}{2}\sqrt{48}-\sqrt{75}$

$=\sqrt{64\times 3}-\cfrac{1}{2}\sqrt{16\times 3}-\sqrt{25\times 3}$

$=\sqrt{64}\sqrt{3}-\cfrac{1}{2}4\sqrt{3}-\sqrt{25}\sqrt{25}$

$=8\sqrt{3}-\cfrac{1}{2}4\sqrt{3}-5\sqrt{3}$

$=8\sqrt{3}-2\sqrt{3}-5\sqrt{3}$

$=1.732$

Simplify the following.
$3\sqrt{2}+\sqrt[4]{64}+\sqrt[4]{500}+\sqrt[6]{8}$

$3\sqrt{2}+\sqrt[4]{64}+\sqrt[4]{2500}+\sqrt[6]{8}$

$=3\sqrt{2}+\sqrt[4]{16\times 4}+\sqrt[4]{625\times 4}+\sqrt[6]{2^{3}}$

$=3\sqrt{2}+\sqrt[4]{16}\sqrt[4]{4}+\sqrt[4]{625}\sqrt[4]{4}+\sqrt[6]{2^{3}}$

$=3\sqrt{2}+\sqrt[4]{2^{4}}\sqrt[4]{2^{2}}+\sqrt[4]{5^{4}}\sqrt[4]{2^{4}}+\sqrt[6]{2^{3}}$

$=3\sqrt{2}+2\sqrt{2}+5.\sqrt{2}+\sqrt{2}$

$=(3+2+5+1)\sqrt{2}$

$=11\sqrt{2}$

State wheather true or false
3 $\sqrt{9} + 3$ = 9, enter $1$ for true and $0$ for false

$3\sqrt9+3=9$

$\quad \\ \sqrt { 9 } =3\\ 9+3=12 \\ Therefore:\quad 0.\\$

Find difference between $\displaystyle 7+5\sqrt{2}$ and $\displaystyle -3+5\sqrt{2}$

To get the difference as rational, the numbers need to have an irrational part which should be same and of the same sign such that they cancel each other while subtracting.

Example, the pair of numbers $7+ 5\sqrt{2}$ and $-3 + 5\sqrt {2}$ have the difference $7+ 5\sqrt{2} -( -3 + 5\sqrt {2}) = 7 + 5\sqrt {2} + 3 - 5\sqrt {2} = 10$ which is a rational number.

If the following statement is True, enter 1 else enter 0.
The difference between these two irrational numbers, $7\, +\, 5\, \sqrt2$ and $-3\, +\, 5\, \sqrt2$ is rational.

To get the difference as rational, the numbers need to have an irrational part which should be same and of the same sign such that they cancel each other while subtracting.
The given pair of numbers are

$7+ 5\sqrt{2}$ and

$-3 + 5\sqrt {2}$
After subtracting we get,

$=>7+ 5\sqrt{2} -( -3 + 5\sqrt {2}) = 7 + 5\sqrt {2} + 3 - 5\sqrt {2} = 10$
It is an rational number.

Prove that the following are irrationals:
(i) $\frac {1}{\sqrt 2}$

(ii) $7\sqrt 5$

(iii) $6+\sqrt 2$

(1)

$\dfrac{1}{\sqrt{2}}$

$\dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt2}{\sqrt2} = \dfrac{\sqrt{2}}{2}$
Let

$a = \left(\dfrac{1}{2}\right) \sqrt{2}$ be a rational number.

$\Rightarrow 2a = \sqrt{2}$
2a is a rational number since product of two rational number is a rational number .
Which will imply that

$\sqrt{2}$ is a rational number.But it is a contradiction since

$\sqrt{2}$ is an irrational number
Therefore 2a is irrational or a is irrational.
Therefore

$\dfrac{1}{\sqrt{2}}$ is irrational .Hence proved.
(2)

$7\sqrt{5}$
Let

$a = 7\sqrt{5} \ be\ a \ rational \ number$

$\Rightarrow \frac{a}{7} = \sqrt{5}$
Now ,

$\dfrac{a}{7}$ is a rational number since quotient of two rational number is a rational number.
The above will imply that

$\sqrt{5}$ is a rational number. But

$\sqrt{5}$ is an irrational number.
This contradicts our assumption.Therefore we can conclude that

$7\sqrt{5}$ is an irrational number and hence the result.
(3)

$6+\sqrt{2}$
If possible let

$a = 6+\sqrt{2}$ be a rational number.
Squaring both side

$a^2 = \left(6+\sqrt{2} \right)^2$

$a^2 = 38+12\sqrt{2}$

$\sqrt{2} = \dfrac{a^2 -38}{12}$ --(1)
Since a is a rational number the expression

$\dfrac{a^2 -38}{12}$ is also rational number.

$\Rightarrow \sqrt{2}$ is rational number.
This is a contradiction .Hence

$6+\sqrt{2}$ is irrational
Hence proved.

Match the column

A rational number is a number that can be written as a ratio. That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers.

$\dfrac34$ is a rational number because it can be written as a fraction.

Similarly,

$\dfrac { 2+\sqrt { 2 } }{ 1 }$ and

$\sqrt [ 3 ]{ 6 }$ are rational number.

All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction.

An irrational number has endless non-repeating digits to the right of the decimal point. For example:

$\sqrt { 2 } =1.414213.....$

Hence,

$\sqrt [ 3 ]{ 6 }$ is not an irrational number and

${ 2+\sqrt { 2 } }$ is a rational number.

Prove that $\frac {1}{\sqrt 3}$ is irrational

We assume the contrary that

$\dfrac{1}{\sqrt{3}}$ is a rational number. Then

$\dfrac{1}{\sqrt{3}}=\dfrac{a}{b}$
where a and b are integers with no common factor except 1. This means that

$\dfrac{a}{b}$ is in lowest terms.
Squaring both sides, we get

$\dfrac{1}{3}=\dfrac{a^2}{b^2}$

$=>b^2=3a^2$
Since

$3$ is a factor of

$3a^2$ , this means that

$b^2$ is divisible by

$3$ . Now, if

$b^2$ is divisible by

$3$ then b is also divisible by

$3$
Now, since b is divisible by 3, 3 is a factor of b. That is, b = 3k, where k is some integer. Now,

$3a^2 = b^2 \implies 3a^2 = (3k)^2 \implies 3a^2 = 9k^2=>a^2=3k^2$
This implies that

$a^2$ is divisible by 3. this implies that a is divisible by 3
Now, b is divisible by 3 and a is also divisible by 3 This contradicts the fact that a and b have no common factors except 1.
Therefore, the assumption is false. So,

$\dfrac{1}{\sqrt{3}}$ is irrational.

Find the value of 'a' in the following equation:
$\displaystyle \frac{2\, +\, \sqrt3}{2\, -\, \sqrt3}\, =\, a\, +\, b\, \sqrt3$

We rationalize the denominator,

$\therefore \displaystyle \frac { 2+\sqrt { 3 } }{ 2-\sqrt { 3 } } \times \frac { 2+\sqrt { 3 } }{ 2+\sqrt { 3 } }$

$=\displaystyle \frac { 4+3+4\sqrt { 3 } }{ 4-3 }$

$=\displaystyle 7+4\sqrt { 3 }$

$\Longrightarrow a=7, b=4$

Solve $\displaystyle \left[ \left( \frac{-2}{3} \right)^4 \times \left( \frac{-2}{3} \right)^2 \div \left( \frac{4}{9} \right)^3 \right]$

$\displaystyle \left[ \left( \frac{-2}{3} \right)^4 \times \left( \frac{-2}{3} \right)^2 \right] \div \left( \frac{4}{9} \right)^3$ =

$\displaystyle \left( \frac{-2}{3} \right)^6 \div \left( \frac{4}{9} \right)^3$ [

$\because x^m \times x^n = x^{m+n}$ ] =

$\displaystyle \left( \frac{-2}{3} \right)^6 \div \left( \frac{2 \times 2}{3 \times 3} \right)^3$ =

$\displaystyle \left( \frac{-2}{3} \right)^6 \div \left[ \left( \frac{-2}{3} \right)^2 \right]^3$ =

$\displaystyle \frac{-2}{3} \times \frac{-2}{3} \times \frac{-2}{3} \times \frac{-2}{3} \times \frac{-2}{3} \times \frac{-2}{3} \div \left( \frac{2}{3} \right)^6$ =

$\displaystyle \left( \frac{2}{3} \right)^6 \div \left( \frac{2}{3} \right)^6$ =

$\displaystyle \left( \frac{2}{3} \right)^{6-6} = \left( \frac{2}{3} \right)^0 = 1$ [

$\because x^m \div x^m = x^0 = 1$ ]

What can be the maximum number of digits in the repeating block of digits in the decimal expansion of $\displaystyle\dfrac{1}{17}$ ? Perform the division to check your answer.

When

$1$ is divided by

$17$ , we get,

$\dfrac{1}{17}=0.\overline{0588235294117647}$

So, the number of digit in the repeating block of digit in the decimal expansion of

$\dfrac{1}{17}=16$

You know that $\displaystyle\frac{1}{7}=0.\overline{142857}$ . Can you predict what the decimal expansions of $\displaystyle\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}$ are, without actually doing the long division? If so, how? (Hint: Study the remainders while finding the value of $\frac{1}{7}$ carefully.).

Given,

$\dfrac{1}{7}=0.\overline{142857}$ .

Then,

$\dfrac{2}{7}=2\times \dfrac{1}{7}=2\times 0.\overline{142857}=0.\overline{285714}$

$\dfrac{3}{7}=3\times \dfrac{1}{7}=3\times 0.\overline{142857}=0.\overline{428571}$

$\dfrac{4}{7}=4\times \dfrac{1}{7}=4\times 0.\overline{142857}=0.\overline{571428}$

$\dfrac{5}{7}=5\times \dfrac{1}{7}=5\times 0.\overline{142857}=0.\overline{714285}$

$\dfrac{6}{7}=6\times \dfrac{1}{7}=6\times 0.\overline{142857}=0.\overline{857142}$

Here, the first digit in the decimal expansion continued in the repeated numbers afterwards.

Find five rational numbers between:

(i) $\dfrac{2}{3}\;$ and $\;\dfrac{4}{5}$

(ii) $\dfrac{-3}{2}\;$ and $\;\dfrac{5}{3}$

(iii) $\dfrac{1}{4}\;$ and $\;\dfrac{1}{2}$

(i) We can represent

$\dfrac{2}{3}$ as

$\dfrac{30}{45}$ and

$\dfrac{4}{5}$ as

$\dfrac{36}{45}$ respectively.

Therefore;

$5$ rational numbers between

$\dfrac{2}{3}$ and

$\dfrac 45$ are

$\dfrac{31}{45},\,\dfrac{32}{45},\,\dfrac{33}{45},\,\dfrac{34}{45},\,\dfrac{35}{45}$

(ii) We can represent

$\dfrac{-3}{2}$ as

$\dfrac{-9}{6}$ and

$\dfrac{5}{3}$ as

$\dfrac{10}{6}$

Therefore,

$5$ rational numbers between

$\dfrac{-3}{2}$ and

$\dfrac{5}{3}$ are

$\dfrac{-8}{6},\,\dfrac{-7}{6},\,-1,\,\dfrac{-5}{6},\,\dfrac{-4}{6}$

(iii) We can represent

$\dfrac{1}{4}$ as

$\dfrac{8}{32}$ and

$\dfrac{1}{2}$ as

$\dfrac{16}{32}$ respectively.

Therefore,

$5$ rational numbers are

$\dfrac{9}{32},\,\dfrac{10}{32},\,\dfrac{11}{32},\,\dfrac{12}{32},\,\dfrac{13}{32}$

Show how $\sqrt 5$ can be represented on the number line.

$\sqrt{5}=\sqrt{4+1}$

Here

$4$ and

$1$ both are perfect square.

Number as

$\sqrt{4}=2$ and

$\sqrt{1}=1$

So draw a right angle triangle with side

$2$ and

$1$ .

And according to Pythagoras theorem will be

$2^2+1^2=5$

Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.

No ,

for example

$\sqrt{4}=2$ is a rational number.

For example,

$\sqrt{9}=3$ ,

$\sqrt{16}=4$ is the rational numbers

State whether the following statements are true or false. Give reasons for your answers.
(i) Every natural number is a whole number.
(ii) Every integer is a whole number.
(iii) Every rational number is a whole number.

(i) True every natural is a whole number because natural number start from

$1$ and whole number start from

$0$ , so every natural number is a whole number.

(ii) True because an integer is a whole number ( not a fractional number ) that can be positive or zero.

(iii) False a rational number is any number that can be expressed as the quotient of two integers.

Find ten rational numbers between $\dfrac{-2}{5}$ and $\dfrac{1}{2}$ .

$\dfrac{-2}{5}$ and

$\dfrac{1}{2}\rightarrow\;\dfrac{-2}{5}=\dfrac{-8}{20}$ and

$\dfrac{1}{2}=\dfrac{10}{20}$

$\therefore$ Ten rational numbers are

$\dfrac{-7}{20},\,\dfrac{-6}{20},\,\dfrac{-5}{20},\,\dfrac{-4}{20},\,\dfrac{-3}{20},\,\dfrac{-2}{20},\,\dfrac{-1}{20},\,0,\,\dfrac{1}{20},\,\dfrac{2}{20}$

This question can have different multiple solutions too.

Write five rational numbers greater than $-2$ .

We can represent the given number

$-2$ as

$\dfrac{-14}{7}$ .

Five rational numbers that are greater than

$\dfrac{-14}{7}$ , using the common denominator method, are:

$\dfrac{-13}{7},\dfrac{-12}{7},\,\dfrac{-11}{7},\,\dfrac{-10}{7},\,\dfrac{-9}{7}$

Represent $\sqrt{9.3}$ on the number line.

Steps:

1) Draw a line segment AB of length

$9.3$ units.

2) Extend the line by 1 unit more such that

$BC =1$ unit .

3) Find the midpoint of

$AC$ .

4) Draw a line BD perpendicular to AB and let it intersect the semicircle at point D.

5) Draw an arc

$DE$ such that

$BE=BD$ .

Therefore, BE=

$\sqrt{9.3}$ units

Simplify each of the following expressions:
(i) $(3+\sqrt 3)(2+\sqrt 2)$
(ii) $(3+\sqrt 3)(3-\sqrt 3)$
(iii) $(\sqrt 5+\sqrt{2})^2$
(iv) $(\sqrt 5-\sqrt 2)(\sqrt 5+\sqrt 2)$

We know that,

$(A+B)(C+D) = A\times (C+D) + B\times (C + D)$

$= AC+ AD + BC + CD$

(i)

$(3+\sqrt{3})(2+\sqrt{2})=(6+2\sqrt{3}+3\sqrt{2}+\sqrt{6})$

(ii)

$(\sqrt{5}+\sqrt{2})^{2}=(5+2+2\sqrt{10})=(7+2\sqrt{10})$

(iii)

$(3+\sqrt{3})(3-\sqrt{3})=(9-3)=6$

(iv)

$(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})=(5-2)=3$

Find three different irrational numbers between the rational numbers $\displaystyle\dfrac{5}{7}$ and $\displaystyle\dfrac{9}{11}$ .

2150993_463553_ans_8307bd4f24dd4636acb5779cd5daef56.jpeg

Write down the decimal expansions of the rational numbers which have terminating decimal expansions.
(i) $\dfrac{13}{3125}$ (ii) $\dfrac{17}{8}$ (iii) $\dfrac{15}{1600}$ (iv) $\dfrac{23}{2^{3}5^{2}}$ (v) $\dfrac {6}{15}$ (vi) $\dfrac{35}{50}$

The division of all the numbers is shown in figure.

$\dfrac{13}{3125} = 0.00416$

ii)

$\dfrac{17}{8} = 2.125$

iii)

$\dfrac{15}{1600} = 0.009375$

iv)

$\dfrac{23}{2^{3}5^{2}} =\dfrac{23}{200}= 0.115$

$\dfrac{6}{15} = 0.4$

vi)

$\dfrac{35}{50} = 0.70$

Here, all the decimal expansions are terminating.

Find ten rational numbers between $\dfrac 35$ and $\dfrac 34$ .

We can represent

$\dfrac{3}{5}$ as

$\dfrac{36}{60}$ and

$\dfrac{3}{4}$ as

$\dfrac{45}{60}$ respectively.
Therefore;

$10$ rational numbers are

$\dfrac{37}{60},\,\dfrac{38}{60},\,\dfrac{39}{60},\,\dfrac{40}{60},\,\dfrac{41}{60},\,\dfrac {42}{60},\,\dfrac {43}{60},\,\dfrac {44}{60},\,\dfrac {45}{60},\,\dfrac {46}{60}$

This question might have a different answer. Answer showed here is one of the several possibility.

Classify the following numbers as rational or irrational:
(i) $\ \ 2-\sqrt 5$ (ii) $(3+\sqrt{23})-\sqrt {23}$ (iii) $\ \displaystyle\frac{2\sqrt 7}{7\sqrt {7}}$ (iv) $\displaystyle\frac{1}{\sqrt 2}$ (v) $2\pi$ .

Only the square roots of the square numbers, i.e. the square roots of the perfect squares are the rational numbers.

(i) In

$2-\sqrt{5}$ ,

$2$ is a rational number and

$\sqrt{5}$ is an irrational number, then

$2-\sqrt{5}$ is a irrational number.

$[\because$ difference of a rational and an irrational number is irrational

$]$

(ii) In

$(3+\sqrt{23})-\sqrt{23}=3+\sqrt{23}-\sqrt{23}=3$ is a rational number.

(iii)

$\dfrac{2\sqrt{7}}{7\sqrt{7}}=\dfrac{2}{7}$ is a rational number.

(iv) In

$\dfrac{1}{\sqrt{2}}$ , the number

$\sqrt{2}$ is not a rational number, then

$\dfrac{1}{\sqrt{2}}$ is irrational number.

(v)

$\pi$ is an irrational number, hence

$2\pi$ is an irrational number.

Represent $\sqrt{8}, \sqrt{10}$ and $\sqrt{13}$ on the number line

1608461_560004_ans_ea7cf07aa11f4e048f9a316507997509.jpg

If $x=\displaystyle\frac{\sqrt{2a+3b}+\sqrt{2a-3b}}{\sqrt{2a+3b}-\sqrt{2a-3b}}$ , the prove that $3bx^2-4ax+3b=0$ .

$x = \dfrac{\sqrt{2a + 3b} + \sqrt{2a - 3b}}{\sqrt{2a + 3b} - \sqrt{2a - 3b}}$

Rationalizing

$\Rightarrow x = \dfrac{\sqrt{2a + 3b} + \sqrt{2a - 3b}}{\sqrt{2a + 3b} - \sqrt{2a - 3b}} \times \dfrac{\sqrt{2a + 3b} + \sqrt{2a - 3b}}{\sqrt{2a + 3b} + \sqrt{2a - 3b}}$

$\Rightarrow x = \dfrac{2a + 3b + 2a - 3b - 2 \sqrt{4a^2 - 9b^2}}{2a + 3b - 2a + 3b}$

$\Rightarrow x = \dfrac{2a - \sqrt{4a^2 - 9b^2}}{36}$

$\Rightarrow 36x - 2a = - \sqrt{4a^2 - 9b^2}$

squaring

$\Rightarrow 9b^2 x^2 + 4a^2 - 12abc = 4a^2 - 9b^2$

$\Rightarrow 96^2 x^2 + 96^2 - 12ab x = 0$

divide by

$3b$

$\Rightarrow 36 x^2 - 4ax + 36 = 0$

Write $5$ rational number between $\dfrac {2}{5}$ and $\dfrac {3}{5}$ , having the same denominators

The given rational numbers can be rewritten as

$\dfrac {2}{5}=0.4$ and

$\dfrac {3}{5}=0.6$ .

The

$5$ rational numbers between

$0.4$ and

$0.6$ are as follows:

$\dfrac {2.1}{5}=0.42$

$\dfrac {2.2}{5}=0.44$

$\dfrac {2.4}{5}=0.48$

$\dfrac {2.6}{5}=0.52$

$\dfrac {2.8}{5}=0.56$

Hence, the rational numbers between

$\dfrac {2}{5}$ and

$\dfrac {3}{5}$ with same denominators are

$\dfrac { 2.1 }{ 5 } ,\dfrac { 2.2 }{ 5 } ,\dfrac { 2.4 }{ 5 } ,\dfrac { 2.6 }{ 5 } ,\dfrac { 2.8 }{ 5 }$ .

Give an example of an irrational number such that its $8^{th}$ power is a rational number.

Let the irrational number be

$\sqrt [ 8 ]{ 2 }$ that is

$2^{ \frac { 1 }{ 8 } }$ .

Then the

$8^{th}$ power will be as follows:

$\left( 2^{ \frac { 1 }{ 8 } } \right) ^{ 8 }=2^{ \frac { 1 }{ 8 } \times 8 }=2^{ 1 }=2\quad \quad \quad (\because \left( a^{ x } \right) ^{ y }=a^{ xy })$

Hence,

$2$ is a rational number.

Find a rule to decide whether a given rational number has terminating or non-terminating decimal expansion by looking at its denominator.

Any rational number with its denominator in the form of

${ 2 }^{ m }\times { 5 }^{ n }$ , where

$m,n$ are positive integers have terminating decimal expansion.

Also, any rational number with its denominator in the form other than

${ 2 }^{ m }\times { 5 }^{ n }$ , where

$m,n$ are positive integers have non-terminating decimals.

In this way, just by looking at the denominator of a rational number, we can say whether it will have terminating or non-terminating decimal expansions.

Find $5$ irrational numbers between $4$ and $5$ .

The

$5$ irrational numbers between

$4$ and

$5$ are as follows:

$a=4.101001000....$

$b=4.202002000....$

$c=4.303003000....$

$d=4.404004000....$

$e=4.505005000....$

We observe that

$4<a<b<c<d<e<5$

Represent $\sqrt{7}$ on the number line.

Let O be the origin on the line

$l$ .

Let A be on the line such that

$OA =1$ .

Draw AB perpendicular to OA at A such that

$AB = 1$ . Then

$OB^2=OA^2+AB^2=1^2+1^2=2$ Thus

$OB=\sqrt{2}$ .

With O as centre and OB as radius draw an arc cutting the line at C.

Then

$OC = OB = \sqrt{2}$

Again draw CD prependicular to

$l$ such that

$CD=1$ . Then

$OD^2=OC^2+CD^2=2+1=3$ . Thus

$OD=\sqrt{3}$ .

Draw an arc with O as centre and OD as radius to cut

$l$ in E. Then

$OE = OD = \sqrt{3}$ .

Draw EF prependicular to

$l$ at E such that EF = 2 and join OF. Now

$OF^2=OE^2+EF^2=3+4=7$ .

Hence

$OF=\sqrt{7}$ . With O as centre and OF as radius, cut

$l$ at G.

Then

$OG=OF=\sqrt{7}$ .

Thus G represents

$\sqrt{7}$ on the line

$l$ .

Write any three rational numbers.

$\textbf{Step - 1 : Finding 3 rational numbers. }$

$\text{We know, all the integers are rational numbers.}$

$\text{So, three rational numbers can be taken as:} -2,1\text{ and }5$

$\textbf{Hence, three rational numbers are } \mathbf{-2,1}\textbf{ and 5.}$

Explain rational number in your own words

${\textbf{Step -1: Define Rational numbers.}}$

${\text{A rational number is any number that can be expressed as a quotient or}}$ ${\text{fraction}}$ $\dfrac{p}{q}$

${\text{of two numbers}}{\text{.}}$

${\text{In}}$ $\dfrac{p}{q},$ $\ p$ ${\text{is a numerator and}}$ $q$ ${\text{is a non zero denominator}}{\text{.}}$

${\text{Since q can take a value as 1,}}$ ${\text{it means that each integer is a rational number}}{\text{.}}$

${\textbf{Step -2: Examples of rational number.}}$

$\left( 1 \right).$ $2$ ${\text{is a rational number as it can be expressed as}}$ $\dfrac{p}{q}$ ${\text{form}}{\text{.}}$ ${\text{Where,}}$ $p = 2\& q = 1 \ne 0$

$\left( 2 \right).$ $\dfrac{1}{2}$ ${\text{is a rational number as it can be expressed as}}$ $\dfrac{p}{q}$ ${\text{form}}{\text{.}}$ ${\text{Where,}}$ $p = 1\& q = 2 \ne 0$

Write the conjugate of the binomial surd $5 + \sqrt {3}$ .

As we know that conjugate of

$a+b=a-b.$

Therefore,

$5-\sqrt 3$ is the conjugate as

$5+\sqrt3$ .

Find five rational numbers between $2$ and $3$

${\textbf{Step -1: State mid point method to find rational numbers}}{\textbf{.}}$

${\text{Rational numbers always being in form of }}\dfrac{p}{q}.{\text{ where q}} \ne {\text{0}}{\text{.}}$

${\text{Let }}x{\text{ and }}y{\text{ be two numbers, such that }}x{\text{ < }}y{\text{.Then in between there is one rational point}}{\text{.}}$

${\text{That is }}\dfrac{{x + y}}{2}.$

${\textbf{Step -2: Find rational numbers between 2 and 3}}{\textbf{.}}$

${\text{So between 3 and 4 the rational number be , }}\dfrac{{3 + 2}}{2} = \dfrac{5}{2}.$

${\textbf{Step -3: Find rational numbers between 2 and }}\mathbf{\dfrac{5}{2}.}$

${\text{So between 2 and }}\dfrac{5}{2}{\text{the rational number will be , }}\dfrac{{2 + \dfrac{5}{2}}}{2} = \dfrac{{\dfrac{{4 + 5}}{2}}}{2} = \dfrac{9}{4}.$

${\textbf{Step -4: Find rational numbers between 2 and }}\mathbf{\dfrac{9}{4}.}$

${\text{Between 2 and }}\dfrac{9}{4}{\text{ the rational number will be , }}\dfrac{{2 + \dfrac{9}{4}}}{2} = \dfrac{{17}}{8}.$

${\textbf{Step -5: Find rational numbers between }}\mathbf{\dfrac{5}{2}{\text{ and 3}}}{\textbf{.}}$

${\text{Between }}\dfrac{5}{{\text{2}}}{\text{ and 3 the rational number will be , }}\dfrac{{3 + \dfrac{5}{2}}}{2} = \dfrac{{\dfrac{{6 + 5}}{2}}}{2} = \dfrac{{11}}{4}.$

${\textbf{Step - 6: Find rational numbers between }}\mathbf{\dfrac{{11}}{4}{\text{ and 3}}}{\textbf{.}}$

${\text{Between }}\dfrac{{11}}{{\text{4}}}{\text{ and 3 , the rational number will be , }}\dfrac{{\dfrac{{11}}{4} + 3}}{2} = \dfrac{{23}}{8}.$

${\textbf{Hence , rational numbers between 3 and 4 are }}\mathbf{\dfrac{5}{2},\dfrac{9}{4},\dfrac{{11}}{4},\dfrac{{23}}{8},\dfrac{{17}}{8}.}$

Find $10$ rational numbers between $-\dfrac {3}{11}$ and $\dfrac {8}{11}$

We know that

$-3<-2<-1<0<1<2<3<4<5<6<7<8$

If we divide the above numbers by

$11$ we get,

$-\dfrac { 3 }{ 11 } <-\dfrac { 2 }{ 11 } <-\dfrac { 1 }{ 11 } <\dfrac { 0 }{ 11 } <\dfrac { 1 }{ 11 } <\dfrac { 2 }{ 11 } <\dfrac { 3 }{ 11 } <\dfrac { 4 }{ 11 } <\dfrac { 5 }{ 11 } <\dfrac { 6 }{ 11 } <\dfrac { 7 }{ 11 } <\dfrac { 8 }{ 11 }$

Hence, the

$10$ rational numbers between

$-\dfrac { 3 }{ 11 }$ and

$\dfrac { 8 }{ 11 }$ are as follows:

$-\dfrac { 2 }{ 11 } ,-\dfrac { 1 }{ 11 },\dfrac { 0 }{ 11 } ,\dfrac { 1 }{ 11 } ,\dfrac { 2 }{ 11 } ,\dfrac { 3 }{ 11 },\dfrac { 4 }{ 11 } ,\dfrac { 5 }{ 11 } ,\dfrac { 6 }{ 11 },\dfrac { 7 }{ 11 }$

Find the value of ${(-5)}^{2}$

${\textbf{Step - 1: Simplify the given number}}{\text{.}}$

${\text{So, }}{\left( -5 \right)^{ 2}}$

$= {\left( -1 \right)^{ 2}} \times {\left( 5 \right)^{ 2}}$

${\text{As,}} {\left( -1 \right)^{ 2}}=1 , \Rightarrow {\left( -5 \right)^{ 2}}=+1 \times {\left( 5 \right)^{ 2}}$

${\textbf{Step - 2: Apply exponent formula to deduce the value}}{\text{.}}$

${\text{Now we know that , }}{a^n} = a \times a \times a \times .........n{\text{ times}}{\text{.}}$

${\text{So , }}{5^2} = 5 \times 5 = 25.$

${\text{So, }}{\left(- 5 \right)^{ 2}} =+25$

${\textbf{Hence Value of }}\mathbf{{\left( -5 \right)^{ 2}} =+25.}$

Find an irrational number between $\dfrac {5}{7}$ and $\dfrac {7}{9}$ . How many more there may be?

$\dfrac{5}{7}$ and

$\dfrac{7}{9}$ are two rational numbers.

To find, an irrational number between

$\dfrac{5}{7}$ and

$\dfrac{7}{9}$ .

Square root of given number

$\dfrac{5}{7}$ multiplied by square root of

$\dfrac{7}{9}$ we get,

an irrational number

$=$

$\dfrac{\sqrt{5}\times\sqrt{7}}{\sqrt{7}\times\sqrt{9}}$

$= \sqrt{\dfrac{5}{9}}=\dfrac{\sqrt{5}}{3}$ .

Then, the irrational number is

$\dfrac{\sqrt{5}}{3}$ .

And we get infinite number of irrational numbers between two rational numbers.

Which of the following fractions have terminating decimal expansion?
$\displaystyle \frac{1}{16}, \frac{4}{25}, \frac{22}{625}, \frac{1}{1080}$ .

Write fractions into their decimal form

$\displaystyle \frac{1}{16} = 0.0625$

$\displaystyle \frac{4}{25} = 0.26$

$\displaystyle \frac{22}{625} = 0.0352$

$\displaystyle \frac{1}{1080} = 0.000925925...$

$\Rightarrow$ In this case, the digits

$925$ are recurring. So, the fraction

$\displaystyle \frac{1}{1080}$ doesn't have terminating decimal expansion, whereas the first three terms have terminating decimal expansion as they don't have any recurring terms in their decimal expansion.

Find $12$ rational number between $-1$ and $2$ .

$-1=\dfrac{-1}{1}=\dfrac{-5}{5}$

$2=\dfrac{2}{1}=\dfrac{10}{5}$

12 rational numbers between

$-1$ and

$2$ :

$-\dfrac{4}{5}, -\dfrac{3}{5}, -\dfrac{2}{5}, -\dfrac{1}{5}, \dfrac{0}{5}, \dfrac{1}{5}, \dfrac{2}{5}, \dfrac{3}{5}, \dfrac{4}{5}, \dfrac{5}{5}, \dfrac{6}{5}, \dfrac{7}{5}$

Find five rational numbers between $\dfrac {2}{3}$ and $\dfrac {3}{5}$

The

$LCM$ of the denominators of the given rational numbers

$\dfrac { 2 }{ 3 }$ and

$\dfrac {3 }{ 5 }$ is

$LCM(3,5)=15$ .

Therefore, with denominator

$15$ , the rational numbers becomes

$\dfrac { 10 }{ 15 }$ and

$\dfrac {9 }{ 15 }$

The difference between the numerators

$9$ and

$10$ is very less, so we multiply both the numerators and denominators of the rational numbers

$\dfrac { 10 }{ 15 }$ and

$\dfrac {9 }{ 15 }$ by

$10$ to obtain the five rational numbers between them.

Now, the rational numbers becomes

$\dfrac { 100 }{ 150 }$ and

$\dfrac {90 }{ 150 }$

Hence, the

$5$ rational numbers between

$\dfrac {90 }{ 150 }$ and

$\dfrac { 100 }{ 150 }$ are as follows:

$\dfrac { 91 }{ 150 } ,\dfrac { 93 }{ 150 },\dfrac { 94 }{ 150 } ,\dfrac { 97 }{ 150 } ,\dfrac { 99 }{ 150 }$

Simplify the following expressions:
$(\sqrt {5} + \sqrt {2})^{2}$

We need to simplify

$(\sqrt{5}+\sqrt{2})^{2}$

We know that

$(a+b)^{2}=a^{2}+2ab+b^{2}$

$(\sqrt{5}+\sqrt{2})^{2}$

$=(\sqrt{5})^{2}+2\times \sqrt{5}\times \sqrt{2}+(\sqrt{2})^{2}$

$=5+2\sqrt{10}+2$

$=7+2\sqrt{10}$

Locate $\sqrt {2}$ on number line.

$O$ draw a unit square

$OABC$ on number line with each side

$1$ unit in length.

By Pythagoras theorem

$OB = \sqrt {1^{2} + 1^{2}} = \sqrt {2}$

We have seen that

$OB = \sqrt {2}$ . Using a compass with centre

$O$ and radius

$OB$ , draw an arc on the right side to

$O$ intersecting the

number line at the point

$K$ . Now

$K$ corresponds to

$\sqrt {2}$ on the number line.

1058533_569274_ans_bccaea278a514a1391549f73a93f291a.png

Multiply $6\sqrt {3}$ with $13\sqrt {3}$

We need to multiply

$6\sqrt 3$ and

$13\sqrt 3$

Thus

$6\sqrt {3} \times 13\sqrt {3} = 6\times 13\times \sqrt {3}\times \sqrt {3}$

$= 78\times 3 = 234$

Simplify:
(a) $\sqrt [4]{81} - \sqrt [5]{32}$
(b) If $2$ is zero of the polynomial $P(x) = 2x^2 - 3x + 5k$ , find $K$ ?

$\sqrt[4]{81}-8\sqrt[3]{343}+15\sqrt[5]{32}+\sqrt{332}$

$=\sqrt[4]{3\times 3\times 3\times 3}-8(\sqrt[3]{7\times 7\times 7})+15(\sqrt[5]{2\times 2\times 2\times 2\times 2})+\sqrt{332}$

$=\sqrt[4]{3^{4}}-8(\sqrt[3]{3^{7}})+15\sqrt[5]{2^{5}}+18.22$

$=3-8\times 7+15\times 2+18.22$

$=3-56+30+18.22=-4.78$

Simplify the following expressions:
$(3 + \sqrt {3})(2 + \sqrt {2})$

By using identity

$(a+b)(c+d)=ac+ad+bc+bd$

$(3+\sqrt{3})(2+\sqrt{2})$

$=3\times 2+\sqrt{3}\times 2+3\sqrt{2}+\sqrt{3}\times \sqrt{2}$

$=6+2\sqrt{3}+3\sqrt{2}+\sqrt{6}$

Subtract $5\sqrt {3} + 7\sqrt {5}$ from $3\sqrt {5} - 7\sqrt {3}$

$(3\sqrt {5} - 7\sqrt {3}) - (5\sqrt {3} + 7\sqrt {5})$

$= 3\sqrt {5} - 7\sqrt {3} - 5\sqrt {3} - 7\sqrt {5}$

$= -4\sqrt {5} - 12\sqrt {3}$

$= -(4\sqrt {5} + 12\sqrt {3})$

Examine, whether the following number is rational or irrational:
$(3 + \sqrt {3}) + (3 - \sqrt {3})$ .

$(3+\sqrt{3})+(3-\sqrt{3})$

$=3+\sqrt{3}+3-\sqrt{3}$

$=3+3=6$ .

Since the number

$6$ is a rational number,

$(3+\sqrt{3})+(3-\sqrt{3})$ is a rational number.

Locate $\sqrt {3}$ on number line

Construct

$BD$ of

$1$ unit length perpendicular to

$OB$ as fig. Join

$OD$

By Pythagoras theorem,

$OD = \sqrt {(\sqrt {2})^{2} + 1^{2}} = \sqrt {2 + 1} = \sqrt {3}$

Using a compass, with centre

$O$ and radius

$OD$ , draw an arc which

intersects the number line at the point

$L$ right side to

$0$ . Then

$'L'$ corresponds to

$\sqrt {3}$ . From this we can conclude that many

points on the number line can be represented by irrational numbers also.

In the same way, we can locate

$\sqrt {n}$ for any positive integers

$n$ , after

$\sqrt {n - 1}$ has been located.

1034807_569281_ans_fede63535c9449fdaa91dc66ce060dd9.jpg

Simplify the following expressions:
$(\sqrt {5} - \sqrt {2})(\sqrt {5} + \sqrt {2})$

$(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})$

We know that

$(a+b)(a-b)=a^{2}-b^{2}$

$=(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})$

$=(\sqrt{5})^{2}-(\sqrt{2})^{2}$

$=5-2$

$=3$

Simplify $\dfrac {1}{7 + 4\sqrt {3}} + \dfrac {1}{2 + \sqrt {5}}$

$\frac{1}{7+4\sqrt{3}}+\frac{1}{2+\sqrt{5}}$

The rationalizing factor of

$7+4\sqrt{3}$ is

$7-4\sqrt{3}$ and the rationalizing factor of

$2+\sqrt{5}$ is

$2-\sqrt{5}$ .

$\frac{1(7-4\sqrt{3})}{(7+4\sqrt{3})(7-4\sqrt{3})}+\frac{1(2+\sqrt{5})}{(2+\sqrt{5})(2-\sqrt{5})}$

$\frac{7-4\sqrt{3}}{49-48}+\frac{2-\sqrt{5}}{4-5}$

$\frac{7-4\sqrt{3}}{1}+\frac{2-\sqrt{5}}{-1}$

$7-4\sqrt{2}-2+\sqrt{5}$

$5-4\sqrt{3}+\sqrt{5}$

Multiply the surd $(\sqrt 5 + \sqrt 2)$ by _____ to get a rational number.

Solution:

Rationalisation of surd

$(\sqrt5+\sqrt2)$

$(\sqrt5+\sqrt2)\times (\sqrt5-\sqrt2)$

$=(\sqrt5)^2-(\sqrt2)^2$

$=5-2$

$=3$

Simplify
$2^{\frac {2}{3}} . 2^{\frac {1}{3}}$

$\begin{aligned}{}{2^{\frac{2}{3}}}\cdot{2^{\frac{1}{3}}} &= {2^{\frac{2}{3} + \frac{1}{3}}}\quad\quad\quad\because[a^m\cdot a^n=a^{m+n}]\\& = {2^{\frac{{2 + 1}}{3}}}\\& = {2^{\frac{3}{3}}}\\ &= {2^1}\\& = 2\end{aligned}$

Insert $4$ rational numbers between $\cfrac{3}{4}$ and $1$ without using $\left (\cfrac { a+b }{ 2 }\right )$ formula.

To find:

$4$ rational number between

$\dfrac34$ and

$1.$

Solution:

We can write

$\dfrac34$ as

$\dfrac{30}{40}$ and

$1$ as

$\dfrac{40}{40}.$

Since, denominators are same for

$\dfrac{30}{40}$ and

$\dfrac{40}{40}.$ So, we can keep increasing the numerator between

$30$ and

$40$ to get rational numbers between them.

Hence,

$\dfrac{31}{40},\dfrac{33}{40},\dfrac{37}{40}$ and

$\dfrac{39}{40}$ are the four required rational numbers between

$\dfrac34$ and

$1.$

Simplify: $(2^{3})^{0}$

$(2^{3})^{0} = 8^{0} = 1 [\because a^{0} = 1]$

Find any two rational numbers between $\left (-\displaystyle\frac{5}{7} \right )$ and $\left (-\displaystyle\frac{2}{7} \right )$ .

Given rational numbers are

$\dfrac{-5}{7}$ and

$\dfrac{-2}{7}$

To find out: Any two rational numbers between them.

The given rational numbers already have the same denominator.

Hence, we can directly find the rational numbers between them by simply adding

$1$ to the numerator of the smaller number.

$\therefore \$ The numerators will be:

$-5+1=-4$

and

$-4+1=-3$

$\dfrac{-5}{7}$ <

$\dfrac{-4}{7}$ <

$\dfrac{-3}{7}$ <

$\dfrac{-2}{7}$

Hence,

$\dfrac{-4}{7}$ and

$\dfrac{-3}{7}$ are two rational numbers between

$\dfrac{-5}{7}$ and

$\dfrac{-2}{7}$ .

Simplify:
(i) $(-4)^{5}\div (-4)^{8}$

(ii) $\left (\dfrac {1}{2^{3}}\right )^{2}$

(iii) $(-3)^{4} \times \left (\dfrac {5}{3}\right )^{4}$

(iv) $\left (\dfrac {2}{3}\right )^{5}\times \left (\dfrac {3}{4}\right )^{2}\times \left (\dfrac {1}{5}\right )^{2}$

(v) $(3^{-7} \div 3^{10})\times 3^{-5}$

(vi) $\dfrac {2^{6}\times 3^{2} \times 2^{3} \times 3^{7}}{2^{8}\times 3^{6}}$

(vii) $y^{a - b} \times y^{b- c}\times y^{c - a}$

(i)

$(-4)^{5}\div (-4)^{8}$

$\Rightarrow (-4)^{5}\times \dfrac{1}{(-4)^{8}}$

$\Rightarrow (-1)\times (4)^{5}\times \dfrac{1}{(4)^{8}}$

$\Rightarrow (-1)\times \dfrac{1}{(4)^{3}}$

$\Rightarrow -\dfrac{1}{64}$

(ii)

$\left (\dfrac {1}{2^{3}}\right )^{2}$

$\Rightarrow \left (\dfrac {1}{8}\right )^{2}$

$\Rightarrow \dfrac {1}{64}$

(iii)

$(-3)^{4} \times \left (\dfrac {5}{3}\right )^{4}$

$\Rightarrow (3)^{4} \times \left (\dfrac {5^4}{3^4}\right )$

$\Rightarrow 5^4$

$\Rightarrow 625$

(iv)

$\left (\dfrac {2}{3}\right )^{5}\times \left (\dfrac {3}{4}\right )^{2}\times \left (\dfrac {1}{5}\right )^{2}$

$\Rightarrow \left (\dfrac {2^5}{3^5}\right )\times \left (\dfrac {3^2}{4^2}\right )\times \left (\dfrac {1}{5^2}\right )$

$\Rightarrow \left (\dfrac {32}{3^3\times 3^2}\right )\times \left (\dfrac {3^2}{16}\right )\times \left (\dfrac {1}{25}\right )$

$\Rightarrow \left (\dfrac {2}{3^3}\right )\times \left (\dfrac {1}{25}\right )$

$\Rightarrow \left (\dfrac {2}{27}\right )\times \left (\dfrac {1}{25}\right )$

$\Rightarrow \dfrac {2}{675}$

(v)

$(3^{-7} \div 3^{10})\times 3^{-5}$

$\Rightarrow \left(\dfrac{1}{3^{7}} \div 3^{10}\right)\times \dfrac{1}{3^{5}}$

$\Rightarrow \left(\dfrac{1}{3^{7}} \times \dfrac{1}{3^{10}}\right)\times \dfrac{1}{3^5}$

$\Rightarrow \left(\dfrac{1}{3^{17}} \times \dfrac{1}{3^{5}}\right)$

$\Rightarrow \dfrac{1}{3^{22}}$

(vi)

$\dfrac {2^{6}\times 3^{2} \times 2^{3} \times 3^{7}}{2^{8}\times 3^{6}}$

$\Rightarrow \dfrac {2^{6}\times 3^{2} \times 2^{2}\times 2 \times 3^{6}\times 3}{2^{6}\times 2^2\times 3^{6}}$

$\Rightarrow \dfrac {1\times 3^{2} \times 2^{2}\times 2 \times 3}{1\times1}$

$\Rightarrow 216$

(vii)

$y^{a - b} \times y^{b- c}\times y^{c - a}$

$\Rightarrow \dfrac{y^a}{ y^ b} \times \dfrac{y^b}{y^c}\times \dfrac{y^c}{ y^a}$

$\Rightarrow 1$

Hence, this is the answer.

Insert any two irrational numbers between $\dfrac{4}{7}$ and $\dfrac{5}{7}$ .

$\dfrac{5}{7}=0.714285714285...=0.\overline{714285}$

and

$\dfrac{4}{7}=0.571428571428...=0.\overline{571428}$ .

Here, the two decimal expansions are non-terminating recurring.

Hence,

$\frac {5} {7}$ and

$\frac {4}{7}$ are two rational numbers.

We know, between any two rational numbers, there are infinitely many irrational numbers.

An irrational number has non-terminating non-recurring decimal expansions.

Then two different irrational number between

$\dfrac{5}{7}$ and

$\dfrac{4}{7}$ can be:

$0.5918264201001...$ and

$0.601001000100001...$

NOTE:

This is two of the various possible answers. It might have different solution.

If $\displaystyle\frac{\sqrt{7}-1}{\sqrt{7}+1}+\frac{\sqrt{7}+1}{\sqrt{7}-1}=a+b\sqrt{7}$ , find the values of a and b.

$\displaystyle\frac{\sqrt{7}-1}{\sqrt{7}+1}+\frac{\sqrt{7}+1}{\sqrt{7}-1}=\frac{\sqrt{7}-1}{\sqrt{7}+1}\times \frac{\sqrt{7}-1}{\sqrt{7}-1}+\frac{\sqrt{7}+1}{\sqrt{7}-1}\times \frac{\sqrt{7}+1}{\sqrt{7}+1}$

$=\displaystyle\frac{(\sqrt{7}-1)^2}{(\sqrt{7})^2-1}+\frac{(\sqrt{7}+1)^2}{(\sqrt{7})^2-1}$

$=\displaystyle\frac{7+1-2\sqrt{7}}{7-1}+\frac{7+1+2\sqrt{7}}{7-1}$

$=\displaystyle \frac{8-2\sqrt{7}}{6}+\frac{8+2\sqrt{7}}{6}$

$=\displaystyle \frac{8-2\sqrt{7}+8+2\sqrt{7}}{6}$

$=\displaystyle\frac{16}{6}=\frac{8}{3}+0\sqrt{7}$

$\therefore \displaystyle\frac{8}{3}+0\sqrt{7}=a+b\sqrt{7}\Rightarrow a=\frac{8}{3}, b=0$ .

Find the value of:
(i) $(3^{0} + 4^{-1})\times 2^{2}$

(ii) $(2^{-1}\times 4^{-1}) \div 2^{-2}$

(iii) $\left (\dfrac {1}{2}\right )^{-2} + \left (\dfrac {1}{3}\right )^{-2} + \left (\dfrac {1}{4}\right )^{-2}$

(iv) $(3^{-1} + 4^{-1} + 5^{-1})^{0}$

(v) $\left [\left (\dfrac {-2}{3}\right )^{-2}\right ]^{2}$

(vi) $7^{-20} - 7^{-21}$

(i)

$(3^{0} + 4^{-1})\times 2^{2}$

$=\left(1 + \dfrac{1}{4}\right)\times 4$

$=\left( \dfrac{5}{4}\right)\times 4$

$=5$

(ii)

$(2^{-1}\times 4^{-1}) \div 2^{-2}$

$=\left(\dfrac{1}{2}\times \dfrac{1}{4}\right) \times 2^{2}$

$=\left(\dfrac{1}{2}\times \dfrac{1}{4}\right) \times 4$

$=\dfrac{1}{2}$

(iii)

$\left (\dfrac {1}{2}\right )^{-2} + \left (\dfrac {1}{3}\right )^{-2} + \left (\dfrac {1}{4}\right )^{-2}$

$=2^2+ 3^2+ 4^2$

$=4+ 9+ 16$

$=29$

(iv)

$(3^{-1} + 4^{-1} + 5^{-1})^{0}$

$=\left(\dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5}\right)^{0}$

$=1$

$\because\ a^0=1$

(v)

$\left [\left (\dfrac {-2}{3}\right )^{-2}\right ]^{2}$

$=\left (\dfrac {-2}{3}\right )^{-4}$

$=\left (\dfrac {-3}{2}\right )^{4}$

$=\dfrac {81}{16}$

(vi)

$7^{-20} - 7^{-21}$

$=\dfrac{1}{7^{20}} - \dfrac{1}{7^{21}}$

$=\dfrac{1}{7^{20}}\left(1 - \dfrac{1}{7}\right)$

$=\dfrac{1}{7^{20}}\left(\dfrac{6}{7}\right)$

$=\dfrac{6}{7^{21}}$

Find any two irrational numbers between $0.15$ and $0.16$ .

$0.15$ and

$0.16$ are two rational numbers.

We know, between any two rational numbers, there are infinitely many irrational numbers.

An irrational number has non-terminating non-recurring decimal expansions.

Then two irrational numbers between

$0.15$ and

$0.16$ can be:

$0.1517682318965401001...$ and

$0.1557826432987921098...$

NOTE:

This is two of the various possible answers. It might have different solution.

Find any two irrational numbers between $3$ and $3.5$ .

$3$ and

$3.5$ are two rational numbers.

We know, between any two rational numbers, there are infinitely many irrational numbers.

An irrational number has non-terminating non-recurring decimal expansions.

Then two irrational numbers between

$3$ and

$3.5$ can be:

$3.124689010010001...$ and

$3.246891787265401001...$

NOTE:

This is two of the various possible answers. It might have different solution.

Find two rational numbers between
(i) $\dfrac {2}{7}$ and $\dfrac {3}{5}$
(ii) $\dfrac {6}{5}$ and $\dfrac {9}{11}$
(iii) $\dfrac {1}{3}$ and $\dfrac {4}{5}$
(iv) $\dfrac {-1}{6}$ and $\dfrac {1}{3}$

1718126_616175_ans_1ef54973909e442882783d109af820b1.jpg

Classify the following numbers as rational or irrational.
(i) $\sqrt{11}$
(ii) $\sqrt{81}$
(iii) $0.0625$
(iv) $0.8\overline{3}$
(v) $1.505500555$ ....

(i)

$\sqrt{11}$ is an irrational number.(

$11$ is not a perfect square number)
(ii)

$\sqrt{81}=9=\displaystyle\frac{9}{1}$ , a rational number.
(iii)

$0.0625$ is a terminating decimal.

$\therefore 0.0625$ is a rational number.
(iv)

$0.8\overline{3}=0.8333$ ...
The decimal expansion is non-terminating and recurring.

$\therefore 0.8\overline{3}$ is a rational number.
(v) The decimal number is non-terminating and non-recurring.

$\therefore 1.505500555$ .... is an irrational number.

Simplify the following using law of exponents.
$(-7)^7\times (-7)^8$ = $(-7)^{m}$ .Find m

we know,

$a^{m}*a^{n}=(a)^{m+n}$

so,

$(-7)^{7}*(-7)^{8}=(-7)^{7+8}$

$=(-7)^{15}$

$\implies m=15$

In the following equations determine whether x, y, z represent rational or irrational numbers.
(i) $x^3=8$
(ii) $x^2=82$
(iii) $y^2=3$
(iv) $z^2=0.09$

(i)

$x^3=8=2^3$ (

$8$ is a perfect cube)

$\rightarrow x=2,$ a rational number.
(ii)

$x^2=81=9^2$ (

$81$ is a perfect square)

$\rightarrow x=9$ , a rational number.
(iii)

$y^2=3\rightarrow y=\sqrt{3}$ , an irrational number.
(iv)

$z^2=0.09=\displaystyle\frac{9}{100}=\left(\displaystyle\frac{3}{10}\right)^2$

$\rightarrow z=\displaystyle\frac{3}{10}$ , a rational number.

Solve the following equations.
$\displaystyle\, 2^x \cdot 5^{x - 1} = 0.2 \cdot 10^{2 - x}$

$2^{x}.5^{x-1}=0.2.10^{2-x}$

$2^x.5^{x-1}=2.10^{1-x}$

$2^x.5^{x-1}=2^{2-x}.5^{1-x}$

comparing powers on both sides

$x=2-x \implies x=1$

Simplify the following:
a) ${\left( {{a^8} \times {a^5}} \right)^0}$
b) ${\left( {{b^2}} \right)^4} \times {b^0}$
c) $- \dfrac{{{a^{ - 3}}{b^{10}}{c^8} \times b{c^8}}}{{{{\left( {{b^{ - 10}} \times {c^6}} \right)}^4}}}$
d) ${\left[ {\dfrac{{ - {a^{ - 5}}}}{{ - {a^{10}}{b^{ - 9}}{c^{ - 7}} \times \left( {{a^{ - 15}} \times {c^0}} \right)}}} \right]^2}$

$a)\,\,{\left( {{a^8} \times {a^{ - 5}}} \right)^0} = 1\,\,\,\,\,\,\,\,\left[ {\because \,\,{a^0} = 1} \right]$

$b)\,\,{\left( {{b^2}} \right)^4} \times {b^0} = {b^8} \times {b^0}\,\,\,\,\,\left[ {{{\left( {{a^n}} \right)}^m} = {a^{mn}}} \right]$

$= {b^8} \times 1\,\,\,\,\,\,\,\,\,\left[ {{a^0} = 1} \right]$

$c)\,\,\cfrac{{ - {a^{ - 3}}{b^{10}}{c^8} \times b{c^8}}}{{{{\left( {{b^{ - 10}} \times {c^6}} \right)}^4}}}$

$= \,\,\cfrac{{ - {a^{ - 3}} \cdot {b^{10}} \cdot {c^8} \times {b^1} \times {c^8}}}{{{b^{ - 40}} \times {c^{24}}}}\,\,\,\,\,\,\,\,\,\,\left[ {\because \,\,{{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right]$

$= \,\,\cfrac{{ - {a^{ - 3}} \times {b^{11}} \times {c^{16}}}}{{{b^{ - 40}} \times {c^{24}}}}\,\,\,\,\,\,\,\,\,\,\left[ {\because \,\,\cfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}} \right]$

$= - {a^{ - 3}} \times {b^{11 + 40}} \times {c^{16 - 24}}$

$= - {a^{ - 3}} \times {b^{51}} \times {c^{ - 8}}$

$= \cfrac{{ - {b^{51}}}}{{{a^3}{c^8}}}$

$d)\,\,{\left[ {\cfrac{{ - {a^{ - 5}}}}{{ - {a^{10}}{b^{ - 9}}{c^{ - 7}} \times \left( {{a^{ - 15}} \times {c^0}} \right)}}} \right]^2}$

$= \cfrac{{ - {a^{ - 10}}}}{{ - {a^{20}} \times {b^{ - 18}} \times {c^{ - 14}} \times {a^{ - 30}} \times {c^0}}}$

$= \cfrac{{ - {a^{ - 10}}}}{{ - {a^{20 - 30}} \times {b^{ - 18}} \times {c^{ - 14 + 0}}}}$

$= \cfrac{{{a^{ - 10}}}}{{{a^{ - 10}} \times {b^{ - 18}} \times {c^{ - 14}}}}$

$= {b^{18}}{c^{14}}$

Find (i)six (ii)sixty (iii) six hundred rational between $\dfrac{-5}{8}$ and $\dfrac{3}{8}$ .

1694858_1000637_ans_ee547520579747079031172fa3ef693d.jpg

Find four rational numbers between $\dfrac{3}{7}$ & $\dfrac{5}{7}$

1244827_1000359_ans_59e82daa449944579d8bc0e338d88a9e.jpeg

If $a = \frac{1}{{\sqrt 5 - \sqrt 3 }},\,\,b = \frac{1}{{\sqrt 7 + \sqrt 5 }}$ and $c = \frac{1}{{\sqrt 9 - \sqrt 7 }},$ then $a + b + c$ equals

$a = \dfrac{1}{{\sqrt 5 - \sqrt 3 }} \times \dfrac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} = \dfrac{{\sqrt 5 + \sqrt 3 }}{{5 - 3}} = \dfrac{{\sqrt 5 + \sqrt 3 }}{2}$

$b = \dfrac{1}{{\sqrt 7 + \sqrt 5 }} \times \dfrac{{\sqrt 7 - \sqrt 5 }}{{\sqrt 7 - \sqrt 5 }} = \dfrac{{\sqrt 7 - \sqrt 5 }}{{7 - 5}} = \dfrac{{\sqrt 7 - \sqrt 5 }}{2}$

$c = \dfrac{1}{{\sqrt 9 + \sqrt 7 }} \times \dfrac{{\sqrt 9 - \sqrt 7 }}{{\sqrt 9 - \sqrt 7 }} = \dfrac{{\sqrt 9 - \sqrt 7 }}{{9 - 7}} = \dfrac{{\sqrt 9 - \sqrt 7 }}{2}$

$\therefore \,\,a + b + c$

$= \dfrac{{\sqrt 5 + \sqrt 3 }}{2} + \dfrac{{\sqrt 7 - \sqrt 5 }}{2} + \dfrac{{\sqrt 9 - \sqrt 7 }}{2}$

$= \dfrac{{\sqrt 5 + \sqrt 3 + \sqrt 7 - \sqrt 5 + \sqrt 9 - \sqrt 7 }}{2}$

$= \dfrac{{\sqrt 3 + 3}}{2}$

$\therefore \,\,\,\dfrac{{3 + \sqrt 3 }}{2}$

$(\sqrt{5 \, + \, 2 \sqrt{6}}) ^x \, + \, (\sqrt{5 \, - \, 2 \sqrt{6}}) ^x \, = \, 10$

left hand side is equale to right hand side

after puting x=2

LHS = 10

=RHS

Represent $\sqrt {10.5}$ on the number line

Instructions:

1) On a straight line mark

$A$ and

$B$ such that

$AB=x$ .

2) Mark point

$C$

$1\ unit$ away from

$B(BC=1\ unit)$

3) Draw the

$\bot$ bisector of

$AC$ , mark the pt of intersection as

$O$

4) Taking

$O$ as centre and

$OC$ as radius draw a semicircle.

5) Draw perpendicular from

$B$ and mark its pt of intersection with semicircle as

$D$

6) Now

$BD=\sqrt x$

7) Taking

$B$ as centre and

$BD=\sqrt x$ as radius construct an arc to make it intersect at

$E$ on the line.

$BE=\sqrt x$

9) Start the number line from

$B$

1438561_995659_ans_5aa6496bfd2e4087b53ecf7de1ecc4b7.png

If $\dfrac{\sqrt 5 + \sqrt 3 } {\sqrt 5 - \sqrt 3 } = a + b\sqrt {15}$ , find $a$ and $b$ .

Consider the given equation

$\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=a+b\sqrt{15}$

$LHS$ is,

$\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$

$\Rightarrow\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \times \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

$\Rightarrow\dfrac{\left ( \sqrt{5}+\sqrt{3} \right )^2}{\left ( \sqrt{5}\right )^2-\left ( \sqrt{3} \right )^2}$

$\Rightarrow\dfrac{5+3+2\sqrt{15}}{2}$

$\Rightarrow\dfrac{8+2\sqrt{15}}{2}$

$\Rightarrow4+\sqrt{15}$

$RHS$ is,

$\Rightarrow a+b\sqrt{15}$

On comparing with the

$RHS$ , we get,

$a=4,b=1$

Five rational numbers between $2$ and $1$ are
Four rational numbers between $\dfrac{2} {3}$ and $\dfrac{3} {5}$ are

(i) as we have to five rational number ,we multiply number by

$\dfrac66$

i.e

$1\times\dfrac66= \dfrac66$ and

$2\times\dfrac66=\dfrac{12}{6}$

thus 5 rational number between 1and 2 are

$\dfrac76,\dfrac86,\dfrac96,\dfrac{10}{6},\dfrac{11}{6}$

(ii) as in

$\dfrac23 ,\dfrac35$ isn't denominator same , to make same denominator take LCM of 3 and 5

i.e LCM(3,5)= 15

now,

fractions are

$\dfrac{10}{15}$ and

$\dfrac{9}{15}$

multiply

$\dfrac{10}{15}$ and

$\dfrac{9}{15}$ by

$\dfrac55$

$\Rightarrow\dfrac{50}{75},\dfrac{45}{75}$

thus 4 rational number between

$\dfrac{2}{3}$ and

$\dfrac35$ are

$\dfrac{46}{75},\dfrac{47}{75},\dfrac{48}{75},\dfrac{49}{75}$

How many rational numbers are there between $- \dfrac{3}{2}$ and 0 with denominator as 1?

$-\cfrac{3}{2}$ is equal to

$-1.5$ so the integers comes between

$-1.5$ and

$0$ is

$-1$

Write the multiplicative ${\left( {{{ - 7} \over {13}}} \right)^{ - 6}}$ with a positive exponent and also with a negative exponent

$\left(\dfrac{-7}{13}\right)^{-6}$

Positive exponent

$\left(\dfrac{13}{7}\right)^{6}=\dfrac{(13)^{6}}{(-7)^{6}}=\dfrac{4826809}{-117649}$

Negative exponent

$\left(\dfrac{-7}{13}\right)^{-6}=\dfrac{(-7)^{6}}{(13)^{-6}}=\dfrac{1/-118649}{1/4826809}$

Insert a rational number between $\frac{1}{3}$ and $\frac{4}{5}$ and arrange in descending order

By taking LCM of 3 and 5 which is 15 and converting these rational numbers into rational numbers with common denominator we get,

$\dfrac{1}{3}=\dfrac{5}{15}$ and

$\dfrac{4}{5}=\dfrac{12}{15}$

the rational number between

$\dfrac{1}{3}$ and

$\dfrac{4}{5}$ is

$\dfrac{7}{15}$

The decending order is

$\dfrac{4}{5},\dfrac{7}{15},\dfrac{1}{3}$

Expand ${\left( {\sqrt 3 + \sqrt 7 } \right)^2}$

${ (\sqrt { 3 } +\sqrt { 7 } ) }^{ 2 }\\ =3+7+2\sqrt { 3 } \sqrt { 7 } \\ =10+2\sqrt { 21 }$

Insert three rational number between $4$ and $4.5$ .

A rational number is a number which can be expressed in the form

$\dfrac{p}{q}$ .

$4.1, 4.2$ and

$4.3$ can be expressed as

$\dfrac{42}{10}$ ,

$\dfrac{43}{10}$ and

$\dfrac{41}{10}$ respectively.

Thus three rational numbers between

$4$ and

$4.5$ is

$4.1,4.2,4.3$

Find the value of m and n if
${4^{2m}} = {\left( {\root 3 \of {16} } \right)^{ - {6 \over n}}} = {\left( {\sqrt 8 } \right)^2}$

$4^{2{m}}=(\sqrt[3]{16})^{-\dfrac{6}{n}}=(\sqrt{8})^{2}$

$2^{4{m}}=2^{-\dfrac{8}{n}}=2^{3}$

$\implies m=\dfrac{3}{4},n=-\dfrac{8}{3}$

Find the 7 rational number between -7/8 and 6/8.

the

$7$ rational numbers between

$\dfrac{-7}{8}$ and

$\dfrac{6}{8}$ are

$\dfrac{-6}{8},\dfrac{-5}{8},\dfrac{-4}{8},\dfrac{-3}{8},\dfrac{-2}{8},\dfrac{-1}{8},0$

Find $3$ rational numbers between $\dfrac{1}{6}$ and $\dfrac{1}{7}$ .

Given rational numbers are

$\dfrac{1}{6}$ and

$\dfrac{1}{7}$ :

L.C.M of

$(6,7)=6\times 7= 42$

Now making the denominators equal, we multiply with

$70$ and

$60$ respectively.

$\Rightarrow \dfrac{1}{6} \times \dfrac{70}{70} = \dfrac{70}{420}$

$\Rightarrow \dfrac{1}{7} \times \dfrac{60}{60} = \dfrac{60}{420}$

the three rational numbers between

$\dfrac{60}{420}$ and

$\dfrac{70}{420}$ are

$\dfrac{64}{420},\dfrac{67}{420},\dfrac{69}{420}$

$(2\sqrt{2}+5\sqrt {3})+(\sqrt{2}-3\sqrt{3})$

$(2\sqrt{2}+5\sqrt{3})+(\sqrt{2}-3\sqrt{3})=3\sqrt{2}+2\sqrt{3}$

Solve
$\left( {\sqrt 8 - \sqrt 2 } \right)\left( {\sqrt 8 + \sqrt 2 } \right)$

$=(\sqrt { 8 } -\sqrt { 2 } )(\sqrt { 8 } +\sqrt { 2 } )\\ ={ (\sqrt { 8 } ) }^{ 2 }-{ (\sqrt { 2 } ) }^{ 2 }\\ =8-2\\ =6$

Insert a rational number between $\frac{2}{9}$ and $\frac{3}{8}$ and arrange in ascending order.

$\dfrac{2}{9}=\dfrac{16}{72}$ and

$\dfrac{3}{8}=\dfrac{27}{72}$

the rational number between

$\dfrac{2}{9}$ and

$\dfrac{3}{8}$ is

$\dfrac{17}{72}$

The ascending order is

$\dfrac{2}{9},\dfrac{17}{72},\dfrac{3}{8}$

Let, $f(x)=2^{nx+1}$ then show that $f(a).f(b).f(c)=4f(a+b+c)$ .

Given,

$f(x)=2^{nx+1}$ .

Now

$f(a).f(b).f(c)$

$=2^{na+1}.$

$2^{nb+1}.$

$2^{nc+1}$

$=2^{n(a+b+c)+1+2}$

$=2^2.2^{n(a+b+c)+1}$

$=2^2.f(a+b+c)$

$=4f(a+b+c)$ .

If ${x^2} = 5$ , then $x$ is a

$x^2=5$

$x=\sqrt5\ or \ -\sqrt5$

$\therefore\ x$ is a

$\text{irrational number}$

The value of $x$ in
$\left(\dfrac{1}{2}\right)^3 \times \left(\dfrac{1}{2}\right)^5 = \left(\dfrac{1}{4}\right)^x$

$\Big(\dfrac 12\Big)^3 \times \Big(\dfrac 12\Big)^5=\Big(\dfrac 14\Big)^x$

$\implies \Big(\dfrac 12\Big)^{3+5}=\Big(\dfrac 12\Big)^{2x}$

$\implies \Big(\dfrac 12\Big)^{8}=\Big(\dfrac 12\Big)^{2x}$

If bases are equal then powers are equated

Therefore,

$8=2x$

$\implies x=4$

Find four rational numbers between $\dfrac{{ - 2}}{5}$ and $\dfrac{{ - 3}}{4}$ .

Given fractions have denominators of

$5$ and

$4$ .

$L.C.M$ of

$5$ and

$4$ is

$5\times 4 =20$

Making the denominators of both the fractions same :

multiply

$-\dfrac{2}{5}$ by

$\dfrac{4}{4}$

$\Rightarrow -\dfrac{2}{5} \times \dfrac{4}{4} = -\dfrac{8}{20}$

multiply

$-\dfrac{3}{5}$ by

$\dfrac{5}{5}$

$\Rightarrow -\dfrac{3}{4} \times \dfrac{4}{4}=-\dfrac{15}{20}$

Rational numbers between

$-\dfrac{2}{5}$ and

$-\dfrac{3}{4}$ are

$-\dfrac{10}{20}, -\dfrac{12}{20}, -\dfrac{13}{20}, -\dfrac{14}{20}$

Evaluate: $\left( \dfrac {1}{2}\times \dfrac {\sqrt {3}}{2}\right)+\left( \dfrac {1}{\sqrt {2}}\times \dfrac {1}{2}\right)$

$(\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}) + (\dfrac{1}{\sqrt 2}\times \dfrac{1}{2})$

$(\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}) + (\dfrac{1}{\sqrt 2}\times \dfrac{1}{2}\times \dfrac{\sqrt 2}{\sqrt 2})$

$\dfrac{\sqrt 3}{4}+\dfrac{\sqrt{2}}{4}=\dfrac{\sqrt 3+\sqrt 2}{4}$

Hence the correct answer is

$\dfrac{\sqrt 3+\sqrt 2}{4}$

Find six rational number between $\dfrac{1}{2}$ and $\dfrac{2}{3}$ .

$\dfrac{1}{2}$ and

$\dfrac{2}{3}$

$\dfrac{1}{2}\times\dfrac{10}{10}=\dfrac{10}{20}$ and

$\dfrac{2}{3}\times\dfrac{10}{10}=\dfrac{20}{30}$

L.C.M of

$20$ and

$30$ is

$60$

The

$6$ rational numbers between

$\dfrac{30}{60}$ and

$\dfrac{40}{60}$ is

$\dfrac{31}{60},\dfrac{32}{60},\dfrac{33}{60},\dfrac{34}{60},\dfrac{35}{60},\dfrac{36}{60}$

$\dfrac{31}{60},\dfrac{8}{15},\dfrac{11}{20},\dfrac{17}{30},\dfrac{7}{12},\dfrac{6}{10}$

Find ten rational number between $\dfrac{-2}{5}$ and $\dfrac{1}{7}$ .

1698433_1015828_ans_2e3649519bd748048367a982152846d8.jpg

Simplify: $3\sqrt [ 3 ]{ 40 } -4\sqrt [ 3 ]{ 320 } -\sqrt [ 3 ]{ 5 }$

$3\sqrt[3]{40} - 4\sqrt[3]{320} - \sqrt[3]{5}$

$= 3\sqrt[3]{8\times5} - 4\sqrt[3]{64\times5} - \sqrt[3]{5}$

$= 3\times 2\sqrt[3]{5} - 4\times 4\sqrt[3]{5} - \sqrt[3]{5}$

$= 6\sqrt[3]{5} - 16\sqrt[3]{5} - \sqrt[3]{5}$

$= -11\sqrt[3]{5}$

Solve the following :
$\left( i \right)\,{\left( {\dfrac{{27}}{{125}}} \right)^{\dfrac{2}{3}}} \times {\left( {\dfrac{9}{{25}}} \right)^{ - \dfrac{3}{2}}}$
$\left( {ii} \right)\,{7^0} \times {\left( {25} \right)^{ - \dfrac{3}{2}}} - {5^{ - 3}}$
$\left( {iii} \right)\,{\left( {\dfrac{{16}}{{81}}} \right)^{ - \dfrac{3}{4}}} \times {\left( {\dfrac{{49}}{9}} \right)^{\dfrac{3}{2}}} \div {\left( {\dfrac{{343}}{{216}}} \right)^{\dfrac{2}{3}}}$

$(i)\ \left( \dfrac{27}{125} \right)^{\dfrac{2}{3}} \times \left( \dfrac{9}{25} \right)^{\dfrac{3}{2}}$

$\left( \left( \dfrac{3}{5} \right)^{3} \right)^{\dfrac{2}{3}} \times \left( \dfrac{3}{5} \right)^{\dfrac{-3}{2}}$

$\left( \dfrac{3}{5} \right)^{2} \times \left( \dfrac{3}{5} \right)^{-3} = \dfrac{3^{2}}{5^{2}} \times \dfrac{5^{3}}{3^{3}}= \dfrac{5}{3}$

$(ii)\ 7^{o} \times (25)^{\dfrac{-3}{2}}- 5^{-3}$

$1 \times (5^{2})^{-3/2} \times \left( \dfrac{49}{9} \right)^{3/2} \div \left( \dfrac{343}{216} \right)^{\dfrac{2}{3}}$

$\left( \left( \dfrac{2}{3} \right)^{4} \right)^{-3/4} \times \left( \left( \dfrac{7}{3} \right)^{2} \right)^{3/2} \div \left( \left( \dfrac{3}{6} \right)^{3} \right)^{2/3}$

$\dfrac{2^{-3}}{3^{-3}} \times \dfrac{7^{3}}{3^{3}} \times \dfrac{3^{2}}{7^{2}}= 7 \times 2^{2} \times 3^{2} \times 2^{3} =\dfrac{9 \times 9}{2}= \dfrac{63}{2}$

Find the value of $x$
i) $\left(\dfrac{4}{3} \right)^{-4} \times \left(\dfrac{4}{3} \right)^{-5} = \left(\dfrac{4}{3} \right)^{-3x}$
ii) $7^x \div 7^3 = 7^5$
iii) $4^{2x + 1} \div 16 = 64$

(i)

$(\dfrac{4}{3})^{-4}\times (\dfrac 43)^{-5}=(\dfrac 43)^{-3x}$

$\implies (\dfrac{4}{3})^{-4+5}=(\dfrac 43)^{-3x}$ (since,

$a^m\times a^n=a^{m+n}$ )

$\implies (\dfrac{4}{3})=(\dfrac 43)^{-3x}$

bases are equal then equate the powers.

$\implies 1=-3x$

$\implies x=\dfrac{-1}{3}$

(ii)

$\dfrac{7^x}{7^3}=7^5$

$\implies 7^{x-3}=7^5$ (since,

$a^m/ a^n=a^{m-n}$ )

bases are equal then equate the powers.

$\implies x-3=5$

$\implies x=8$

(ii)

$\dfrac{4^{2x+1}}{4^2}=4^3$

$\implies 4^{2x+1-2}=4^3$ (since,

$a^m/ a^n=a^{m-n}$ )

bases are equal then equate the powers.

$\implies 2x-1=3$

$\implies 2x=4$

$\implies x=2$

Carry out the following additions of rational numbers.
$\dfrac {5}{36}+\dfrac {6}{42}$

Simplifying $\dfrac{5}{{36}} + \dfrac{6}{{42}}$ .

$\dfrac{5}{{36}} + \dfrac{6}{{42}} = \dfrac{5}{{36}} + \dfrac{1}{7}$

Taking the LCM,

$\dfrac{5}{{36}} + \dfrac{1}{7} = \dfrac{{5\left( 7 \right) + 1\left( {36} \right)}}{{252}}$

$= \dfrac{{71}}{{252}}$

Find three rational numbers between $5$ and $-2$ .

calculate first average

$\cfrac{{ - 2 + 5}}{2} = \cfrac{3}{2}$

Again,

$\cfrac{{ - 2 + \cfrac{3}{2}}}{2} = - \cfrac{1}{4}$

and,

$\cfrac{{\cfrac{3}{2} + 5}}{2} = \cfrac{{13}}{4}$

Hence the three rational numbers between -2 and 5 is

$\cfrac{{ - 1}}{4},\cfrac{3}{2},\cfrac{{13}}{4}$

Find any ten rational numbers between $\frac{-5}{6}$ and $\frac{5}{8}$ .

Making their denominators same = $\dfrac{-5\times 8}{6\times 8}= \dfrac{-40}{48}$ and $\dfrac{5\times 6}{8\times 6}= \dfrac{30}{48}$
We can pick any ten integers as numerator in between $-40$ and $30$ with denominator as $48$ :
$(\dfrac{-35}{48},\dfrac{-31}{48},\dfrac{-7}{48},\dfrac{-23}{48},\dfrac{9}{48},\dfrac{1}{48},\dfrac{19}{48},0,\dfrac{27}{48},\dfrac{24}{48}=\dfrac{1}{2})$

If $\dfrac{{{{25}^x} \times {{10}^{2x}}}}{{{4^x}}} = {5^8}$ , find $x$ .

$\dfrac{{25}^{x}\times{10}^{2x}}{{4}^{x}}={5}^{8}$

$\Rightarrow\,\dfrac{{5}^{2x}\times{2}^{2x}\times{5}^{2x}}{{2}^{2x}}={5}^{8}$

$\Rightarrow\,{5}^{4x}\times{2}^{2x-2x}={5}^{8}$

$\Rightarrow\,{5}^{4x}\times{2}^{0}={5}^{8}$

$\Rightarrow\,{5}^{4x}={5}^{8}$

Since bases are same, we can equate the powers.

$\Rightarrow\,4x=8$

$\therefore\,x=\dfrac{8}{4}=2$

Solve for $x$ such that $4^{2\log_2 x}=81$ .

Now,

$4^{2\log_2 x}=81$

or,

$2^{4\log_2 x}=81$

or,

$2^{\log_2 x^4}=81$

or,

$x^4=81$ [ Since

$\log_a a=1$ for

$a>0$ ]

or,

$x^4=3^41$

or,

$x=3$

Find an irrational number between $0.1$ and $0.19$ .

$0.1$ and

$0.19$ are two rational numbers, as it is terminating.

We know, between any two rational numbers, there are infinitely many irrational numbers.

An irrational number has non-terminating non-recurring decimal expansions.

Then irrational number between

$0.1$ and

$0.19$ can be

$0.1010010001...$ .

Similarly,

$0.13030030003...$ is also an irrational number between

$0.1$ and

$0.19$ .

NOTE:

These are some of the possible answers. It might have different solution.

Carry out the following additions of rational numbers.
$\dfrac {11}{17}+\dfrac {13}{19}$

Taking the LCM in $\dfrac{{11}}{{17}} + \dfrac{{13}}{{19}}$ .

$\dfrac{{11}}{{17}} + \dfrac{{13}}{{19}} = \dfrac{{11\left( {19} \right) + 13\left( {17} \right)}}{{323}}$

$= \dfrac{{430}}{{323}}$

Find the value to three places of decimals of each of the following. It is given that $\sqrt {2}=1.414, \sqrt {3}=1.732, \sqrt {5}=2.236$ and $\sqrt {10}=3.162$
$\dfrac {\sqrt {5}+1}{\sqrt {2}}$

Given that:-

$\sqrt{2} = 1.414$

$\sqrt{3} = 1.732$

$\sqrt{5} = 2.236$

$\sqrt{10} = 3.162$

$\therefore \; \cfrac{\sqrt{5} + 1}{\sqrt{2}} = \cfrac{2.236 + 1}{1.414} = \cfrac{3.236}{1.414} = 2.2885431 \approx 2.288$

Hence, the correct answer is

$2.288$ .

Find $3$ irrational numbers between $\dfrac {3} {8}$ and $\dfrac {2}{5}$ .

Given rational numbers an be written in decimal form as,

$\dfrac{3}{8}=0.375$

$\dfrac{2}{5}=0.4$

We know, between any two rational numbers, there are infinitely many irrational numbers. An irrational number has non-terminating non-recurring decimal expansions.

Then three different irrational number between

$\dfrac{3}{8}$ and

$\dfrac{2}{5}$ can be:

$0.3750750075000075\dots$

$0.387038700387\dots$

$0.3909009000900009\dots$

Simplify the expression:
$\cfrac{{7}^{n+3}-9\times {7}^{n+1}}{31\times {7}^{n+1}-3\times {7}^{n+2}}$

$\dfrac{7^{n+3}-9\times 7^{n+1}}{31\times 7^{n+1}-3\times 7^{n+2}}$

$\dfrac{7^{n+1}(7^{2}-9)}{7^{n+1}(31-3\times 7)}$

$=\dfrac{40}{10}$

$=4$

Solve :
$\dfrac{7 + 3\sqrt{5}}{3+\sqrt{5}} - \dfrac{7 - 3\sqrt{5}}{3- \sqrt{5}} = a + \sqrt{5}b$

$\cfrac{7+3\sqrt{5}}{3+\sqrt{5}}-\cfrac{7-3\sqrt{5}}{3-\sqrt{5}}$

$=\cfrac{(7+3\sqrt{5})(3-\sqrt{5})-(7-3\sqrt{5})(3+\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}$

$=\cfrac{(6+2\sqrt{5})-(6-2\sqrt{5})}{3^{2}-(\sqrt{5})^{2}}$

$=\cfrac{4\sqrt{5}}{4}$

$=\sqrt{5}$

Now,

$\sqrt{5}=a+b\sqrt{5}$

$\Rightarrow a=0,\; b=1$

If $x = 8 + 3\sqrt 7$ , then find the value of ${x^2} + \dfrac{1}{{{x^2}}}$

$x=8+3\sqrt{7}$

$x^{2}=(8+3\sqrt{7})^{2}$

$x^{2}=8^{2}+(3\sqrt{7})^{2}+2\times 8\times3\sqrt{7}$

$x^{2}=64+63+48\sqrt{7}$

$x^{2}=127+48\sqrt{7}$ .......................(1)

$\dfrac{1}{x^{2}}=\dfrac{1}{127+48\sqrt{7}}$

Rationalising the denominator on the RHS,

$\dfrac{1}{x^{2}}=\dfrac{1}{127+48\sqrt{7}}\times\dfrac{127-48\sqrt{7}}{127-48\sqrt{7}}$

$\dfrac{1}{x^{2}}=\dfrac{127-48\sqrt{7}}{127^{2}-(48\sqrt{7})^{2}}$

$\dfrac{1}{x^{2}}=\dfrac{127-48\sqrt{7}}{16129-16128}$

$\dfrac{1}{x^{2}}=127-48\sqrt{7}$ .................(2)

Adding (1) and (2)

$x^{2}+\dfrac{1}{x^{2}}=127+48\sqrt{7}+127-48\sqrt{7}$

$x^{2}+\dfrac{1}{x^{2}}=127+127=254$

$x^{2}+\dfrac{1}{x^{2}}=254$

Divide the sum of $-\cfrac { 13 }{ 5 }$ and $\cfrac { 12 }{ 7 }$ by the product of $-\cfrac { 13 }{ 7 }$ and $-\cfrac { 1 }{ 2 }$

Sum

$=\cfrac { -13 }{ 5 } +\cfrac { 12 }{ 7 } =\cfrac { -91+60 }{ 35 } =\cfrac { -31 }{ 35 }$

Product

$=\cfrac { -31 }{ 7 } \times \left( \cfrac { -1 }{ 2 } \right) =\cfrac { 13 }{ 14 }$

Division result

$=\cfrac { \cfrac { -31 }{ 35 } }{ \cfrac { 13 }{ 14 } } =\cfrac { -31 }{ 35 } \times \cfrac { 14 }{ 13 }$

$=\cfrac { -62 }{ 65 }$

Given $x = 9 + 4\sqrt 5$ find $\sqrt x - \dfrac{1}{{\sqrt x }} =$ ?

Given that ,

$x=9+4\sqrt5$

Now we can write that

$\sqrt x=\sqrt(9+4\sqrt5)$

$=>\sqrt(\sqrt 5+\sqrt4+2\times \sqrt(5\times 4))$

$=>\sqrt x=[\sqrt(\sqrt(5)+\sqrt(4))^2]$

Now we can write

$\sqrt x=\sqrt 5+2$

and

$\dfrac{1}{\sqrt x}=\dfrac{1}{\sqrt 5+2}$

now rationalising by

$\sqrt 5-2$

now

$=>\dfrac{1}{\sqrt5+2}=\dfrac{1}{\sqrt5+2}\times \dfrac{\sqrt5-2}{\sqrt 5-2}$

$=>\sqrt5-2$

Now

$\sqrt x=\sqrt5+2$ and

$\dfrac{1}{\sqrt x}=\sqrt5-2$

Now

$\sqrt x-\dfrac{1}{\sqrt x}=\sqrt5+2 -(\sqrt 5-2)$

$=>4$

$\sqrt{2+\sqrt3}+\sqrt{2-\sqrt3}=?$
simplify with steps.

$let \sqrt {2+\sqrt3}+\sqrt{2-\sqrt3}=a\\ then\> a^2=2+\sqrt3+2-\sqrt3+2\sqrt{(2+\sqrt3)(2-\sqrt3)}\\=4+2\sqrt{4-3}\\=4+2=6\\\therefore a=\sqrt6$

Rationalise the denominators of the following: $\dfrac{1}{{\sqrt 7 }}$ , $\dfrac{1}{{\sqrt 7 - \sqrt 6 }}$ , $\dfrac{1}{{\sqrt 5 + \sqrt 2 }}$ , $\dfrac{1}{{\sqrt 7 - 2}}$

Part (A)

$\dfrac{1}{\sqrt{7}}$

$=\dfrac{1}{\sqrt{7}}\times \dfrac{\sqrt{7}}{\sqrt{7}}$ Rationalize

$=\dfrac{\sqrt{7}}{7}$ Ans.

Part (B)

$\dfrac{1}{\sqrt{7}-\sqrt{6}}$

$=\dfrac{1}{\sqrt{7}-\sqrt{6}}\times \dfrac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}$ Rationalize

$=\dfrac{\sqrt{7}+\sqrt{6}}{{{\left( \sqrt{7} \right)}^{2}}-{{\left( \sqrt{6} \right)}^{2}}}$

$=\dfrac{\sqrt{7}+\sqrt{6}}{7-6}$

$=\sqrt{7}+\sqrt{6}$ Ans.

Part (C)

$\dfrac{1}{\sqrt{5}+\sqrt{2}}$

$=\dfrac{1}{\sqrt{5}+\sqrt{2}}\times \dfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ Rationalize

$=\dfrac{\sqrt{5}-\sqrt{2}}{{{\left( \sqrt{5} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}}$

$=\dfrac{\sqrt{5}-\sqrt{2}}{5-2}$

$=\dfrac{\sqrt{5}-\sqrt{2}}{3}$ Ans.

Part (D)

$\dfrac{1}{\sqrt{7}-2}$

$=\dfrac{1}{\sqrt{7}-2}\times \dfrac{\sqrt{7}-2}{\sqrt{7}-2}$

$=\dfrac{\sqrt{7}-2}{{{\left( \sqrt{7} \right)}^{2}}-{{\left( 2 \right)}^{2}}}$

$=\dfrac{\sqrt{7}-2}{7-4}$

Rationalize

$=\dfrac{\sqrt{7}-2}{3}$ Ans.

Find 5 rational numbers between $- \dfrac{1}{3}$ and $\dfrac{1}{2}$ .

$- \cfrac{1}{3}$ and

$\cfrac{1}{2}$

L.C.M

$= 6$

$- \cfrac{1}{3} \times \cfrac{2}{2} = \cfrac{{ - 2}}{6}$

$\cfrac{1}{2} \times \cfrac{3}{3} = \cfrac{3}{6}$

$= 5 + 1 = 6$

multiply

$\cfrac{6}{6}$ to both numbers

$\cfrac{{ - 2}}{6} \times \cfrac{6}{6} = \cfrac{{ - 12}}{{36}}$

$\cfrac{3}{6} \times \cfrac{6}{6} = \cfrac{{18}}{{36}}$

Five numbers are :-

$\cfrac{{ - 11}}{{36}},\cfrac{{ - 10}}{{36}},\cfrac{{ - 9}}{{36}},\cfrac{{ - 8}}{{36}},\cfrac{{ - 7}}{{36}}$

Simplify the following
$\dfrac {\sqrt {5}-1}{\sqrt {5}+1}+\dfrac {\sqrt {5}+1}{\sqrt {5}+1}$

1325373_1088047_ans_59d74d55ca2a40608521ede5d1c5edb2.jpg

Solve
$\left( \cfrac { -5 }{ 4 } +\cfrac { 3 }{ 8 } \right) +\cfrac { -7 }{ 6 }$

$\displaystyle \left ( \frac{-5}{4}+\frac{3}{8} \right )+\frac{-7}{6}$

$\displaystyle =\left ( \frac{-5\times 2}{4\times 2}+\frac{3}{8} \right )+\frac{-7}{6}$

$\displaystyle =\left ( \frac{-10+3}{8} \right )+\frac{-7}{6}$

$\displaystyle =\frac{-7}{8}+\frac{-7}{6}=-7\left ( \frac{1}{8}+\frac{1}{6} \right )$

$\displaystyle =-7\left ( \frac{1\times 3}{8\times 3}+\frac{1\times 4}{6\times 4} \right )$

$\displaystyle =-7\times \frac{7}{24}=\frac{-49}{24}$

If $\dfrac{{3 + 2\sqrt 2 }}{{3 - \sqrt 2 }} = a + b\sqrt 2$ , then find the values of $a$ and $b$ .

$\dfrac {3+2\sqrt 2}{3-\sqrt 2}=a+b\sqrt 2$

$\Rightarrow \ \dfrac {3+2\sqrt 2}{3-\sqrt 2}\times \dfrac {3+\sqrt 2}{3+\sqrt 2}=a+b\sqrt 2$

$\Rightarrow \ \dfrac {9+3\sqrt 2+6\sqrt 2+4}{9-2}=a-b\sqrt 2$

$\Rightarrow \ \dfrac {13}{7}+\dfrac {9\sqrt 2}{7}=a+b\sqrt 2$

On comparing both the sides of

$a=\dfrac {13}{7}$ and

$b=\dfrac {9}{7}$

If $a = 2 + \sqrt 3$ , then find the value of a $a - \frac{1}{a}$

Given that,

$a=2+\sqrt{3}$

Then, Value of

$a-\dfrac{1}{a}=2+\sqrt{3}-\dfrac{1}{2+\sqrt{3}}$

$=2+\sqrt{3}-\dfrac{1}{2+\sqrt{3}}\times \dfrac{2-\sqrt{3}}{2-\sqrt{3}}\,\,\,\,\,on\,\,rationalize\,$

$=2+\sqrt{3}-\dfrac{\left( 2-\sqrt{3} \right)}{4-3}$

$=2+\sqrt{3}-2+\sqrt{3}$

$=2\sqrt{3}$

Hence, this is the answer.

Find $4$ rational numbers between $\dfrac { 1 }{ 6 }\ and\ \dfrac { 3 }{ 8 }$

$\dfrac{1}{6}=\dfrac{8}{48}$ and

$\dfrac{3}{8}=\dfrac{24}{48}$

Hence, 4 rational numbers between

$\dfrac{1}{6}$ and

$\dfrac{3}{8}$ are :

$\dfrac{9}{48}, \dfrac{10}{48}, \dfrac{11}{48}, \dfrac{12}{48}$

Find five rational numbers between $\frac{2}{5}$ and $\frac{3}{5}$

Since, We want five rational numbers, we write $\dfrac{2}{5}$ and $\dfrac{3}{5}$ .

So multiply in numerator and denominator by $5+1=6$ and we get,

$\dfrac{2\times 6}{5\times 6}=\dfrac{12}{30}$ and $\dfrac{3}{5}=\dfrac{3\times 6}{5\times 6}=\dfrac{18}{30}$

We know that, $13<14<15<16<17$

$\Rightarrow \dfrac{12}{30}<\dfrac{13}{30}<\dfrac{14}{30}<\dfrac{15}{30}<\dfrac{16}{30}<\dfrac{17}{30}<\dfrac{18}{30}$

Hence, $5$ rational numbers between $\dfrac{2}{5}$ and $\dfrac{3}{5}$ are:

$\Rightarrow \dfrac{13}{30},\dfrac{14}{30},\dfrac{15}{30},\dfrac{16}{30},\dfrac{17}{30}$ ans.

List five rational numbers between: $-1$ and $0$ .

$5$ rational numbers between

$-1$ and

$0$

As we want

$5$ rational number

$5+1=6$

Now multiply with

$\dfrac6 6$

$-1\times \dfrac {6}{6}\quad \, and\,\, 0\times \dfrac {6}{6}$

$=\quad \dfrac {-6}{6} \, and\,\, \dfrac {0}{6}$

So rational number between

$\dfrac {-6}{6}$ &

$\dfrac {0}{6}$ are

$\dfrac {-5}{6},\ \dfrac {-4}{6},\ \dfrac {-3}{6},\ \dfrac {-2}{6},\ \dfrac {-1}{6}$

After how many decimal places will the decimal expansion of the number $\dfrac{5^3}{ 2^5^4}$ terminate? Justify your answer without performing the actual division.

The decimal expansion of the number

$\dfrac{5^{3}}{2^{3}.5^{4}}$ will terminate after

$3$ decimal places.

$\dfrac{5^{3}}{2^{3}.5^{4}}=\dfrac{1}{2^{3}.5^{4-3}}=\dfrac{1}{8 \times 5}$

$=\dfrac{1}{40}=0.025$ .

Hence, the decimal will terminate after

$3$ decimal place.

List five rational numbers between: $-2$ and $-1$ .

$5$ rational numbers between

$-2$ and

$-1$

As we want

$5$ rational number

$-5+1=6$

$-2\times \dfrac {6}{6}\quad -1\times \dfrac {6}{6}$

$\quad \dfrac {-12}{6}\quad \quad \dfrac {-6}{6}$

So rational number between

$\dfrac {-12}{6}$ &

$\dfrac {-6}{6}$

$\dfrac {-11}{6},\ \dfrac {-10}{6},\ \dfrac {-9}{6},\ \dfrac {-8}{6},\ \dfrac {-7}{6}$

List five rational numbers between: $\dfrac {-4}{5}$ and $\dfrac {-2}{3}$ .

Given: -4/5 and-2/3

= -4×3/5×3 and -2×5/3×5 ( making denominator same-LCM)
= -12/15 and -10/15
= -12×3/15×3 and -10×3/15×3
= -36/45 and -30/45
Five rational nos. between -4/5 and -2/3 are -31/45, -32/45, -33/45, -34/45 and -35/45

Represent the real numbers on the number-line
(i) $\sqrt {10}$
(ii) $\sqrt {13}$
(iii) $\sqrt {2}$
(iv) $\sqrt {5}$
(v) $\sqrt {3}$

Solution-

$\sqrt{1}<\sqrt{2}<\sqrt{4}, \sqrt{1}< \sqrt{3}<\sqrt{4}, \sqrt{4}<\sqrt{5}<\sqrt{9}, \sqrt{9}< \sqrt{10}, \sqrt{16},\sqrt{9}<\sqrt{13},\sqrt{16}.$

1130843_1103752_ans_ae3251818be94d4382d14f232b708f64.png

Insert three rational numbers between: $\dfrac{3}{5}$ to $\dfrac{7}{8}$

Given rational numbers are

$\dfrac{3}{5}$ and

$\dfrac{7}{8}$

Denominators of the rational numbers are

$5$ and

$8$

$L.C.M (5,8) = 5\times 8 =40$

Convert them to rational numbers with same denominators

$\dfrac{3 \times 8 }{5 \times 8}$ ,

$\dfrac{7 \times 5}{8 \times 5}$

$\dfrac{24}{40}$ ,

$\dfrac{ 35}{40}$

Between

$24$ and

$35$ we can choose any three integers as numerators.

Therefore,

$\dfrac{25}{40}, \dfrac{26}{40}, \dfrac{30}{40}$ lie between

$\dfrac{3}{5}$ to

$\dfrac{7}{8}$

List five rational numbers between :
$\dfrac{1}{2}$ and $\dfrac{2}{3}$

To find 5 rational b/w :

$\dfrac{1}{2} and \dfrac{2}{3}$

LCM

$(2,3) \Rightarrow 2\times 3\Rightarrow 6$

$\therefore \dfrac{1\times 3}{2\times 3} = \dfrac{3}{6}$

and

$\dfrac{2\times 2}{3\times 2} = \dfrac{4}{6}$

$\therefore$ we know multiply the no by

$\dfrac{6}{6}$

$\therefore \dfrac{3\times 6}{6\times 6} = \dfrac{18}{36}$

and

$\dfrac{4\times 6}{6\times 6} = \dfrac{24}{36}$

$\therefore$ 5 rational no.are :-

$\left \{ \dfrac{23}{36},\dfrac{22}{36},\dfrac{21}{36},\dfrac{20}{36},\dfrac{19}{36} \right \}$

If $10^{x}=64$ , what is the value of $10^{\dfrac{x}{2}+1}$ ?

Given

$10^x=64$

Then,

$10^{\dfrac{x}{2}+1}=10^{\dfrac{x}{2}}\times10$

$=[10^x]^{\dfrac{1}{2}}\times10$

$=[64]^{\dfrac{1}{2}}\times10$

$=8\times10$

$=80$

Simplify
$\sqrt {27}+\sqrt {48}-\sqrt {12}+\sqrt {75}-\sqrt {108}$

Solving given eq.

$\sqrt{27}+\sqrt{48}-\sqrt{12}+\sqrt{75}-\sqrt{108}$

$\implies 3\sqrt{3}+4\sqrt{3}-2\sqrt{3}+5\sqrt{3}-6\sqrt{3}$

$\implies 4\sqrt3$

$\dfrac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times 6^{-5}}$ .

$\begin{matrix} \dfrac { { { 3^{ -5 } }\times { { 10 }^{ -5 } }\times 125 } }{ { { 5^{ -7 } }\times { 6^{ -5 } } } } =? \\ \Rightarrow \dfrac { { (1)\times { 5^{ 3 } }\times { 6^{ 5 } }\times { 5^{ 7 } } } }{ { { 3^{ 5 } }\times { { 10 }^{ 5 } } } } \\ \Rightarrow \dfrac { { { 5^{ 10 } }\times { { \left( { 2\times 3 } \right) }^{ 5 } } } }{ { { 3^{ 5 } }\times { { \left( { 2\times 5 } \right) }^{ 5 } } } } =\dfrac { { { 5^{ 10 } }\times { 2^{ 5 } }\times { 3^{ 5 } }^{ } } }{ { { 3^{ 5 } }\times { 2^{ 5 } }\times { 5^{ 5 } } } } = \\ { 5^{ \left( { 10-5 } \right) } }={ 5^{ 5 } }=3125\, \, \, \, \, Ans. \\ \end{matrix}$

Simplify
$\left[\dfrac{4}{7}\right]^{-5}\times \left[\dfrac{7}{4}\right]^{-7}$ by giving reasons.

$\left(\dfrac{4}{7}\right)^{-5}\times\left(\dfrac{7}{4}\right)^{-7}$

$=\left(\dfrac{7}{4}\right)^{5}\times\left(\dfrac{4}{7}\right)^{7}$

$=\left(\dfrac{7}{4}\right)^{5}\times\left(\dfrac{4}{7}\right)^{5}\left(\dfrac{4}{7}\right)^2$

$=\left(\dfrac{4}{7}\right)^2$

$=\left(\dfrac{16}{49}\right)$

Solve :
$(\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}$

${\left(\surd {3}+\surd {2}\right)}^{6}-{\left(\surd {3}+\surd {2}\right)}^{6}$

${ \left( { \left( \surd { \quad 3 } +\surd { 2 } \right) }^{ 3 } \right) }^{ 2 }-{ \left( { \left( \surd { \quad 3 } +\surd { 2 } \right) }^{ 3 } \right) }^{ 2 }$

$\left( { \left( \surd { 3 } +\surd { 2 } \right) }^{ 3 }-{ \left( \surd { 3 } +\surd { 2 } \right) }^{ 3 } \right) +\left( { \left( \surd { 3 } +\surd { 2 } \right) }^{ 3 }-{ \left( \surd { 3 } +\surd { 2 } \right) }^{ 3 } \right)$

$\left[ 3\surd { 3 } +2\surd { 2 } +3\surd { 56} \left( \surd { 3 } +\surd { 2 } \right) -\left( 3\surd { 3 } +2\surd { 2 } +3\surd { 6 } \left( \surd { 3 } -\surd { 2 } \right) \right) \right]$

$\left[ 3\surd { 3 } +2\surd { 2 } +3\surd { 6} \left( \surd { 3 } +\surd { 2 } \right) +\left( 3\surd { 3 } +2\surd { 2 } +3\surd { 6 } \left( \surd { 3 } -\surd { 2 } \right) \right) \right]$

$\left[ 3\surd { 3 } +2\surd { 2 } +3\surd { 6 } \left( \surd { 3 } +\surd { 2 } \right) +\left( 3\surd { 3 } +2\surd { 2 } +3\surd { 6 } \left( \surd { 3 } -\surd { 2 } \right) \right) \right]$

$\left[ 3\surd { 3 } +2\surd { 2 } +3\surd { 6 } \left( \surd { 3 } +\surd { 2 } \right) +\left( 3\surd { 3 } -2\surd { 2 } -3\surd { 6 } \left( \surd { 3 } -\surd { 2 } \right) \right) \right]$

$\left[ 4\surd { 2 } +5\surd { 6 } \left( \surd { 3 } +\surd { 2 } +\surd { 3 } -\surd { 2 } \right) \right] \left[ 6\surd { 3 } +2.6.\surd { 3 } \right]$

$\left( 4\surd { 2 } +6.3\surd { 2 } \right) \left( 6\surd { 3 } +12\surd { 3 } \right)$

$\left( 4\surd { 2 } +18\surd { 2 } \right) \left( 6\surd { 3 } +12\surd { 3 } \right)$

$2\surd { 2 } \left( 2+9 \right) 6\surd { 3 } \left( 1+2 \right)$

$2\surd { 2 } \left( 11 \right) 6\surd { 3 } \left( 3 \right)$

$6\times 6\surd { 6 } \quad \Rightarrow \quad 396\surd { 6 }$ Ans.

List five rational numbers between:
$-1$ and $0$

To find 5 rational no: b/w -1 & 0

So, we multiply the no: by

$\dfrac{6}{6}$

$\therefore -1\times \dfrac{6}{6} = \dfrac{-6}{6}$

and

$\dfrac{0}{1}\times \dfrac{6}{6} = \dfrac{0}{6}$

$\therefore$ 5 rational no:-

$\left \{ \dfrac{-5}{6},\dfrac{-4}{6},\dfrac{-3}{2},\dfrac{-2}{2},\dfrac{-1}{2} \right \}$

1126101_1140171_ans_bf9508e2c6cb4e49b3f13daf93f5be26.jpg

Represent $\sqrt{10}$ on number line by direct method.

R.E.F image

$AC$

$= \sqrt{10}$

$= \sqrt{9+1}$

$= \sqrt{3^{2}+1^{2}}$

1119609_1134718_ans_60604d59b11a4b79830a711906e6014a.png

Write $5$ rational numbers between $\dfrac{3}{5}$ and $\dfrac{6}{5}$ .

$\dfrac{3}{5}=\dfrac{3\times 6}{5\times 6}=\dfrac{18}{30}$

$\dfrac{6}{5}=\dfrac{6\times 6}{5\times 6}=\dfrac{36}{30}$

we know that

$18<22<23<26<30<34<36$

$\Rightarrow \dfrac{18}{30}<\dfrac{22}{30}<\dfrac{23}{30}<\dfrac{26}{30}<\dfrac{30}{30}<\dfrac{34}{30}<\dfrac{36}{30}$

Hence 5 rational numbers between

$\dfrac{3}{5},\dfrac{6}{5}$ are

$\dfrac{22}{30},\dfrac{23}{30},\dfrac{26}{30},\dfrac{30}{30},\dfrac{34}{30}$

Simplify $12\sqrt{18}-6\sqrt{20}-3\sqrt{50}+8\sqrt{45}$

$12\sqrt{18}-6\sqrt{20}-3\sqrt{50}+8\sqrt{45}$

$=12\sqrt{9\times 2}-6\sqrt{4\times 5}-3\sqrt{25\times 2}+8\sqrt{9\times 5}$

$=36\sqrt{2}-12\sqrt{5}-15\sqrt{2}+24\sqrt{5}$

$=\sqrt{2}(36-15)+\sqrt{5}(24-12)$

$=21\sqrt{2}+12\sqrt{5}$

Find an irrational number between two numbers $\dfrac{1}{7}$ and $\dfrac{2}{7}$ and justify your answer.

$\dfrac{1}{7}=0.142857142857...=0.\overline{142857}$

and

$\dfrac{2}{7}=0.28571428571428...=0.\overline{285714}$ .

Here, the two decimal expansions are non-terminating recurring.

Hence,

$\frac {1} {7}$ and

$\frac {2}{7}$ are two rational numbers.

We know, between any two rational numbers, there are infinitely many irrational numbers.

An irrational number has non-terminating non-recurring decimal expansions.

Then an irrational number between

$\dfrac{1}{7}$ and

$\dfrac{2}{7}$ is

$0.15015001500015...$ .

Similarly,

$0.21020020002...$ is another irrational number between

$\dfrac{1}{7}$ and

$\dfrac{2}{7}$ .

NOTE:

This is some of the various possible answers. It might have different solution.

Simplify:
$\dfrac{{25 \times {t^{ - 4}}}}{{{5^{ - 2}} \times 10 \times {t^{ - 8}}}}\left( {t \ne 0} \right)$

$\dfrac{{25 \times {t^{ - 1}}}}{{{5^{ - 2}} \times 10 \times {t^{ - 1}}}}$

$\dfrac{{5^2 \times {t^{ - 1}}}}{{{5^{ - 2}} \times 10 \times {t^{ - 1}}}}$

Since,

$x^a\cdot x^{b} = x^{a+b } \,\&\, \dfrac{x^a}{x^b} = x^{a - b}$

$\dfrac{{5^2 \times {t^{ - 1}}}}{{{5^{ - 2}} \times 5\times 2 \times {t^{ - 1}}}}$

$\dfrac{{5^2 \times {t^{ - 1}}}}{{{5^{ - 2+1}} \times 2\times {t^{ - 1}}}}$

$\dfrac{{5^2 \times {t^{ - 1}}}}{{{5^{ - 1}} \times 2\times {t^{ - 1}}}}$

$\dfrac{5^2 \times t^{-1}\times t^1 \times 5^1} {2}$

$\dfrac{5^3 \times t^{0}} {2}$

$\dfrac{5^3 \times 1} {2}$

$\dfrac{5^3 } {2}$

$\dfrac{125 } {2}$

find the value of 'x' by using following equations
$- \sqrt {3x} = 3 + 2$
$- \sqrt {3x} = 5$

$=> - √3x = 3 + 2$

$=> - √3x = 5$

$=> √3x = - 5$

$=> (√3x)^2 = (-5)^2$ (squaring both side)

$=> 3x = 25$

$=> x = 25/3$ (Ans)

Add $2 + \sqrt { 3 }$ and $2 - \sqrt { 3 }$

Add

$2+\sqrt{3}$ and

$2-\sqrt{3}$

$= (2+\sqrt{3})+(2-\sqrt{3})$

$=2+\sqrt{3}+2-\sqrt{3}$

$= 4$

$2^{5}\times2^{3}$

$\dfrac { { 2 }^{ 5 }\times { 2 }^{ 3 }\\ ={ 2 }^{ 5+3 }={ 2 }^{ 8 } }{ }$

Rationalise the denominator and simplify:
$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}$

$\dfrac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}*\dfrac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}=({\sqrt 3 - \sqrt 2 })^2=3+2-2\sqrt6=5-2\sqrt6$

Show that $7-\sqrt 5$ is irrational .given that $\sqrt 5$ is irrational.

Let us assume to the contrary that

$7 -\sqrt 5$ is rational number

i.e, we can find co-prime

$a$ &

$b$ where

$b\neq 0$ such that:-

$\Rightarrow$

$7- \sqrt 5= \dfrac {a} {b}$

$\Rightarrow$

$7- \dfrac {a} {b}= \sqrt 5$

by rearranging this equation we get

$\Rightarrow$

$\sqrt 5=7- \dfrac {a} {b}$

$\Rightarrow$

$\sqrt 5b={7b-a}$

since

$a$ &

$b$ are integers,we get

$\dfrac {7b-a} {b}$ is a rational number & so

$\sqrt 5$ is rational But, this contradicts the fact that

$\sqrt 5$ is irrational number.

$\therefore$ our assumption i.e,

$7-\sqrt 5$ is rational number is incorrect.

Hence,

$7 -\sqrt 5$ is irrrational

What number should ${5}^{3}$ be multiplied so that the product may be equal to ${5}^{5}$ ?

Let the number be multiplied is

$x$

So,

$\Rightarrow x\times5^3=5^5$

$\Rightarrow x=\dfrac{5^5}{5^3}$

$\Rightarrow x=5^{5-3}$

$\Rightarrow x=5^2=25$

Is this a negative rational number?
$\dfrac{5}{7}$

A rational number is negative if its numerator & denominator are opposite in signs.

So,

$\dfrac{5}{7}$ is not a negative rational number

Write the number whose expanded form is as below:
$3 \times {10^4} + 6 \times {10^3} + 5 \times {10^2} + 8 \times {10^0}$

$\begin{array}{l} 3\times { 10^{ 4 } }+6\times { 10^{ 3 } }+5\times { 10^{ 2 } }+8\times { 10^{ 0 } } \\ \Rightarrow 30000+6000+500+8 \\ \Rightarrow 36508 \end{array}$

$2^{-3}\times (-7)^{-3}$ .

Consider the given expression,

$\Rightarrow {{2}^{-3}}\times {{\left( -7 \right)}^{-3}}$

$\Rightarrow \dfrac{1}{8}\times \dfrac{1}{\left( -343 \right)}$

$\Rightarrow -\dfrac{1}{2744}$

Hence, this is the answer.

Simplify: $9\sqrt {5}-3\sqrt {5}+\sqrt {125}$

Now,

$9\sqrt {5}-3\sqrt {5}+\sqrt {125}$

$9\sqrt {5}-3\sqrt {5}+\sqrt {25\times 5}$

$=9\sqrt {5}-3\sqrt {5}+5\sqrt {5}$

$=11\sqrt{5}$ .

Is this a negative rational number?
$\dfrac{6}{11}$

A rational number is negative if its numerator and denominator are opposite in signs.

So,

$\dfrac{6}{11}$ is not a negative rational number.

Is this a negative rational number?
$\dfrac{3}{-5}$

A rational number is negative if its numerator and denominator are opposite in sings.

So,

$\dfrac{3}{-5}$ is a negative rational number.

Insert two rational numbers between $\dfrac{3}{8}$ and $\dfrac{7}{12}$ .

1970907_1233072_ans_6799e61ec7cc45aa995e2683775b2f10.jpg

Is this a negative rational number?
$\dfrac{-2}{-9}$

A rational number is negative if its numerator and denominator are opposite in sings.

So,

$\dfrac{-2}{-9}$ is not a negative rational number.

Is this a negative rational number?
$0$

Zero is a rational number because it can be written in

$\dfrac{p}{q}$ form but neither it is positive or negative Rational number.

Is this a negative rational number?
$\dfrac{-2}{3}$

A rational number is negative if its numerator and denominator are opposite in signs.

So,

$\dfrac{-2}{3}$ is a negative rational number.

Write whether the rational number $\cfrac {11}{125}$ will have a terminating decimal expansion or a non terminating repeating decimal expansion.

If the denominator can be in the powers of either

$2$ or

$5$ or both, then it will be a terminating decimal, otherwise non-terminating.

Now, the denominator

$125=5^3$ , i.e. is in the form

$2^m\times5^n$ ; where

$n,m$ are non-negative.

Hence, the rational number has terminating decimal expansion.

Insert three rational numbers between $\dfrac {9}{11}$ and $\dfrac {5}{7}.$

We are making denominator of both numbers same,so that we can easily compare both numbers.
1187875_1241598_ans_c31d8a76e99946c19d41cd4a449c7576.jpg

1187875_1241598_ans_c31d8a76e99946c19d41cd4a449c7576.jpg

Simplify the following using laws of exponent
$\dfrac{(7xy\, z)^0}{[12x^2(y^8)^{-3}(z^3)^2]^{-1}}$

We have,

$=\dfrac{(7xyz)^0}{[12x^2(y^8)^{-3}(z^3)^2]^{-1}}$

$=\dfrac{1}{[12x^2(y^8)^{-3}(z^3)^2]^{-1}}$

$=12x^2(y^8)^{-3}(z^3)^2$

$=12x^2y^{-24}z^6$

$=\dfrac{12x^2z^6}{y^{24}}$

Hence, this is the answer.

Simplify the following using laws of exponent
$-[(2)^7]^4 \div 4^4$

We have,

$-[(2)^7]^4\div 4^4$

$=-(2)^{28}\div (2^2)^4$

$=-\dfrac{(2)^{28}}{2^8}$

$=-2^{20}$

Hence, this is the answer.

Evaluate $p^5\times p^8$ .

${ p }^{ 5 }\times { p }^{ 8 }={ p }^{ 5+8 }$

$={ p }^{ 13 }$

Solve: $\dfrac{3}{{\sqrt 5 - \sqrt 3 }}$

Given,

$\dfrac {3}{\sqrt {5} - \sqrt {3}}$

$\Rightarrow \dfrac {3}{\sqrt{5} - \sqrt {3}}\times \dfrac {\sqrt {5} + \sqrt {3}}{\sqrt {5} + \sqrt {3}}$ (Rationizing)

$= \dfrac {3(\sqrt {5} + \sqrt {3})}{5 - 3}$

$= \dfrac {3}{2} (\sqrt {5} + \sqrt {3})$ .

Simplify the following using laws of exponent
$(3^2)^2 \times 3^{10}$

We have,

$(3^2)^{2}\times 3^{10}$

$=3^4\times 3^{10}$

$=3^{4+10}$

$=3^{14}$

Hence, this is the answer.

Is the product of two irrational numbers always an irrational number ?

“The product of two irrational number is “sometime” irrational”

Its is possible that some irrational number may multiply to from a rational product.

$\sqrt{5} \times \sqrt{2}= \sqrt{10}$ ; Irrational

$\sqrt{8} \times \sqrt{2} =\sqrt{16}$

$=4$ Rational.

Hence this is the answer.

Find three rational numbers between $4$ and $5$

1351264_1248706_ans_b09a1f6c21574a12b6bcf297e781ceee.PNG

Solve $\left( 2 ^ { 3 } \cdot 3 \right) \div 2 ^ { 2 } \cdot 5 ^ { 2 }$

$\dfrac { { 2 }^{ 3 }.3 }{ { 2 }^{ 2 }.{ 5 }^{ 2 } } =\dfrac { 24 }{ { \left( 10 \right) }^{ 2 } } =\dfrac { 24 }{ 100 } =0.24$

Classify the following number as rational or irrational :
$(3+\sqrt{23})-\sqrt{23}$ .

$(3+\sqrt{23})-\sqrt{23}$

$= 3+\sqrt{23}-\sqrt{23}$

$3$ , which is a rational number.

Therefore, the given number is a rational number.

$49\times \left(-7\right)^ {m}=-343$

$\begin{array}{l} 49\times { \left( { -7 } \right) ^{ m } }=-343 \\ { \left( { -7 } \right) ^{ m } }=-\frac { { 343 } }{ { 49 } } \\ { \left( { -7 } \right) ^{ m } }=-7 \\ { { Implies, } } \\ m=1 \end{array}$

Solve: $5 - 3\dfrac{3}{4}$

$5-3\frac{3}{4}=5-\frac{3\times 4+3}{4}=\frac{5}{1}-\frac{15}{4}$

$LCM : 1\times 4=4\Rightarrow \frac{5\times 4-15}{4}=\frac{5}{4}$

1117403_1278065_ans_7ee4a5cf402a4e81892f281fb4c9f9d3.JPG

For any positive real number x, prove that there exists an irrational number y such that 0 < y < x.

$x$ is irrational, then

$y=\dfrac{x}{2}$ is also an irrational number such that

$0<y<x$ . If

$x$ is irrational, then

$\dfrac{\sqrt{2}}{x}$ is an irrational number such that

$\dfrac{\sqrt{x}}{2}<x$ as

$\sqrt{2}>1$ .

hence , for any positive real number x, there exists an irrational number y such that

$0<y<x$ .

Simplify the following using laws of exponents:-
$2^{3}\times 2^{10}\times 2^{5}$

As we know that,

$a^m\times a^n=a^{m+n}$

Hence,

$\begin{aligned}{} 2^3\times2^{10}\times 2^5&={ 2^{ 3+10+5 } } \\ &={ 2^{ 18 } } \end{aligned}$

So, the given expression is equal to

${2^{18}}.$

Rationalize the denominator $\frac{1}{{\sqrt 5 + \sqrt 6 - \sqrt {11} }}$

$\dfrac{1}{\sqrt5+\sqrt6-\sqrt11}\times\dfrac{\sqrt5+\sqrt6+\sqrt11}{\sqrt5+\sqrt6+\sqrt11}$

$=\dfrac{\sqrt5+\sqrt6+\sqrt11}{2\sqrt30}=\dfrac{\sqrt30}{60}(\sqrt5+\sqrt6+\sqrt11)$

Find two irrational number between $1$ and $2$ .

We have,

To find the irrational number 1 and 2.

So, we know that,

$1<\dfrac{1+2}{2}<2$

$\Rightarrow 1<\dfrac{3}{2}<2$

Do square roots and we get,

$\sqrt{1}<\sqrt{\dfrac{3}{2}}<\sqrt{2}$

So, $\sqrt{\dfrac{3}{2}}$ is one rational number.

And other number can be $1.101110$

The repeating decimal.

Hence, this is the answer.

$n$ is a two digit number. $P(n)$ is the product of the digit of $n$ and $S(n)$ , is the sum of the digits of $n$ . If $n=P(n)+S(n)$ then find the units digit of $n$ .

A] Let a,b be tens & unit digits of n

Then n = 10 a + b

p(n) + s(n) = 10 a + b

= a b + a+b = 10 a +b

= a b = 9 a

$\therefore b = 9$ as

$a \neq 0$

$\therefore$ units digit is 9

1186802_1292308_ans_394a6c98fced4b8bb19bccf106ebfc2e.jpg

Write any two irrational number lying between $3$ and $4$ .

To find: Irrational numbers between 3 and 4.

$3^2 = 9$

$9\le x\le 16$

$4^2 = 16$

$3\le \sqrt x\le4$

where

$\sqrt x$ is an irrational number between 3 and 4.

$\therefore$ the irrational numbers can be

$\sqrt{10}, \sqrt{11}, \sqrt {12} ....$

solve : $\left\{ \left( \dfrac { - 3 } { 2 } \right) ^ { 2 } \right\} ^ { - 3 }$

$\begin{array}{l} We\, have \\ { \left\{ { { { \left( { \dfrac { { -3 } }{ 2 } } \right) }^{ 2 } } } \right\} ^{ -3 } } \\ ={ \left\{ { \dfrac { { -3 } }{ 2 } \times \dfrac { { -3 } }{ 2 } } \right\} ^{ -3 } } \\ ={ \left\{ { \dfrac { 9 }{ 4 } } \right\} ^{ -3 } } \\ ={ \left\{ { \dfrac { 4 }{ 9 } } \right\} ^{ 3 } } \\ =\dfrac { 4 }{ 9 } \times \dfrac { 4 }{ 9 } \times \dfrac { 4 }{ 9 } \\ =\dfrac { { 64 } }{ { 729 } } \\ It\, is\, the\, required\, answer. \end{array}$

Simplify and evaluate :
$[2^{5}\times 2^{4}]\div 2^{3}$

Given simplify :

$= \dfrac{{\left[ {{2^5} \times {2^4}} \right]}}{{{2^3}}}$

$= \dfrac{{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}}{{2 \times 2 \times 2}}$

$= 64$

Hence,

We get

$=64$

Solve:
$\sqrt{12}+5\sqrt{4}+\sqrt{4}-\sqrt{81}$

$\sqrt{12}+5\sqrt{4}+\sqrt{4}-\sqrt{81}$

$=\sqrt{4\times 3}+5\times 2+2-9$

$=2\sqrt{3}+10+2-9$

$=2\sqrt{3}+3$

Find four rational Numbers between $\dfrac{1}{2}$ and $\dfrac{2}{3}$

$\textbf{Step 1: Calculating rationals between a pair of numbers.}$

$\dfrac{1}{2} = \dfrac{3}{6} = \dfrac{30}{60}$

$\dfrac{2}{3} = \dfrac{4}{6} = \dfrac{40}{60}$

$\text{Thus, common denominator of both the fractions is 60}$ .

$\text{So, four rational numbers are :}$

$\dfrac{31}{60}, \dfrac{32}{60}, \dfrac{33}{60}, \dfrac{34}{60}$

$\textbf{Hence, four rationals }\boldsymbol{\dfrac{31}{60}, \dfrac{32}{60}, \dfrac{33}{60}, \dfrac{34}{60}}\textbf{ lies between }\boldsymbol{\dfrac{1}{2}}\textbf{ and }\boldsymbol{\dfrac{2}{3}.}$

Evaluate : $2\sqrt[3]{4}+7\sqrt[3]{32}-\sqrt[3]{500}$

$We\, have \\ 2\sqrt [ 3 ]{ 4 } +7\sqrt [ 3 ]{ { 32 } } -\sqrt [ 3 ]{ { 500 } } \\ =2\sqrt [ 3 ]{ 4 } +7\sqrt [ 3 ]{ { 2\times 2\times 2\times 4 } } -\sqrt [ 3 ]{ { 5\times 5\times 5\times 4 } } \\ =2\sqrt [ 3 ]{ 4 } +7\times 2\sqrt [ 3 ]{ 4 } -5\sqrt [ 3 ]{ 4 } \\ =\sqrt [ 3 ]{ 4 } \left( { 2+14-5 } \right) \\ =11\sqrt [ 3 ]{ 4 } \\ =11\times 1.5874 \\ =17.461$

Hence, this is the answer.

What is the value of $\pi$ ?

$\textbf{Step 1: Find the required value . }$

$\text{For a circle of any radius, the ratio between it's circumference and diameter is fixed,}$

$\text{this value is }\pi.$

$\Rightarrow$

$\dfrac{\text{Circumference}}{\text{Diameter}}$

$=\pi=3.14159$

$\textbf{Hence, the value of}$

$\mathbf{pi}$

$\textbf{is}$

$\bf{3.14.}$

Is $\left(\pi-\dfrac{22}{7}\right)$ a rational number, an irrational number or zero.

$\pi$ is an irrational number, and

$\dfrac{22}{7}$ is a rational number, so, the difference of the two is an irrational number.

Simplify : $\dfrac{1}{1 + \sqrt{2}}$ + $\dfrac{1}{\sqrt{2} + \sqrt{3}}$ + $\dfrac{1}{\sqrt{3} + \sqrt{5}}$

$\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}$

$=\dfrac{1}{1+\sqrt{2}}\times{\dfrac{1-\sqrt{2}}{1-\sqrt{2}}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}\times{\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}\times{\dfrac{\sqrt{3}-\sqrt{5}}{\sqrt{3}-\sqrt{5}}}$

$=\dfrac{1-\sqrt{2}}{-1}+\dfrac{\sqrt{2}-\sqrt{3}}{-1}+\dfrac{\sqrt{3}-\sqrt{5}}{-2}$

$=\dfrac{\sqrt{2}-1}{1}+\dfrac{\sqrt{3}-\sqrt{2}}{1}+\dfrac{\sqrt{5}-\sqrt{3}}{2}$

$=\dfrac{2\sqrt{2}-2+2\sqrt{3}-2\sqrt{2}+\sqrt{5}-\sqrt{3}}{2}$

$=\dfrac{\sqrt{5}+\sqrt{3}-2}{2}$

Simplify the following :
$2\sqrt{5}+\sqrt{125}$

$\begin{array}{l} Given, \\ 2\sqrt { 5 } +\sqrt { 125 } \\ Now, \\ 2\sqrt { 5 } +\sqrt { 125 } \\ \sqrt { 125 } =5\sqrt { 5 } \\ =2\sqrt { 5 } +5\sqrt { 5 } \\ Add\, \, similar\, \, elements\, \, :2\sqrt { 5 } +5\sqrt { 5 } =7\sqrt { 5 } \\ =7\sqrt { 5 } \end{array}$

Find three rational number between $\frac{-3}{14}$ and $\frac{6}{14}$

3 rational numbers betwwen

$-\dfrac{3}{14}$ and

$\dfrac{6}{14}$ are

$\dfrac{1}{14},\dfrac{2}{14},\dfrac{3}{14}$

Find three rational numbers between $-\dfrac {3}{2}$ and $\dfrac {4}{5}$ .

1226833_1397329_ans_10b1aaff26fb41eca94750990c6e15e3.jpg

Prove that the following are irrational.
$\dfrac{1}{\sqrt{2}}$

Let us assume that

$\dfrac{1}{\sqrt{2}}$ is rational number

Hence

$\dfrac{1}{\sqrt{2}}$ can be written in the form of

$\dfrac{a}{b}$ where

$a,b(b\ne 0)$ are co-prime

$\implies \dfrac{1}{\sqrt{2}}=\dfrac{a}{b}$

$\implies \dfrac{b}{a}=\sqrt{2}$

But here

$\sqrt{2}$ is irrational and

$\dfrac{b}{a}$ is rational

$\mathrm{Rational\ne Irrational}$

This is a contradiction so

$\dfrac{1}{\sqrt{2}}$ is a irrational number

Find three rational numbers between $\dfrac {4}{3}$ and $\dfrac {5}{6}$ .

We have,

$\dfrac{4}{3}=1.33333......$ and

$\dfrac{5}{6}=0.83333.......$ .

Now three rational between them are

$-.9,1,1.1$ .

Rationalise: $5-\sqrt{3}$

$5-\sqrt{3}$

$=5-\sqrt{3}\times\dfrac{5+\sqrt{3}}{5+\sqrt{3}}$

$=\dfrac{25-3}{5+\sqrt{3}}$

$=\dfrac{22}{5+\sqrt{3}}$

Prove that the following is irrational.
$6+\sqrt{2}$

Let us assume that

$6+\sqrt{2}$ is rational number

Hence

$6+\sqrt{2}$ can be written in the form of

$\dfrac{a}{b}$ where

$a,b(b\ne 0)$ are co-prime

$\implies 6+\sqrt{2}=\dfrac{a}{b}$

$\implies \sqrt{2}=\dfrac{a}{b}-6$

$\implies \dfrac{a-6 b}{b}=\sqrt{2}$

But here

$\sqrt{2}$ is irrational and

$\dfrac{a-6{b}}{b}$ is rational

$\mathrm{Rational\ne Irrational}$

This is a contradiction

$6+\sqrt{2}$ is a irrational number

Prove that the following is irrational.
$7\sqrt{5}$

Let us assume that

$7\sqrt{5}$ is rational number

Hence

$7\sqrt{5}$ can be written in the form of

$\dfrac{a}{b}$ where

$a,b(b\ne 0)$ are co-prime

$\implies 7\sqrt{5}=\dfrac{a}{b}$

$\implies \sqrt{5}=\dfrac{a}{7 b}$

But here

$\sqrt{5}$ is irrational and

$\dfrac{a}{7 b}$ is rational

$\mathrm{Rational\ne Irrational}$

This is a contradiction

$7\sqrt{5}$ is a irrational number

Find a rational number between $-\dfrac{2}{3}$ and $\dfrac{1}{4}$ .

$-\dfrac{2}{3}=\dfrac{-8}{12}$

$\dfrac{1}{4}=\dfrac{3}{12}$

Hence, a rational number between

$\dfrac{-2}{3}$ and

$\dfrac{1}{4}$ is

$\dfrac{1}{12}$ .

$\dfrac{1+\sqrt7}{1-\sqrt7}+\dfrac{1-\sqrt7}{1+\sqrt7}$

$\dfrac{1+\sqrt7}{1-\sqrt7}+\dfrac{1-\sqrt7}{1+\sqrt7}\\\dfrac{(1+\sqrt7)^2+(1-\sqrt7)^2}{1-7}\\\dfrac{1+7+2\sqrt7+1+7-2\sqrt7}{-6}\\\dfrac{16}{-6}=-\dfrac{8}{3}$

Rationalise: $4-\sqrt{5}$

given,

$4-\sqrt{5}$

$=4-\sqrt{5}\times\dfrac{4+\sqrt{5}}{4+\sqrt{5}}$

$=\dfrac{16-5}{4+\sqrt{5}}$

$=\dfrac{11}{4+\sqrt{5}}$

Prove that the following is irrational.
$3+2\sqrt{5}$

Let us assume that

$3+2\sqrt{5}$ is rational number

Hence

$3+2\sqrt{5}$ can be written in the form of

$\dfrac{a}{b}$ where

$a,b(b\ne 0)$ are co-prime

$\implies 3+2\sqrt{5}=\dfrac{a}{b}$

$\implies 2\sqrt{5}=\dfrac{a}{b}-3$

$\implies \dfrac{a-3 b}{2 b}=\sqrt{5}$

But here

$\sqrt{5}$ is irrational and

$\dfrac{a-3{b}}{2 b}$ is rational

$\mathrm{Rational\ne Irrational}$

This is a contradiction

$3+2\sqrt{5}$ is a irrational number

Construct $\sqrt{17}$ using a number line

We know that

$\sqrt{16}=4$ and

$\sqrt{17}=\sqrt{{4}^{2}+{1}^{2}}$

Mark point

$A$ representing

$4$ on number line .

Now construct

$AB$ of unit length perpendicular to

$OA$ .

Then taking

$O$ as centre and

$OB$ as radius.

Draw an arc intersecting number line at

$C$ .

$C$ is represented as

$\sqrt{17}$

1310980_1406655_ans_ee49fcdf2b584d9bb92f0e641587e893.PNG

Simplify:

$\dfrac { a+\sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ a-\sqrt { { a }^{ 2 }-{ b }^{ 2 } } } + \dfrac { a-\sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ a+\sqrt { { a }^{ 2 }-{ b }^{ 2 } } }$

$\dfrac { a+\sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ a-\sqrt { { a }^{ 2 }-{ b }^{ 2 } } } + \dfrac { a-\sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ a+\sqrt { { a }^{ 2 }-{ b }^{ 2 } } }$

$=\dfrac { a^2 + { { a }^{ 2 }-{ b }^{ 2 } }+2a\sqrt {a^2-b^2} +a^2+a^2-b^2-2a\sqrt {a^2-b^2} }{ a^2- { { a }^{ 2 }+{ b }^{ 2 } } }$

$=\dfrac { a^2 + { { a }^{ 2 }-{ b }^{ 2 } } +a^2+a^2-b^2 }{ a^2- { { a }^{ 2 }+{ b }^{ 2 } } }$

$=\dfrac { 2( { 2{ a }^{ 2 }-{ b }^{ 2 } )}}{ { b }^{ 2 } }$

Write the simplest rationalising factor of : $\left( i \right)\sqrt {32} \;\left( {ii} \right)\sqrt {72} \;\left( {iii} \right)\root 3 \of 5 \;\left( {iv} \right)\root 5 \of 9 \left( v \right)\root 3 \of {135}$

1236697_1501982_ans_1fa6f2bb06794b66a3115101945d69e0.jpeg

Rationalise:
$\dfrac { 1 }{ x-\sqrt { x } }$

$\dfrac{1}{x-\sqrt{x}}$

$=\dfrac{1}{x-\sqrt{x}}\times\dfrac{x+\sqrt{x}}{x+\sqrt{x}}$

$=\dfrac{x+\sqrt{x}}{{x}^{2}-x}$

Expand ${a^3}{b^2},{a^2}{b^3},{b^2}{a^3},{b^3}{a^2}.$ Are they all same?

Now,

$a^3b^2=a\times a\times a\times b\times b$ .

$a^2b^3=a\times a\times b\times b\times b$ .

$b^2a^3=b\times b\times a\times a\times a$ .

$b^3a^2=b\times b\times b\times a\times a$ .

It is clear that all they are not same.

Find the value of $a$ and $b$ if $\dfrac { \sqrt { 11 } -\sqrt { 7 } }{ \sqrt { 11 } +\sqrt { 7 } } =a-b\sqrt { 77 }$ .

$\dfrac { \sqrt { 11 } -\sqrt { 7 } }{ \sqrt { 11 } +\sqrt { 7 } } =\dfrac { \sqrt { 11 } -\sqrt { 7 } }{ \sqrt { 11 } +\sqrt { 7 } } \times \dfrac { \sqrt { 11 } -\sqrt { 7 } }{ \sqrt { 11 } -\sqrt { 7 } }$ ........... Rationalization

$=\dfrac { { \left( \sqrt { 11 } -\sqrt { 7 } \right) }^{ 2 } }{ 11-7 } =\dfrac { 11+7-2\sqrt { 77 } }{ 4 }$

$=\dfrac { 9 }{ 2 } -\dfrac { \sqrt { 77 } }{ 2 }$

$\therefore$

$a=\dfrac { 9 }{ 2 } , b=\dfrac { 1 }{ 2 }$

Solve:
$\dfrac { 7+3\sqrt { 5 } }{ 2+\sqrt { 5 } } -\dfrac { 7-3\sqrt { 5 } }{ 2-\sqrt { 5 } }$

$\cfrac{7 + 3 \sqrt{5}}{2 + \sqrt{5}} - \cfrac{7 - 3 \sqrt{5}}{2 - \sqrt{5}}$

$= \cfrac{\left( 7 + 3 \sqrt{5} \right) \left( 2 - \sqrt{5} \right) - \left( 7 - 3 \sqrt{5} \right) \left( 2 + \sqrt{5} \right)}{\left( 2 + \sqrt{5} \right) \left( 2 - \sqrt{5} \right)}$

$= \cfrac{\left( 14 - 7 \sqrt{5} + 6 \sqrt{5} - 15 \right) - \left( 14 + 7 \sqrt{5} - 6 \sqrt{5} - 15 \right)}{4 - 5}$

$= \cfrac{\left( -1 - \sqrt{5} \right) - \left( -1 + \sqrt{5} \right)}{-1}$

$= 1 + \sqrt{5} - 1 + \sqrt{5}$

$= 2 \sqrt{5}$

$\dfrac { 4 }{ 5 } \times p\times p\times p\times p\times r\times r\times r\times r$

Given expression is

$\dfrac { 4 }{ 5 } \times p\times p\times p\times p\times r\times r\times r\times r$

Let

$A=\dfrac { 4 }{ 5 } \times p\times p\times p\times p\times r\times r\times r\times r$

$= \dfrac{4}{5}( p\times p\times p\times p)\times (r\times r\times r\times r)$

$A=\dfrac { 4 }{ 5 } { p }^{ 4 }{ r }^{ 4 }$

Write two irrational numbers between '3' and '4'

Two irrational numbers between

$3$ and

$4$ are :

$3^2 = 9$ and

$4^2 = 16$

$\therefore \sqrt{11}$ and

$\sqrt{13}$ are required numbers.

List three rational numbers between $6$ and $8 .$

${\textbf{Step -1: Find a common number which when multiplied with numerator and denominator}}$

${\textbf{gives}}$

${\mathbf{6}}$

${\textbf{and}}$

${\mathbf{8.}}$

${\text{Let}}$

$6$

${\text{ be a common number when multiplied}}$

${\text{would give}}$

${\text{as follows, }}$

$= \dfrac{6}{1} \times \dfrac{6}{6}$

${\text{and}}$

$= \dfrac{8}{1} \times \dfrac{6}{6}$

${\text{It would give,}}$

$\dfrac{{36}}{6}$

${\text{and}}$

$\dfrac{{48}}{{6{\text{ }}}}{\text{}}{\text{.}}$

${\textbf{Step -2: Find rational numbers between the found common fractions. }}$

${\text{Rational numbers between}}$

$\dfrac{{36}}{6}$

${\text{and}}$

$\dfrac{{48}}{{6{\text{ }}}}$

${\text{are,}}$

$\dfrac{{37}}{6},{\text{ }}\dfrac{{38}}{6},{\text{ }}\dfrac{{39}}{6},{\text{ }}\dfrac{{40}}{6},{\text{ }}\dfrac{{41}}{6},{\text{ }}\dfrac{{42}}{6},{\text{ }}\dfrac{{43}}{6},{\text{ }}\dfrac{{44}}{6},{\text{ }}\dfrac{{45}}{6},{\text{ }}\dfrac{{46}}{6},{\text{ }}\dfrac{{47}}{6}{\text{.}}$

${\textbf{ Hence, the rational numbers between}}$

${\mathbf{6}}$

${\textbf{and}}$

${\mathbf{8}}$

${\textbf{are,}}$

${\mathbf{\dfrac{{37}}{6},{\text{ }}\dfrac{{38}}{6},{\text{ }}\dfrac{{39}}{6},{\text{ }}\dfrac{{40}}{6},{\text{ }}\dfrac{{41}}{6},{\text{ }}\dfrac{{42}}{6},{\text{ }}\dfrac{{43}}{6},{\text{ }}\dfrac{{44}}{6},{\text{ }}\dfrac{{45}}{6},{\text{ }}\dfrac{{46}}{6},{\text{ }}\dfrac{{47}}{6}}}{\textbf{.}}$

$\dfrac{3}{2\sqrt{5}}$ is $\dfrac {3\sqrt{5}} {b}$ then value of b is ___ .

Let's rationalize

$\dfrac{3}{2 \sqrt{5}}$

Multiply numerator and denominator by

$\sqrt{5}$ as

$\sqrt{5} \sqrt{5} = 5$

$\therefore \dfrac{3 \times \sqrt{5}}{2 \sqrt{5} \times \sqrt{5}} = \dfrac{3 \sqrt{5}}{2 \sqrt{25}} = \dfrac{3 \sqrt{5}}{10}$

$= \dfrac{3 \sqrt{5}}{10}$

Comparing with

$\dfrac{3 \sqrt{5}}{b}$

we get value of '

$b$ ' as

$10$

1396445_1668256_ans_218ac976acc34df6b2a015c3d1db3645.jpg

Is 5.131131113.... a rational number or irrational number?

A number that can be expressed as

$\dfrac{p}{q}$ , where

$p,q$ are integers and

$q\neq 0$ , is called a rational number.

Irrational numbers:A number that cannot be expressed in the form

$\dfrac{p}{q}$ , where

$p,q$ are integers and

$q\neq 0$ , is called irrational number.

These numbers neither terminates nor repeat the particular number in the process of division is said to be a non terminating non repeating decimal expansion.

Here,

$5.131131113…$ has non terminating non repeating decimal expansion.So it is an irrational number.

Find three rational numbers between $\dfrac {1}{5}$ and $\dfrac {1}{4}$ .

$\dfrac {1}{4} = \dfrac {1}{4} \times \dfrac {20}{20} = \dfrac {20}{80}$ and

$\dfrac {1}{5} = \dfrac {1}{5} \times \dfrac {16}{16} = \dfrac {16}{80}$

Now,

$\dfrac {17}{80}, \dfrac {9}{40}, \dfrac {19}{80}$ are three rational numbers lying between

$\dfrac {1}{4}$ and

$\dfrac {1}{5}$ .

Simplify: $a\left ( a^0 + b^0 \right )^3$

Given :

$a(a^0+b^0)^3$

$a^{\circ} = 1 , b^{\circ} =1$

$\therefore (a) (1+1)^{3} = a(2)^{3} = 80$

Write down a rational number whose numerator is the largest number of two digits and denominator is the smallest number of four digits .

We know that the largest two digit number is

$99$
So the smallest four digit number is

$1000$
Numerator

$= 99$
Denominator

$= 1000$
Rational number

$= \dfrac{99}{1000}$

Simplify : ${p^{\frac{1}{2}}} \times {p^{\frac{1}{2}}}$

$={ p }^{\frac {1}{2} }\times { p }^{\frac{1}{2 }}$

$={ p }^{ \frac { 1 }{ 2 } +\frac { 1 }{ 2 } }={ p }^{ \frac { 2 }{ 2 } }={ p }^{ 1 }=p$

Find six rational numbers between 3 and 4.

$\textbf{Step-1: Definition}$

$\text{A rational number is a number of the form}$

$\dfrac{p}{q}$

$\text{where}$

$p\neq 0.$

$\textbf{Step-2: Rational numbers between 3 and 4}$

$\text{The following are rational numbers between 3 and 4}$

$3.1,3.2,3.3,3.4,3.5,3.6$

$\textbf{Hence the rational numbers between 3 and 4 are}$

$\mathbf{3.1,3.2,3.3,3.4,3.5,3.6}$

List five rational numbers between
$\cfrac{-4}{5}$ and $\cfrac{-2}{3}$

$\cfrac{-4}{5}$ and

$\cfrac{-2}{3}$
Let us write

$\cfrac{-4}{5}$ and

$\cfrac{-2}{3}$ as rational numbers with same denominators

$\Rightarrow$

$\cfrac{-4}{5}=\cfrac{-36}{45}$ and

$\cfrac{-2}{3}=\cfrac{-30}{45}$

$\therefore$

$\cfrac{-36}{45}< \cfrac{-35}{45}< \cfrac{-34}{45}< \cfrac{-33}{45}< \cfrac{-32}{45}< \cfrac{-31}{45}< \cfrac{-30}{45}$

$\Rightarrow$

$\cfrac { -4 }{ 5 } <\cfrac { -7 }{ 9 } <\cfrac { -34 }{ 45 } <\cfrac { -11 }{ 15 } <\cfrac { -32 }{ 45 } <\cfrac { -31 }{ 45 } <\cfrac { -2 }{ 3 }$
Therefore, five rational numbers between

$\cfrac{-4}{5}$ and

$\cfrac{-2}{3}$

$\cfrac { -7 }{ 9 } ,\cfrac { -34 }{ 45 } ,\cfrac { -11 }{ 15 } ,\cfrac { -32 }{ 45 } ,\cfrac { -31 }{ 45 } ,\cfrac { -2 }{ 3 }$

Prove that :
$(a^{-1}+b^{-1})^{-1} = \dfrac{ab}{a+b}$

Given,

$\left(a^{-1}+b^{-1}\right)^{-1}$

$=\dfrac{1}{\left(a^{-1}+b^{-1}\right)}$

$=\dfrac{1}{\frac{1}{a}+\frac{1}{b}}$

$=\dfrac{1}{\frac{b+a}{ab}}$

$=\dfrac{ab}{b+a}$

$[hence \, proved]$

$\dfrac{3\sqrt{2}}{\sqrt{5}}$ is $\dfrac{3\sqrt{a}}{5}$ then value of a is ___ .

Let's rationalize

$\dfrac{3 \sqrt{2}}{\sqrt{5}}$

Multiply numerator and denominator by

$\sqrt{5}$ as

$\sqrt{5} \times \sqrt{5} = 5$

$\therefore \dfrac{3 \sqrt{2}}{\sqrt{5}} = \dfrac{3 \sqrt{2} \sqrt{5}}{\sqrt{5} \sqrt{5}} = \dfrac{3 \sqrt{2 \times 5}}{\sqrt{25}} = \dfrac{3 \sqrt{10}}{5}$

Comparing

$\dfrac{3 \sqrt{10}}{5}$ with

$\dfrac{3 \sqrt{a}}{5}$ ,

we get value of 'a' as

$10$ .

$\dfrac{\sqrt{2}}{\sqrt{5}}$ is $\dfrac {\sqrt{10}} {c}$ then value of c is ___ .

Let's rationalize

$\dfrac{\sqrt{2}}{\sqrt{5}}$

Multiply numerator and denominator by

$\sqrt{5}$ as

$\sqrt{5} \times \sqrt{5} = 5$

$\therefore \dfrac{\sqrt{2}}{\sqrt{5}} = \dfrac{\sqrt{2} \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} = \dfrac{\sqrt{10}}{\sqrt{5 \times 5}}$

$= \dfrac{\sqrt{10}}{5}$

Now comparing with

$\dfrac{\sqrt{10}}{C}$

we get value of c as

$5$ .

1396477_1668262_ans_f679d581b4a44389ae78141866e8a500.jpg

Define an irrational number.

An irrational number is a number that cannot be expressed as a fraction $\dfrac{p}{q}$ for any integers $p,q$ . Irrational numbers have decimal expansions that neither terminate nor become periodic. Every real transcendental number is irrational.

Examine whether $\sqrt{5}-2$ is rational or irrational.

Given,

$\sqrt 5 -2$

Here

$\sqrt 5$ is an irrational number and

$2$ is a rational number.

We know, difference of a rational and an irrational results in an irrational number.

Since

$\sqrt 5$ is an irrational number, then

$\sqrt 5 -2$ is also an irrational.

Hence, the given number is irrational.

After rationalizing the denominator of $\dfrac{1}{\sqrt{6}-\sqrt{5}}$ and simplifying we get $\sqrt{a}+\sqrt{5}$ then value of $a$ is ___ .

The given fraction is

$\dfrac{1}{\sqrt{6} - \sqrt{5}}$

To rationalize the denominator we can multiply both numerator and denominator by

$\sqrt{6} + \sqrt{5}$ (the conjuate of

$\sqrt{6} - \sqrt{5}$ )

$\therefore \dfrac{1}{\sqrt{6} - \sqrt{5}} = \dfrac{1 \times (\sqrt{6} + \sqrt{5})}{(\sqrt{6} - \sqrt{5}) (\sqrt{6} + \sqrt{5})}$

$= \dfrac{\sqrt{6} + \sqrt{5}}{(\sqrt{6})^2 - (\sqrt{5})^2}$

$.... ((a - b)(a + b) = a^2 - b^2)$

$= \dfrac{\sqrt{6} + \sqrt{5}}{6 - 5}$

$= \dfrac{\sqrt{6} + \sqrt{5}}{1} = \sqrt{6} + \sqrt{5}$

Now, comparing

$\sqrt{6} + \sqrt{5}$ with

$\sqrt{a} + \sqrt{5}$ ,

we get the value of 'a' as

$6$ .

1397088_1668296_ans_6333cceb0441411b8dc615398f199e49.jpg

If rationalization of $\dfrac{\sqrt{2}+\sqrt{5}}{\sqrt{3}}$ is $\dfrac{\sqrt{a}+\sqrt{15}}{3}$ then value of $a$ is ___ .

Let's rationalize

$\dfrac{\sqrt{2}+\sqrt{5}}{\sqrt{3}}$

Multiply numerator and denominator

$\sqrt{3}$ as

$\sqrt{3}\times \sqrt{3}=3$

$\therefore \dfrac{(\sqrt { 2 }+\sqrt { 5 })\times\sqrt { 3 }}{\sqrt { 3 }\times \sqrt { 3 }}=\dfrac{\sqrt { 2 }\sqrt {3 }+\sqrt { 5 }\sqrt { 3 }}{3}=\dfrac{\sqrt { 6 }+\sqrt { 15 }}{3}$

Now, comparing with

$\dfrac{\sqrt {a }+\sqrt { 15 }}{3}$

We get value of

$'a'$ as

$6$ .

$\dfrac{\sqrt{3}+1}{\sqrt{2}}$ is $\dfrac{\sqrt{6}+\sqrt{a}}{2}$ then value of a is ___ .

Let's rationalize

$\dfrac{\sqrt{3} + 1}{\sqrt{2}}$

Multiply numerator and denominator by

$\sqrt{2}$ as

$\sqrt{2} \times \sqrt{2} = 2$

$\therefore \dfrac{(\sqrt{3} + 1) \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \dfrac{(\sqrt{3} \times \sqrt{2}) + \sqrt{2}}{2}$

$= \dfrac{\sqrt{6} + \sqrt{2}}{2}$

Now, comparing with

$\dfrac{\sqrt{6} + \sqrt{a}}{2}$

we get value of 'a' as

$2$ .

$\dfrac{1}{\sqrt{12}}$ is $\dfrac {\sqrt{a}} {6}$ then value of a is ___ .

Let's rationalize

$\dfrac{1}{\sqrt{12}}$

Multiply numerator and denominator by

$\sqrt{12}$ as

$\sqrt{12} \times \sqrt{12} = 12$

$\therefore \dfrac{1}{\sqrt{12}} = \dfrac{\sqrt{12}}{\sqrt{12} \sqrt{12}} = \dfrac{\sqrt{4 \times 3}}{12} = \dfrac{\sqrt{4} \sqrt{3}}{12}$

$= \dfrac{2 \sqrt{3}}{12}$

$= \dfrac{\sqrt{3}}{6}$

Comparing with

$\dfrac{\sqrt{a}}{6}$

we get value of 'a' as

$3$ .

Examine whether the following number is rational or irrational: $0.3796$ .

Given number is

$0.3796$ .

To find out,

Whether the given number is rational or irrational.

We know that a number having a terminating or non-terminating, repeating decimal expansion is a rational number.

The given number has a terminating decimal expansion.

Therefore, it is a rational number.

Hence,

$0.3796$ is a rational number.

Prove that :
$\sqrt{\dfrac{1}{4}}+(0.01)^{-1/2}-(27)^{2/3} = \dfrac{3}{2}$

Given,

$\sqrt{\dfrac{1}{4}}+(0.01)^{-\frac{1}{2}}-(27)^{\frac{2}{3}}$

$=\dfrac{1}{2}+10-9$

$=\dfrac{1}{2}+1$

$=\dfrac{3}{2}$

$[hence \, proved]$

Examine, whether the following numbers are rational or irrational: $7.478478....$

Number has repeating and non-terminating decimal
Hence it is Rational

Complete the following sentence:

Every point on the number line corresponds to a ... number which may be either ... or ...

We know that, every point on the number line is real. Further, all the real numbers are either rational or irrational.

Hence, the three blanks in the given sentence are:

Real, rational, irrational

So, the complete sentence will be:

Every point on the number line corresponds to a real number which may be either rational or irrational.

Find the value of $(3^{-1}+4^{-1}+5^{-1})^0$

We know that

$a^{-m}=\dfrac {1}{a^m}$

Hence,

$(3^{-1}+4^{-1}+5^{-1})^0 = \bigg( \dfrac 13+\dfrac 14 + \dfrac 15 \bigg)^0$

$=\bigg( \dfrac{20+15+12}{60} \bigg)^0$ [Since LCM of

$3,\ 4,\ 5$ is

$60$ ]

$=\bigg(\dfrac{47}{60}\bigg) ^0$

We know

$a^0=1$

So,

$\bigg( \dfrac{47}{60}\bigg)^0 =1$

Therefore

$(3^{-1}+4^{-1}+5^{-1})^0=1$

Give an example of two irrational numbers whose quotient is an irrational number.

1650132_1668380_ans_be3d40f366a14a2bbbb9f58ffa1af581.jpg

Give an example of two irrational numbers whose quotient is a rational number.

$\sqrt{8},\sqrt{2}$ are the irrational numbers whose quotient is

$2$ a rational number

Express each of the following as a rational number with positive denominator:
$\dfrac{6}{-9}$

Given :

$\dfrac{6}{-9}$

this can be written as

$\dfrac{-6}{9}$

$\Rightarrow \dfrac{-6}{9}$ is a rational number with positive denominator

Which of the following rational numbers are negative?
$\dfrac{-3}{7}$

1516048_1676114_ans_c07a4246708e45b5bc280a76af3c29f3.jpg

Express each of the following numbers as a product of powers to their prime factors: 675

1539637_1676135_ans_db122e579f8245b1872aae53066b34e8.jpeg

check the given numbers is positive or not.
$\dfrac{9}{8}$

1590789_1676103_ans_e6618c4711a547739d671ce5ffb0edd1.jpg

Separate positive and negative rational numbers from the following rational numbers:
$\displaystyle \frac{-5}{-7}, \frac{12}{-5}, \frac{7}{4}, \frac{13}{-9}, 0, \frac{-18}{-7}, \frac{-95}{116}, \frac{-1}{-9}$

1590787_1676094_ans_f4ef54613a5b4e8791c6998429719c91.jpg

Which of the following rational numbers are positive:
$\dfrac{-19}{-13}$

1516044_1676106_ans_18d24834025a4a1aa977a06bbecd598d.jpg

Which of the following rational numbers are positive:
$\dfrac{-8}{7}$

$\dfrac{-8}{7}=-\ \left (\dfrac{8}{7} \right ) \rightarrow negative \ rational \ number$

Write down the numerator of each of the following rational numbers:
$5$

1515996_1676037_ans_cd65267ef5f84f2aa0a9665c5a363c5d.jpg

Express each of the following as a rational number with positive denominator:
$\dfrac{-15}{-28}$

1590912_1676152_ans_40911503581a443d909f99dd99c72c68.jpg

Which of the following rational numbers are negative?
$\dfrac{9}{-83}$

1516118_1676119_ans_196d22dd8589446497ef2a5778d2a33a.jpg

Without actual division, show that each of the following rational number is a terminating decimal. Express in decimal form: $\dfrac {23}{ ( 2^3 \times 5^2)}$ .

Clearly either

$2$ or

$5$ is not factor of

$23 ,$ so given fraction is in its simplest form.

Also, the given fraction already in the form of

$(2^m \times 5^n)$ .
Hence, the given rational is a terminating decimal.

We can written it as:

$\dfrac {23}{(2^3 \times 5^2 )} =\dfrac {23}{200} =0.115$ , which is the decimal form.

Fill in the blanks so as to make the statement true:
If $p$ and $q$ are positive integers, then $\dfrac{p}{q}$ is a _ rational number and $\dfrac{p}{-q}$ is a __ rational number.

$p$ and

$q$ are positive integers, then

$\dfrac{p}{q}$ is a positive rational number and

$\dfrac{p}{-q}$ is a negative rational number.

Select those rational numbers which can be written as a rational number with denominator $4$ :
$\displaystyle \frac{7}{8}, \frac{64}{16}, \frac{36}{-12}, \frac{-16}{17}, \frac{5}{-4}, \frac{140}{28}$

1591471_1676310_ans_2c691e0f6ac94f04a8ab3644d206ebdb.jpg

Find $10$ rational numbers between $\dfrac{7}{13}$ and $\dfrac{-4}{13}$ .

We know that, between

$-4$ and

$7$ , below mentioned numbers will lie

$-3, -2, -1, 0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 6$

According to the definition, rational numbers are in the form of

$\dfrac pq$ where

$q$ is not equal to zero.

Therefore, ten rational numbers between

$\dfrac{7}{13}$ and

$\dfrac{-4}{13}$ are :

$-\dfrac3{13}, -\dfrac2{13}, -\dfrac1{13}, 0, \dfrac1{13}, \dfrac2{13}, \dfrac3{14}, \dfrac4{14}, \dfrac5{14}, \dfrac6{14}$

Find six rational numbers between $\dfrac{-4}{8}$ and $\dfrac{3}{8}$ .

We know that, between

$-4$ and

$3$ , below mentioned numbers will lie

$-3, -2, -1, 0, 1, 2.$

According to the definition, rational numbers are in the form of

$\dfrac pq$ where

$q$ is not equal to zero.

Therefore, six rational numbers between

$\dfrac{-4}8$ and

$\dfrac38$ are :

$-\dfrac38, -\dfrac28, -\dfrac18, 0, \dfrac18, \dfrac28, \dfrac38$

Express each of the following as a rational number with positive denominator:
$\dfrac{-28}{-11}$

1590921_1676159_ans_dd564411249f4a3399e97240a0f681ea.jpg

Express $392$ as a product of its prime factors

$392 = 2^3 \times 7^2$
2102875_1676160_ans_0942173dfb324210bd26986bb76764d2.png

Give an example of two irrational whose sum is rational.

Let us consider two irrational number be

$( 5 + \surd{ 3} )$ and

$( 5- \surd {3} )$

Sum =

$( 5 + \sqrt{ 3} )$ +

$( 5- \sqrt {3} )=10$

which is a rational number.

Without actual division, show that the following rational number has a terminating decimal form and also, express in decimal form
$\dfrac {19} { 3125}$

Here, denominator is

$3125 = 5^5$ .

The rational number can be written as:

$\dfrac { 19} {3125} = \dfrac {19}{(2^0 \times 5^5 )}$ , which is similar to the form of

$(2^m \times 5^n)$ in denominator.

Hence, the given rational is terminating,
and its decimal expansion is

$\dfrac {19} { 3125}= 0.00608$ .

Give an example of two irrationals whose product is rational.

Let us consider that the two irrational number be

$( 3 + \sqrt {2} )$ and

$( 3 - \sqrt {2} )$

product =

$( 3 + \sqrt {2} ) ( 3 - \sqrt {2} )$

$= (3)^2- ( \sqrt {2})^2 =9-2= 7$
which is rational number.

Let $x$ and $y$ be rational and irrational numbers, respectively. Is $x + y$ necessarily an irrational number? Given an example in support of your answer.

As we know before,

The sum of rational and irrational number is mainly an irrational number.

Therefore, if we consider m as rational and n as irrational

Then the sum of

$m+n$ is an irrational number

Example:

$m=1$ and

$n=\sqrt{2}$

$m+n=1+\sqrt{2}$ is an irrational number.

Without actual division, show that the following rational number is a terminating decimal. Express in decimal form: $\dfrac {15} {1600}$ .

Given rational number is

$\dfrac{15}{1600}$ .

We can represent the denominator, as a product of its prime factors as,

$1600=2\times 2\times 2\times 2\times 2\times 2\times 5\times 5$

Hence, the denominator is

$1600 =$

$2^6 \times 5^2$ .

We know that a rational number will have a terminating decimal expansion if the denominator is of the form

$(2^m \times 5^n)$ .

Hence, the given rational is terminating.

Now, its decimal expansion is

$\dfrac {15}{1600} = 0.009375$ .

Without actual division, show that each of the following rational number is a terminating decimal. Express in decimal form: $\dfrac {17} { 320}$ .

Here, denominator is

$320 = 2^6 \times 5$ .

The given rational number can be written as:

$\dfrac {17}{320} = \dfrac {17} {(2^6 \times 5 )}$ , which is similar to the form of

$(2^m \times 5^n)$ in denominator.

Hence, the given rational is terminating,
and its decimal expansion is

$\dfrac {17} { 320} = 0.053125$ .

Without actual division, show that each of the following rational number is a terminating decimal. Express in decimal form $\dfrac {171} {800}$ .

Here, denominator is

$800 = 2^5 \times 5^2$ .

Either

$2$ or

$5$ is not a factor of

$171$ , so fraction is in its simplest form.

Then, it can be written as:

$\dfrac {24} {125} =\dfrac {171} {(2^5 \times 5^2 )}$ , which is similar to the form of

$(2^m \times 5^n)$ in denominator.

Hence, the given rational is terminating
and its decimal expansion is

$\dfrac {171} {800} = 0.21375$ .

Without actual division, show that each of the following rational number is a terminating decimal. Express in decimal form $\dfrac {24} {125}$ .

Here, Denominator

$= 125 = 5^3$ .
5 is not a factor of 24, so given fraction is in its simplest form.

It can be written as:

$\dfrac {24} {125} = \dfrac {24} {(2^0 \times 5^3 )}$ , which is similar to the form of

$(2^m \times 5^n)$

Hence, the given rational number is terminating .
and its decimal expansion is

$\dfrac {24}{125}= 0.192$ .

Find four rational numbers between $\dfrac{3}{7}$ and $\dfrac{5}{7}$ .

Let us consider

$\dfrac{3}{7}=\dfrac{3\times 3}{7\times 3}=\dfrac{9}{21}$

And

$\dfrac{5}{7}=\dfrac{5\times 3}{7\times 3}=\dfrac{15}{21}$

As we know

$\dfrac{9}{21} < \dfrac{10}{21} < \dfrac{11}{21} < \dfrac{12}{21} < \dfrac{13}{21} < \dfrac{14}{21} < \dfrac{15}{21}$

Thus the four rational numbers which lies in-between

$\dfrac{3}{7}=\dfrac{9}{21}$ and

$\dfrac{5}{7}=\dfrac{15}{21}$ are

$\dfrac{10}{21}, \dfrac{11}{21}, \dfrac{12}{21}$ and

$\dfrac{13}{21}$ .

Give an example of two irrational numbers whose difference is an irrational number.

Difference is an irrational number

For example :

$(7\sqrt{3}+1)-(5\sqrt{3}-8)=2\sqrt{3}+9$ .

Simplify $\dfrac{2\sqrt{30}}{\sqrt{6}}-\dfrac{3\sqrt{140}}{\sqrt{28}}+\dfrac{\sqrt{55}}{\sqrt{99}}$ .

1673354_1715228_ans_b66c6d02381546c9ac0960fc3160fa51.jpg

Express with rational denominator :
$\dfrac{3\sqrt{-2}+2\sqrt{-5}}{3\sqrt{-2}-2\sqrt{-5}}$ .

1569807_1743022_ans_5c131292801b4662abb70528c8fbee01.jpg

Given, $\sqrt{2}=1.414$ and $\sqrt{6}=2.449$ , find the value of $\dfrac{1}{\sqrt{3}-\sqrt{2}-1}$ correct to $3$ places of decimal.

1578425_1715547_ans_6db69525c41346f3b5a9a8a2859fad81.jpg

Represent $(1+\sqrt{9.5})$ on the number line.

1961113_1715258_ans_45513f8e7b5e455f8b4b3c57c47bbe4f.jpg

Represent $\sqrt{4.7}$ geometrically on the number line.

1581143_1715249_ans_1e0b3b56441c49a0bc5bc448139afa1b.jpg

Represent $\sqrt{7.28}$ geometrically on the number line.

1708476_1715254_ans_6e6d1c61a70a4348ac579f9d7f779c31.jpg

Find two irrational numbers between $0.16$ and $0.17$ .

$0.16$ and

$0.17$ are two rational numbers.

We know, between any two rational numbers, there are infinitely many irrational numbers.

An irrational number has non-terminating non-recurring decimal expansions.

Then two irrational numbers lying between

$0.16$ and

$0.17$ are:

$0.16111611116111116....$ and

$0.169669666966669.....$ .

NOTE:

This is two of the various possible answers. It might have different solution.

Give an example of two irrational numbers whose difference is a rational number.

Difference is a rational number.

$(6-\sqrt{5})-(2-\sqrt{5})=4$ .

Give an example of two irrational numbers, the sum of which is a rational number.

Express with rational denominator :
$\dfrac{(a+\sqrt{-1})^{3}-(a-\sqrt{-1})^{3}}{(a+\sqrt{-1})^{2}-(a-\sqrt{-1})^{2}}$ .

1569852_1743053_ans_f6b1ec5d8bd04f5db04d6a41863515db.jpg

Write the following integer as a rational number. Write the numerator and the denominator in the following case.
$-23$

$-23$

It can be written in a rational form as

$\cfrac{-23}{1}$

$\therefore -23=$ Numerator

and

$1=$ Denominator

Given below is a negative rational number or not?
$0$

$\Rightarrow$ A rational number is a number that can be written as a ratio.

$\Rightarrow$ The number

$0$ is rational number but its neither a positive nor a negative rational number.

Express the following as a rational number:
$(23/25)^0$

By definition, we have

$a^0=1$ for every rational number

$a$ .

Hence,

$(23/25)^0=1$

Mark tick against the correct answer in each of the following:
$(5/6)^0 =$ ?

$(5/6)^0 =1$
By definition, we have

$a^0=1$ for ever integer.

What are rational number? Give examples of five positive and negative rational numbers. Is there any rational number which is neither positive nor negative? Name it.

The numbers of the form

$(p/q)$ , where

$p$ and

$q$ are integers and

$q\neq 0$ , are called rational number.
Examples for positive rational number:-

$(2/3), (-4/-5), (1/2), (20/30), (-50/-40)$
Examples for negative rational numbers:-

$(4/-5), (-2/3), (10/-56), (46/-34), (-23/32)$
Yes. There is a number which is neither positive not negative, i.e., Zero

$(0)$

Given below is a negative rational number or not?
$-6$

1961165_1779848_ans_4f041b1585cf47e9b023828d27579945.jpg

Write a rational number in which the numerator is less than $'-7\times11'$ and the denominator is greater than $'12+4'.$

Let

$-7\times 11=p=-77$

$12+4=q=16$
Rational number

$=\dfrac{p}{q}=\dfrac{-77}{16}$
Hence, it has more than one answer like

$\dfrac{-78}{17},\dfrac{-79}{18},\dfrac{-80}{19}$

List four rational numbers between $\dfrac{5}{7}$ and $\dfrac{7}{8}$

First, make the denominator the same,

For that, the LCM of

$7$ and

$8$ is 56.

Therefore,

$\dfrac{5\times 8}{8\times 8}=\dfrac{40}{46}$

and

$\dfrac{7\times 7}{8\times 7}=\dfrac{49}{56}$

So four rational numbers between

$\dfrac{40}{46}$ and

$\dfrac{49}{56}$ are

$\dfrac{42}{56},\dfrac{44}{56},\dfrac{46}{56},\dfrac{48}{56}.$

In each of the following cases, write the rational number whose numerator and denominator are respectively as under:
$5-39$ and $54-6$

$5-39$ and

$54-6$
Let numerator

$p=5-39=-34$
Denominator

$q=54-6=48$
Hence, rational number,

$\dfrac{p}{q}=\dfrac{-34}{48}$

In each of the following cases, write the rational number whose numerator and denominator are respectively as under:
$(-4)\times6$ and $8\div 2$

$(-4)\times6$ and

$8÷2$
Let numerator

$p=-4\times6=-24$
Denominator

$q=8÷2=4$
Hence, rational number,

$\dfrac{p}{q}=\dfrac{-24}{4}$

In each of the following cases, write the rational number whose numerator and denominator are respectively as under:
$35\div(-7)$ and $35-18$

$35÷(-7)$ and $35-18$
Let numerator $p=35÷(-7)=-5$
Denominator $q=35-18=17$
Hence, rational number, $\dfrac{p}{q}=\dfrac{-5}{17}$

Solve the following:
$2^{-2}\times 2^{-3}$

Use

$a^m \times a^n=a^{m+n}$

$=2^{-2}\times 2^{-3}=(2)^{-2-3}=a^{m+n}$

$=2^{-2}\times 2^{-3}=(2)^{-2-3}=(2)^{-5}$

$a^3 \times a^{-10}$

${\textbf{Step -1: Use the formula.}}$

$\because{a^m} \times {a^n} = {a^{m + n}}$

${\text{Given}}$

${a^m} = {a^3}$

${\text{and}}$

${a^n} = {a^{ - 10}}$

${\textbf{Step -2: Evaluate the values to get the answer}}{\textbf{.}}$

$\Rightarrow {a^3} \times {a^{ - 10}} = {a^{3 + \left( { - 10} \right)}}$

$\Rightarrow {a^{ - 7}}$

${\textbf{Hence, the value of}}$

${\mathbf{{a^3} \times {a^{ - 10}}}}$

${\textbf{is}}$

${\mathbf{{a^{ - 7}}.}}$

Find the value of $x$ so that
$\left( \dfrac 53\right)^{-2}\times \left( \dfrac 53\right)^{-14}=\left( \dfrac 53\right)^{8x}$

$\left( \dfrac 53\right)^{-2}\times \left( \dfrac 53\right)^{-14}=\left( \dfrac 53\right)^{8x}$
Use,

$a^m \times a^n=a^{m+n}$

$\Rightarrow \left( \dfrac 53\right)^{-2}\times \left( \dfrac 53\right)^{-14}=\left( \dfrac 53\right)^{8x}$

$\Rightarrow \left( \dfrac 53\right)^{-2-14}=\left( \dfrac{5}{3}\right)^{8x}$

$\Rightarrow \left( \dfrac{5}{3}\right)^{-16}=\left( \dfrac 53 \right)^{8x}$
Compare both the sides and the result is =

$16=8x$

$x=-2$ .

Simplify :
$(-2)^3 \times (-2)^{-6}=(-2)^{2x-1}$

Given,

$(-2)^3 \times (-2)^{-6}=(-2)^{2x-1}$
Use,

$a^m \times a^n =a^{m+n}$
Then,

$(-2)^3 \times (-2)^{-6}=(-2)^{2x-1}$

$\Rightarrow (-2)^{3-6}=(-2)^{2x-1}$

$\Rightarrow (-2)^{-3}=(-2)^{2x-1}$
Compare both sides and the result is =

$-3=2x-1$

$2x=-3+1$

$2x=-2$

$x=-1$

By what number should we multiply $(-29)^0$ so that the product becomes $(+29)^0$ .

Let

$x$ be multiplied with

$(-29)^0$ to get

$(+29)^0$
So,

$x\times (-29)^0=(29)^0$

$\Rightarrow x \times 1=1$

$\Rightarrow x=1$ .

The value of $\left(\dfrac{1}{2^3}\right)^2$ is equal to _______.

$\left(\dfrac{1}{2^6}\right)$
By using Law of exponent -

$(a^m)^n=(a)^{mn}$

$\because \left(\dfrac{1}{2^3}\right)^2 =\left(\dfrac{1^3}{2^3}\right)^2 = \left(\dfrac{1}{2}\right)^{3 \times 2}= \left(\dfrac{1}{2}\right)^6 = \dfrac{1}{2}^6$

By multiplying $(10)^5$ by $(10)^{-10}$ we get _____.

$(10)^-5$
As,

$a^m \times a^n=a^{m+n}$

$\Rightarrow (10)^5 \times (10)^{-10}= (10)^{5-10}$
So,

$\Rightarrow (10)^5 \times (10)^{-10}= (10)^{-5}$

$5^5 \times 5^{-5}=$ ______.

$1$
As,

$x^m \times x^n=(x)^{m+n}$

$\Rightarrow \ 5^5 \times 5^{-5}=(5)^{5-5}=(5)^0=1$
.

$[2^{-1}+3^{-1}+4^{-1}]^0=$ ______.

$1$

We know that,

$a^0=1$ .

$[2^{-1}+3^{-1}+4^{-1}]^0=1$

Express $3^{-5}\times 3^{-4}$ as a power of $3$ with positive exponent.

Use

$a^m \times a^n =a^{m+n}$ and

$a^{-m}=\dfrac{1}{a^m}$
Therefore,

$\therefore 3^{-5}\times 3^{-4}=(3)^{-5-4}=(3)^{-9}=\dfrac{1}{3^9}$

Find $x$ so that $\left( \dfrac{2}{9}\right)^{3}\times \left( \dfrac{2}{9}\right)^{-6}=\left( \dfrac{2}{9}\right)^{2x-1}$ .

Given,

$\left( \dfrac{2}{9}\right)^{3}\times \left( \dfrac{2}{9}\right)^{-6}=\left( \dfrac{2}{9}\right)^{2x-1}$ .
Using law of exponents,

$a^m \times a^n =a^{m+n}$
Thus,

$\left( \dfrac{2}{9}\right)^{3-6}=\left( \dfrac{2}{9}\right)^{2x-1}$

$\Rightarrow \left( \dfrac{2}{9}\right)^{-3}=\left( \dfrac 29\right)^{2x-1}$
On comparing both the sides we get,

$\Rightarrow -3=2x-1$

$\Rightarrow -2=2x$

$\Rightarrow x=-1$

A half life is the amount of time that it takes for a radioactive substance to decay to one half of its original quantity.
Suppose radioactive decay causes $300$ grams of a substance to decrease to $300\times 2^{-3}$ grams after $3$ half-lives. Evaluate $300\times 2^{-3}$ to determine how many grams of the substance are left.
Explain why the expression $300\times 2^{-n}$ can be used to find the amount of the substance that remains after $n$ half-lives.

A half life is the amount of time that it takes for a radioactive substance to decay to one half of its original quantity.

Since,

$300\ g$ of a substance decreases to

$\cfrac{300}2=300\times 2^{-1}\ g$ after one half life.

So, amount left after one half life will be (original weight-decayed weight) =

$300-\cfrac{300}2=\cfrac{300}2\ g$

So, for

$2^{nd}$ half life its initial weight will be

$\cfrac{300}2\ g=300\times2^{-1}\ g$

So,

$\cfrac12\times\cfrac{300}{2}=\cfrac{300}{2^2}=300\times2^{-2}\ g$ will be decayed or decreased in

$2^{nd}$ half life.

So, amount remaining after two half life will be (original weight at the start of second half life-decayed weight in second half life )

$\cfrac{300}2-\cfrac{300}4=\cfrac{300}4\ g=300\times 2^{-2}\ g$

Now, for

$3^{rd}$ half life its initial weight will be

$\cfrac{300}4\ g=300\times2^{-2}\ g$

So,

$\cfrac12\times\cfrac{300}{4}=\cfrac{300}{2^3}=300\times2^{-3}\ g$ will be decayed or decreased in

$3^{rd}$ half life.

So, amount remaining after

$3$ half lives will be (original weight at the start of third half life-decayed weight in third half life )

$\cfrac{300}4-\cfrac{300}8=\cfrac{300}8\ g=300\times 2^{-3}\ g$

So, after

$3$ half-lives

$300\times 2^{-3}\ g$ of substance is remaining, which is equal to

$\cfrac{300}8=37.5\ g$

We, can observe from above all

$3$ half lives that after

$n^{th}$ half life, amount of substance remaining is

$300\times 2^{-n}$

So, it is clear hat expression

$300\times 2^{-n}$ can be used to find the amount of the substance that remains after

$n$ half-lives.

In a repeater machine with $0$ as an exponent, the base machine is applied $0$ times.
What do these machines do to a piece of chalk?

Repeater Machine is a hypothetical machine which automatically enlarges items several times.

In given machinse

$(\times a)$ gives us base

$a$ and

$0$ is its multiplicity.

So, it will stretch the object by

$(\times a^{0})$

We will use the law of exponent -

$a^0=1$

So, it will stretch the object by

$(\times a^{0})=\times1$

That is object will remain as it is.

Since,

$3^0=1, 13^0 =1, 29^0=1$

So, machine

$(\times 3^0 ), ( \times 13^0)$ and

$( \times 29^0)$ produce nothing on not changing the piece of chalk.

In a repeater machine with $0$ as an exponent, the base machine is applied $0$ times.
What do you think the value of $6^o$ is ?

Repeater Machine is a hypothetical machine which automatically enlarges items several times.

We will use law of exponent -

$a^0=1$

So,

$( \times 6^0)=\times1$ machine does not change the piece.

or, value of

$6^0=1$

For the hook-up, determine whether there is a single repeater machine that will do the same work. If so, describe or draw it.

A machine

$(\times x^y)$ with input

$a$ will give an output

$ax^y$

When, multiple such machines are hooked up their stretching powers are multiplied.

Machine

$1$ is

$(\times 12^2)$

Machine

$2$ is

$(\times 12^3)$

As per the law of exponent-

$a^m \times a^n =a^{m+n}$

So,

The work that can be performed by the repeater machine is equal to

$12^2 \times 0.12^3=12^{5}$ .

It is described in a figure with

$12$ as base and

$5$ as power.

1678947_1793382_ans_9dd3c1a1ee6c42258011b8b5fd6a9922.png

Ajay had a $1\ cm$ piece of gum. He put it through repeater machine given below and it came out $\dfrac{1}{100,000}\ cm$ long. What is the missing value?

If input to

$a$ is given to

$\times x^y$ machine then output is

$ax^y$

Here,

$x$ is given as

$\times \cfrac1{10}$ , which is base, but power

$y$ is missing

But we know input and output, so we can find the power.

The

$1\ cm$ piece of gum that Ajay put came out as

$\dfrac {1}{100000}$

From this we can conclude that the machine is a shrinking machine.

Therefore,

$\cfrac1{100000}=1\times \left(\cfrac1{10}\right)^y$

$\Rightarrow y=5$

So, the type of shrinking machine is

$\left(\times\left( \dfrac{1}{10}\right)^5\right)$

Hence, the value missing is

$5$ .

While studying her family's history. Shikha discovers records of ancestors she has had in the past $12$ generations. She started to make a diagram to help her figure this out. The diagram soon become very complex.
Write an equation for the number of ancestors in a given generation $n$ .

On the basis of generations- ancestor table, the numbers of ancestors in Past generation number

$n$ will

$2^n$ .

1677879_1793222_ans_bce2687c11af4074b448ac1d2340bb28.png

Insert a rational number and an irrational number between the following:
$0.0001$ and $0.001$ .

$0.00011$ is a rational number

$0.0001$ and

$0.001$ , as it is terminating.
Now,

$0.000113133133...$ is an irrational number between

$0.0001$ and

$0.001$ , since it is a non-terminating and non-recurring decimal.

NOTE:

This is some of the various possible answers. It might have different solution.

Simplify :
$4\sqrt {12}\times 7\sqrt {6}$ .

$4\sqrt {12} \times 7\sqrt {6} = 4\sqrt {2\times 2\times 3} \times 7\sqrt {2\times 3}$

$= 8\sqrt {3} \times 7\sqrt {2} \times \sqrt {3}$

$= 24 \times 7\sqrt {2} = 168\sqrt {2}$ .

Represent geometrically the following number on the number line:
$\sqrt {2.3}$ .

Presentation of

$\sqrt {2.3}$ on number line:
Mark the distance

$2.3$ units from a fixed point

$A$ on a given line to obtain a point

$B$ such that

$AB = 2.3$ units. From

$B$ mark a distance of

$1$ unit and mark the new point as

$C$ . Find the mid-point of

$AC$ and mark that point as

$O$ . Draw a semicircle with Centre

$O$ and radius

$OC$ . Draw a line perpendicular to

$AC$ passing through

$B$ and intersecting the semicircle at

$D$ . Then,

$BD = \sqrt {2.3}$ .
Now, draw an arc with centre

$B$ and radius

$BD$ , which intersects the number line in

$E$ . Thus,

$E$ represents

$\sqrt {2.3}$ .
1662118_1794032_ans_76ba0884fd7848a58f23e82fb6d78eed.png

If $z^{2} = 0.04$ , find if $z$ is rational or irrational .

$z^{2} = 0.04 \Rightarrow z = \sqrt {0.04} = 0.2$ ,

which can be written in the form

$\dfrac {p}{q}$ , where

$q\neq0$

i.e

$0.2=\dfrac{2}{10}=\dfrac{1}{5}$

Hence,

$z$ is a rational number.

Let $x$ be rational and $y$ be irrational. Is $xy$ necessarily irrational? Justify your answer by an example.

Let

$x = 0$ (a rational number) and

$y = \sqrt {3}$ be an irrational number. Then,

$xy = 0(\sqrt {3}) = 0$ , which is not an irrational number.

Hence,

$xy$ is not necessarily an irrational number.

Find three rational numbers between:
$\dfrac {5}{6}$ and $\dfrac {6}{7}$ .

$\dfrac {5}{7} = \dfrac {5}{7}\times \dfrac {10}{10} = \dfrac {50}{70}$ and

$\dfrac {6}{7} = \dfrac {6}{7}\times \dfrac {10}{10} = \dfrac {60}{70}$

$\Rightarrow \dfrac {51}{70}, \dfrac {52}{70}, \dfrac {53}{70}$ are three rational numbers lying and between

$\dfrac {50}{70}$ and

$\dfrac {60}{70}$ and therefore lie between

$\dfrac {5}{7}$ and

$\dfrac {6}{7}$ .

Arrange in ascending order. $\sqrt[3]{2}, \sqrt 3, \sqrt[6] 5$

Here we can express the given expressions as:

$\sqrt[3]2 = 2^{1/3}$

$\sqrt3 = 3^{1/2}$

$\sqrt[6] 5 = 5^{1/6}$

Let us make the roots common so,

$2^{1/3}= 2(2\times 1/2 \times 1/3) = 4^{1/6}$

$3^{1/2} = 3(3\times 1/3 \times 1/2) = 27^{1/6}$

$5^{1/6} = 5^{1/6}$

Now, let us arrange in ascending order,

$4^{1/6}< 5^{1/6}< 27^{1/6}$

So,

$2^{1/3}< 5^{1/6}< 3^{1/2}$

So,

$\sqrt[3]2< \sqrt[6] 5< \sqrt 3$

Without actually performing the long division, state whether the following rational number have terminating or non-terminating repeating (recurring) decimal expansion: $\frac{17}{8}$ .

Given, rational number is

$\frac{17}{8}$ .

We know,

$\frac{p}{q}$ is terminating if:

$p$ and

$q$ are co-prime and

$q$ is of the form of

$2^{n}\times5^{m}$ , where

$n$ and

$m$ are non-negative integers.

Firstly, we check whether the numerator and denominator are co-prime,

$17=17\times1$

$8=2\times2\times2$

$\Rightarrow 17$ and

$8$ have no common factors.

Therefore,

$17$ and

$8$ are co-prime.

Now, we check whether

$q$ is in the form of

$2^{n}\times5^{m}$ ,

denominator

$8=2^{3}$

$=1\times2^{3}$

$=5^{0}\times2^{3}$

Since the denominator is of the form

$2^{n}\times5^{m}$ where

$n=3$ and

$m=0$ ,

thus,

$\frac{17}{8}$ has a terminating decimal expansion.

Prove that $3+\sqrt{5}$ is an irrational number.

$\textbf{Step 1: Prove that}$

$3+\sqrt{5}$

$\textbf{is an irrational number}$

$\text{Let's assume}$

$3+\sqrt{5}$

$\text{is a rational number and}$

$3+\sqrt{5}=\dfrac{a}{b}$

$\text{Now,}$

$\Rightarrow 3+\sqrt{5}=\dfrac{a}{b}$

$\Rightarrow \sqrt{5}=\dfrac{a}{b}-3$

$\Rightarrow \sqrt{5}=\dfrac{a-3 b}{b}$

$\text{Since,}$

$\dfrac{a-3 b}{b}$

$\text{is a rational number.}$

$\text{So,}$

$\sqrt5$

$\text{is also a rational number, but it is a contradiction.}$

$\text{Therefore, our assumption is wrong.}$

$\textbf{Hence,}$

$3+\sqrt{5}$

$\textbf{is an irrational number.}$

Insert five
rational numbers between:

(i) $\dfrac{-4}{5}$ and $\dfrac{-2}{3}$

(ii) $\dfrac{-1}{2}$ and $\dfrac{2}{3}$

(i) $\dfrac{-4}{5}$
and $\dfrac{-2}{3}$

LCM of $5,3=15$

$\therefore \dfrac{-4}{5}=\dfrac{-4 \times 3}{5 \times 3}=\dfrac{-12}{15}$

$\text { and } \dfrac{-2}{3}=\dfrac{-2 \times 5}{3 \times 5}=\dfrac{-10}{15}$

$\because$
Between -10 and $-12,$ there are is one rational.

$\therefore$
Multiplying numerator and denominator by $(5+1)=6$

$\therefore \dfrac{-12}{15}=\dfrac{-12 \times 6}{15 \times 6}=\dfrac{-72}{90}$

$\text { and } \dfrac{-10}{15}=\dfrac{-10 \times 6}{15 \times 6}=\dfrac{-60}{90}$

$\therefore 5$ rational numbers between $\dfrac{-72}{90}$ and $\dfrac{-60}{90}$

$\dfrac{-71}{90}, \dfrac{-70}{90}, \dfrac{-69}{90}, \dfrac{-68}{90}, \dfrac{-67}{90}$

$\dfrac{-71}{90}, \dfrac{-7}{9}, \dfrac{-23}{30}, \dfrac{-34}{45}, \dfrac{-67}{90}$

$(i i)-\dfrac{1}{2}$
and $\dfrac{2}{3}$

LCM of $2,3,=6$

$\therefore \dfrac{-1}{2}=\dfrac{-1 \times 3}{2 \times 3}=\dfrac{-3}{6}$ and

$\dfrac{2}{3}=\dfrac{2 \times 2}{3 \times 2}=\dfrac{4}{6}$

Now 5
rational numbers between

$-\dfrac{1}{2} \text { and } \dfrac{2}{3}$

$\dfrac{-2}{6}, \dfrac{-1}{6}, \dfrac{1}{6}, \dfrac{2}{6}, \dfrac{3}{6}$

Find ten rational numbers between $-2/5$ and $1/7$ .

The rational numbers $-2/5$ and $1/7$

By taking LCM for $5$ and $7$ which is $35$

So, $-2/5 = (-2 \times 7) / (5 \times 7) = -14/35$

$1/7 = (1 \times 5) / (7 \times 5) = 5/35$

Now, we can insert any $10$ numbers between $-14/35$ and $5/35$

i.e., $-13/35, -12/35, -11/35, -10/35, -9/35, -8/35, -7/35, -6/35, -5/35, -4/35, -3/35, -2/35, -1/35, 1/35, 2/35, 3/35, 4/35$

Hence, the ten rational numbers between $-2/5$ and $1/7$ are

$-6/35, -5/35, -4/35, -3/35, -2/35, -1/35, 1/35, 2/35, 3/35, 4/35$

Fill in the blanks:
$\bigg(-\dfrac{1}{2}\bigg)^0 + (-2)^0 =$ ........

$\bigg(-\dfrac{1}{2}\bigg)^0 + (-2)^0 = 1 + 1 = 2$

Rationalize the denominator of the following and hence evaluate by taking $\sqrt {2} = 1.414$ , upto three places of decimal.
$\dfrac {\sqrt {2}}{2 + \sqrt {2}}$ .

Given number is

$\dfrac {\sqrt {2}}{2 + \sqrt {2}}$

Rationalizing the denominator of the given number, we get:

$\dfrac {\sqrt {2}}{2 + \sqrt {2}} \times \dfrac {2 - \sqrt {2}}{2 - \sqrt {2}}$

Now, using the algebraic property

$(a+b)(a-b)=a^2-b^2$ in the denominator, we get:

$\dfrac {\sqrt {2} (2 - \sqrt {2})}{(2)^{2} - (\sqrt {2})^{2}}$

$= \dfrac {\sqrt {2}(2 - \sqrt {2})}{4 - 2}$

$= \dfrac {\sqrt {2} (2 - \sqrt {2})}{2}$

$= \dfrac {2\sqrt {2} - 2}{2}$

$= \dfrac {2(\sqrt {2} - 1)}{2}$

$=\sqrt {2} - 1$

Given that,

$\sqrt2=1.414$

Hence,

$\sqrt {2} - 1 = 1.414 - 1$

$= 0.414$

Hence,

$\dfrac {\sqrt {2}}{2 + \sqrt {2}}=0.414$ .

Find six rational numbers between $-1/2$ and $5/4$ .

Six rational numbers between

$-1/2$ and

$5/4$ can be calculate as below

$LCM$ of

$2, 4 = 4$

$-1/2 = (-1\times 2)/ (2\times 2)$
We get,

$= -2/4$
Now, six rational numbers between

$-1/2$ and

$5/4$ are as follows:

$-1/4, 0, 1/4, 2/4, 3/4$ and

$4/4$ .

Simplify :
$\dfrac {3}{\sqrt {8}} + \dfrac {1}{\sqrt {2}}$ .

$\dfrac {3}{\sqrt {8}} + \dfrac {1}{\sqrt {2}} = \dfrac {3}{\sqrt {4\times 2}} + \dfrac {1}{\sqrt {2}} = \dfrac {1}{\sqrt {2}} \left (\dfrac {3}{\sqrt{4}} + 1\right )= \dfrac {1}{\sqrt {2}} \left (\dfrac {3}{2} + 1\right )$

$= \dfrac {5}{2\sqrt {2}} = \dfrac {5}{2\sqrt {2}} \times \dfrac {\sqrt {2}}{\sqrt {2}} = \dfrac {5\sqrt {2}}{4}$ .

Write the decimal expansion of the following number which have terminating decimal expansion:
$\dfrac{2^{2}\times7}{5^{4}}$ .

We know,

$\dfrac{2^{2}\times7}{5^{4}}=\dfrac{2^{2}\times7}{2^{0}\times5^{4}}$

Multiplying and diving by

$2^{4}$ ,

$=\dfrac{2^{2}\times7\times2^{4}}{2^{0}\times5^{4}\times2^{4}}$

$=\dfrac{2^{2}\times7\times2^{4}}{1\times5^{4}\times2^{4}}$

$=\dfrac{7\times2^{6}}{(2\times5)^{4}}$

$=\dfrac{448}{(10)^{4}}$

$=\dfrac{448}{10000}$

$=0.0448$ and it is terminating.

1790227_1815225_ans_e76b5e160d0246aaa1811622a5ff463b.jpg

Write the decimal expansion of the following number which have terminating decimal expansion:
$\dfrac{27}{8}$ .

Given rational number is

$\dfrac{27}{8}$ .

We know,

$\dfrac{p}{q}$ is terminating if:

$p$ and

$q$ are co-prime and

$q$ is of the form of

$2^{n}\times5^{m}$ , where

$n$ and

$m$ are non-negative integers.

Firstly, we check whether the numerator and denominator are co-prime,

Numerator

$=27=3^3$

Denominator

$=8=2^3$

$\Rightarrow 27$ and

$8$ have no common factors.

Therefore,

$27$ and

$8$ are co-primes.

Now, we check whether

$q$ is in the form of

$2^{n}\times5^{m}$ ,

Denominator

$=8=2^3\times5^0$

So, the denominator is in the form

$2^{n}\times5^{m}$ , where

$n=3$ and

$m=0$ .

Thus,

$\dfrac{27}{8}$ is a terminating decimal.

We know,

$\dfrac{27}{8}=\dfrac{3^{3}}{2^{3}\times5^{0}}$ .
Multiplying and dividing by

$5^{3}$ ,

$=\dfrac{27\times5^{3}}{2^{3}\times5^{3}}$

$=\dfrac{27\times125}{2^{3}\times5^{3}}$

$=\dfrac{3375}{(2\times5)^{3}}$

$=\dfrac{3375}{(10)^{3}}$

$=\dfrac{3375}{1000}$

$=3.375$

Therefore,

$\dfrac{27}{8}=3.375$ is the decimal expansion.

Write down a rational number with numerator $(-5) \times (-4)$ and with denominator $(28 - 27) \times (8 - 5)$ .

It is given that
Numerator

$= (-5) \times (-4) = 20$
Denominator

$= (28 - 27) \times (8 - 5) = 1 \times 3 = 3$
So the rational number

$= \dfrac{20}{3}$

Write the denominator of each of the following rational numbers :
$-7$

$-7 = \dfrac{-7}{1}$
Here the denominator

$=1$

Write the decimal expansion of the following number which have terminating decimal expansion:
$\dfrac{6}{15}$ .

Given rational number is

$\dfrac{6}{15}$ .

We know,

$\frac{p}{q}$ is terminating if:

$p$ and

$q$ are co-prime and

$q$ is of the form of

$2^{n}\times5^{m}$ , where

$n$ and

$m$ are non-negative integers.

Firstly, we check whether the numerator and denominator are co-prime,

Numerator

$=6=2\times3$

Denominator

$=15=3\times5$

$\Rightarrow 6$ and

$15$ have a common factor

$3$

Therefore, divide

$6$ and

$15$ by

$3$ .

Then we get,

$\dfrac{6}{15}=\dfrac{2}{5}$ , i.e. here

$p$ and

$q$ are co-primes.

Now, we check whether

$q$ is in the form of

$2^{n}\times5^{m}$ ,

Denominator

$=5=2^0\times5^1$

So, the denominator is in the form

$2^{n}\times5^{m}$ , where

$n=0$ amd

$m=1$ .

Thus,

$\dfrac{6}{15}$ is a terminating decimal and

$\dfrac{6}{15}=0.4$ .

Write the decimal expansion of the following number which have terminating decimal expansion:
$\frac{64}{455}$ .

Given rational number is

$\dfrac{64}{455}$ .

We know,

$\frac{p}{q}$ is terminating if:

$p$ and

$q$ are co-prime and

$q$ is of the form of

$2^{n}\times5^{m}$ , where

$n$ and

$m$ are non-negative integers.

Firstly, we check whether the numerator and denominator are co-prime,

Numerator

$=64=2^6$

Denominator

$=455=5\times7\times13$

$\Rightarrow 6$ and

$15$ have no common factors.

Therefore,

$64$ and

$455$ are co-primes.

Now, we check whether

$q$ is in the form of

$2^{n}\times5^{m}$ ,

Denominator

$=455=5\times7\times13$

So, the denominator is not in the form

$2^{n}\times5^{m}$ .

Thus,

$\dfrac{64}{455}$ is a non-terminating decimal.

Write the decimal expansion of the following number which have terminating decimal expansion:
$\dfrac{129}{2^{2}\times5^{7}\times7^{5}}$ .

Given rational number is

$\dfrac{129}{2^{2}\times5^{7}\times7^{5}}$ .

We know,

$\frac{p}{q}$ is terminating if:

$p$ and

$q$ are co-prime and

$q$ is of the form of

$2^{n}\times5^{m}$ , where

$n$ and

$m$ are non-negative integers.

Firstly, we check whether the numerator and denominator are co-prime,

Numerator

$=129=3*43$

Denominator

$=2^{2}\times5^{7}\times7^{5}$

$\Rightarrow 129$ and

$2^{2}\times5^{7}\times7^{5}$ have no common factors.

Therefore,

$129$ and

$2^{2}\times5^{7}\times7^{5}$ are co-prime.

Now, we check whether

$q$ is in the form of

$2^{n}\times5^{m}$ ,

Denominator

$=2^{2}\times5^{7}\times7^{5}$

So, the denominator is not of the form

$2^{n}\times5^{m}$ .

Thus,

$\dfrac{129}{2^{2}\times5^{7}\times7^{5}}$ is a non-terminating repeating decimal.

Write the decimal expansion of the following number which have terminating decimal expansion:
$\frac{29}{243}$ .

Given, rational number is

$\dfrac{29}{243}$ .

Then,

$\dfrac{29}{243}=\dfrac{29}{3^{5}}$ .

Since the denominator is not of the form

$2^{n}\times5^{m}$ , where

$n,m$ are non-negative integers,

$\therefore\frac{29}{243}$ has a non-terminating repeating decimal expansion.

Write the decimal expansion of the following number which have terminating decimal expansion:
$\dfrac{13}{125}$ .

Given rational number is

$\dfrac{13}{125}$ .

We know,

$\dfrac{p}{q}$ is terminating if:

$p$ and

$q$ are co-prime and

$q$ is of the form of

$2^{n}\times5^{m}$ , where

$n$ and

$m$ are non-negative integers.

Firstly, we check whether the numerator and denominator are co-prime,

Numerator

$=13=13^1$

Denominator

$=125=5^3$

$\Rightarrow 13$ and

$125$ have no common factors.

Therefore,

$13$ and

$125$ are co-primes.

Now, we check whether

$q$ is in the form of

$2^{n}\times5^{m}$ ,

Denominator

$=125=5^3$

So, the denominator is in the form

$2^{n}\times5^{m}$ , where

$n=0$ and

$m=3$ .

Thus,

$\dfrac{13}{125}$ is a terminating decimal.

We know,

$\dfrac{13}{125}=\dfrac{13}{5^{3}}$

Multiplying and dividing by

$2^{3}$ ,

$=\dfrac{13\times2^{3}}{5^{3}\times2^{3}}$

$=\dfrac{13\times8}{2^{3}\times5^{3}}$

$=\dfrac{104}{(10)^{3}}$

$=\dfrac{104}{1000}$

$=0.104$ .

Therefore,

$\dfrac{13}{125}=0.104$ is the terminating decimal.

Write the decimal expansion of the following number which have terminating decimal expansion:
$\dfrac{35}{50}$ .

Given rational number is

$\dfrac{35}{50}$ .

We know,

$\frac{p}{q}$ is terminating if:

$p$ and

$q$ are co-prime and

$q$ is of the form of

$2^{n}\times5^{m}$ , where

$n$ and

$m$ are non-negative integers.

Firstly, we check whether the numerator and denominator are co-prime,

Numerator

$=35=7\times5$

Denominator

$=50=5^{2}\times2^{1}$

$\Rightarrow 35$ and

$50$ have a common factor

$5$

Therefore, divide

$35$ and

$50$ by

$5$ .

Then we get,

$\dfrac{35}{50}=\dfrac{7}{5\times2}$ , i.e. here

$p$ and

$q$ are co-primes.

Now, we check whether

$q$ is in the form of

$2^{n}\times5^{m}$ ,

Denominator

$=5\times2$

So, the denominator is in the form

$2^{n}\times5^{m}$ , where

$n=1$ amd

$m=1$ .

Thus,

$\dfrac{35}{50}$ is a terminating decimal and

$\dfrac{35}{50}=0.7$ .

Simplify, giving answers with positive index
$(-3)^2(3)^3$

$(-3)^2(3)^3$

It can be written as

$=(-1 \times 3)^2.(3)^3$

So we get

$=(-1)^2. 3^{2 + 3}$

$=1. 3^5$

$=3^5$

Which of the following are not rational numbers :
(i) $-3$
(ii) $0$

(iii) $\dfrac{0}{4}$

(iv) $\dfrac{8}{0}$

(v) $\dfrac{0}{0}$

(i)

$-3 = \dfrac{-3}{1}$ is a rrational number.

(ii)

$0 = \dfrac{0}{1}$ is a rational number.

(iii)

$\dfrac{0}{4}$ is a rational number.

(iv)

$\dfrac{8}{0}$ is not a rational number.

(v)

$\dfrac{0}{0}$ is not a rational number as both numerator and denominator are zero.

Simplify, giving answers with positive index
$(-4x)(-5x^2)$

$(-4x)(-5x^2)$

It can be written as

$=(-1 \times 4 \times x). (-1 \times 5 \times x^2)$

On further calculation

$=(-1 \times 4 \times x). (-1 \times 5 x^2)$

So we get

$= -1 \times -1 \times 4 \times 5 \times x^{1 + 2}$

Here

$=-1^{1 + 1}. 4^1. 5^1 x^3$

$=20x^3$

$(-6)^{0}$

$(-6)^{0}=1(a^{0}=1)$

Simplify, giving answers with positive index:
$(-2)^2 \times (0)^3 \times (3)^3$

$(-2)^2 \times (0)^3 \times (3)^3$

It can be written as

$4 \times 0 \times 27$

On further calculation

$=0$

Evaluate
$8^{0}+4^{0}+2^{0}$

$8^{0}+4^{0}+2^{0}$ = 1 + 1 +1 = 3 (

$a^{0}=1)$

Evaluate
$8^3 \times 8^{-5}\times 8^4$

$8^3 \times 8^{-5}\times 8^4$
It can be written as

$=8^{3 + 4 - 5}$
On further calculations

$=8^{7 - 5}$
so we get

$=8^2=64$

Express the following rational numbers in the standard form.

$\dfrac{-7}{-20}$

Here a rational number is in standard form if its denominator is positive in lowest term.

$\dfrac{-7}{-20} = \dfrac{7}{20}$

Simplify giving answers with positive index
$(-a)^5 (a^2)$

$(-a)^5 (a^2)$

It can be written as

$= (-1 \times a)^5 (a^2)$

On further calculations

$= (-1)^5 \times a^{5 + 2}$

So we get

$=-1 \times a^7$

$=-a^7$

Rationalize the denominator.
$\dfrac{2}{3\sqrt{7}}$

$\dfrac{2}{3\sqrt{7}}$

$=\dfrac{2}{3\sqrt{7}}\dfrac{\sqrt{7}}{\sqrt{7}}=\dfrac{2\sqrt{7}}{3\times 7}=\dfrac{2\sqrt{7}}{21}$

Insert two irrational numbers between $5$ and $6.$

$\textbf{Step -1: Calculating the two irrational numbers }$

$\text{We need to find two irrational numbers between 5 and 6}$

$\text{Since 5 and 6 are consecutive numbers the root of every}$

$\text{number between their squares is irrational}$

$\implies 25\leq x^2\leq 36$

$\implies x^2 \text{ can take the values 30 and 33}$

$\implies x=\sqrt{30},\sqrt{33}$

$\textbf{Hence, Two irrational numbers between 5 and 6 are }\mathbf{\sqrt{30},\sqrt{33} .}$

Rationalize the denominator.
$\dfrac{1}{3\sqrt{5}+2\sqrt{2}}$

$\dfrac{1}{3\sqrt{5}+2\sqrt{2}}$

$=\dfrac{1}{(3\sqrt{5}+2\sqrt{2})}\times\dfrac{(3\sqrt{5}-2\sqrt{2})}{(3\sqrt{5}-2\sqrt{2})}$

$=\dfrac{(3\sqrt{5}-2\sqrt{2})}{(3\sqrt{5})^2-(2\sqrt{2})^2} \, \, \,...[(a+b)(a-b)=a^2-b^2]$

$=\dfrac{(3\sqrt{5}-2\sqrt{2})}{45-8}$

$=\dfrac{3\sqrt{5}-2\sqrt{2}}{37}$

Write a pair of irrational numbers whose difference is rational.

$\sqrt{5}-3$ and

$\sqrt{5}+3$ are irrational numbers whose difference is rational.

$(\sqrt{5}-3)-(\sqrt{5}+3)=\sqrt{5}-3-\sqrt{5}-3=-6$
Here, the resultant is rational.

Rationalize the denominator.
$\dfrac{1}{\sqrt{5}}$

$\dfrac{1}{\sqrt{5}}=\dfrac{1}{\sqrt{5}}\times \dfrac{\sqrt{5}}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}$

Given Universal set is
$\left\{-6,-5 \dfrac{3}{4},-\sqrt{4},-\dfrac{3}{5},-\dfrac{3}{8}, 0, \dfrac{4}{5}, 1,1 \dfrac{2}{3}, \sqrt{8}, 3.01, \pi, 8.47\right\}$
From the given set, find:
Set of irrational numbers

Rational numbers are numbers of the form

$\dfrac{p}{q},$ where

$q \neq 0$

$U=\left\{-6,-5 \dfrac{3}{4},-\sqrt{4},-\dfrac{3}{5},-\dfrac{3}{8}, 0, \dfrac{4}{5}, 1,1 \dfrac{2}{3}, \sqrt{8}, 3.01, \quad r, 8.47\right\}$
Clearly,

$-5 \dfrac{3}{4},-\dfrac{3}{5},-\dfrac{3}{8}, \dfrac{4}{5}$ and

$1 \dfrac{2}{3}$ are of the form

$\dfrac{p}{q}$
Hence, they are rational numbers.

since the set of integers is a subset of rational numbers,-6,0 and 1 are also rational numbers. Thus, decimal numbers 3.01 and 8.47 are also rational numbersbecause they are terminating decimals.

Hence, from the above set, the set of rational numbers is

$Q,$ and

$Q=\left\{-6,-5 \dfrac{3}{4},-\dfrac{3}{5},-\dfrac{3}{8}, 0, \dfrac{4}{5}, 1,1 \dfrac{2}{3}, 3.01,8.47\right\}$

We need to find the set of irrational numbers. Irrational numbers are numbers that are not rational.

From above, the set of rational numbers is Q and

$Q=\left\{-6,-5 \dfrac{3}{4},-\dfrac{3}{5},-\dfrac{3}{8}, 0, \dfrac{4}{5}, 1,1 \dfrac{2}{3}, 3.01,8.47\right\}$
The set of irrational numbers is the set of complement of the rational numbers over real numbers.
Here the set of irrational numbers is

$U-Q=\{\sqrt{8}, \pi\}$

Show that $5+\sqrt{7}$ is an irrational number.

Let us assume that

$5+\sqrt{7}$ is a rational number.

Then

$5 + \sqrt{7}=\dfrac{p}{q}$ , where

$p$ and

$q$ are two integers and

$q \neq 0$

$\Rightarrow \sqrt{7}=\dfrac{p}{q}-5=\dfrac{p-5q}{q}$

Since,

$p$ ,

$q$ and

$5$ are integers, so

$\dfrac{p-5q}{q}$ is a rational number.

Therefore

$\sqrt{7}$ is also a rational number.

But this contradicts the fact that

$\sqrt{7}$ is an irrational number.

This contradiction has arisen due to our assumption that

$5+\sqrt{7}$ is a rational number.

Hence,

$5+\sqrt{7}$ is an irrational number.

Are the square roots of all positive integers irrational ? If not, give an example of the square root of a number that is a rational number.

Square root of all positive integers is not irrational number.
E.g.

$\sqrt{4} = 2$ Rational number.

$\sqrt{9} = 3$ Rational number.

Classify the following numbers as rational or irrational:
$\sqrt{225}$

$\sqrt{225}=\sqrt{15^2}=15$

$15$ is a rational number.

$\therefore \sqrt{225}$ is a rational number.

Rationalize the denominator.
$\dfrac{12}{4\sqrt{3}-\sqrt{2}}$

$\dfrac{12}{4\sqrt{3}-\sqrt{2}}$

$=\dfrac{12}{4\sqrt{3}-\sqrt{2}}\times \dfrac{4\sqrt{3}+\sqrt{2}}{4\sqrt{3}+\sqrt{2}}$

$=\dfrac{12(4\sqrt{3}+\sqrt{2})}{(4\sqrt{3})^2-(\sqrt{2})^2} \, \, \,...[(a+b)(a-b)=a^2-b^2]$

$=\dfrac{12(4\sqrt{3}+\sqrt{2})}{48-2}$

$=\dfrac{12(4\sqrt{3}+\sqrt{2})}{46}$

$=\dfrac{6(4\sqrt{3}+\sqrt{2})}{23}$

Write the conjugates of the binomial surd $3\sqrt {7} + 7\sqrt {3}$

We know that the when sum of two terms and the difference of the same two terms are multiplied, the product is always a rational number.

Let us apply this concept to a binomial surd

$(3\sqrt { 7 } +7\sqrt { 3 })$ .

When we multiply this with the difference of the same two terms, that is, with

$(3\sqrt { 7 } -7\sqrt { 3 })$ , the product is:

$(3\sqrt { 7 } +7\sqrt { 3 } )(3\sqrt { 7 } -7\sqrt { 3 } )=\left( 3\sqrt { 7 } \right) ^{ 2 }-\left( 7\sqrt { 3 } \right) ^{ 2 }=(9\times 7)-(49\times 3)=63-147=-84\quad \quad (∵a^{ 2 }-b^{ 2 }=(a+b)(a-b))$

Since

$-84$ is a rational number.

Hence,

$(3\sqrt { 7 } -7\sqrt { 3 })$ is the conjugate of

$(3\sqrt { 7 } +7\sqrt { 3 })$ .

Write the conjugates of binomial surd given as $\sqrt {a} + \sqrt {b}$

We know that the when sum of two terms and the difference of the same two terms are multiplied, the product is always a rational number.

Let us apply this concept to a binomial surd

$(\sqrt { a } +\sqrt { b })$ .

When we multiply this with the difference of the same two terms, that is, with

$(\sqrt { a } -\sqrt { b })$ , the product is:

$(\sqrt { a } +\sqrt { b } )(\sqrt { a } -\sqrt { b } )=\left( \sqrt { a } \right) ^{ 2 }-\left( \sqrt { b } \right) ^{ 2 }=a-b\quad \quad (∵a^{ 2 }-b^{ 2 }=(a+b)(a-b))$

Since

$a-b$ is a rational number.

Hence,

$(\sqrt { a } -\sqrt { b })$ is the conjugate of

$(\sqrt { a } +\sqrt { b })$ .

Prove that $\sqrt{2} + \sqrt{5}$ is irrational

1941827_559961_ans_0d9c216ba1ab458ca6e3f48ec32fbf48.jpg

Find 3 irrational numbers between $\dfrac{2}{3}$ and $\dfrac{3}{4}$ using their decimal expansion.

The decimal expansions of

$\dfrac {2}{3}$ and

$\dfrac {3}{4}$ are as follows:

Here, the one decimal expansions are non-terminating recurring, while the other is terminating.

Hence,

$\frac {2} {3}$ and

$\frac {3}{4}$ are two rational numbers.

We know, between any two rational numbers, there are infinitely many irrational numbers.

An irrational number has non-terminating non-recurring decimal expansions.

Then

$3$ different irrational number between

$\dfrac{2}{3}$ and

$\dfrac{3}{4}$ can be:

$a=0.6707007000$

$b=0.6808008000$

$c=0.6909009000$

Here we observe that

$\dfrac {2}{3}<a<b<c<\dfrac {3}{4}$ .

NOTE:

This is

$3$ of the various possible answers. It might have different solution.

Why is 0.111222333444..., where each number appears 3 times in a row irrational?

Since in

$0.111222333444....$ each number appears 3 times in a row is a non- terminating and non- recurring decimal expansion.

Hence,

$0.111222333444....$ is irrational.

If $x = \sqrt {7} - \sqrt {5}, y = \sqrt {5} - \sqrt {3}, z = \sqrt {3} - \sqrt {7}$ , then find the value of $x^{3} + y^{3} + z^{3} - 2xyz$

1608318_516714_ans_11342e1e55964776aebc947d8f2bbefb.jpg

Write the conjugates of the binomial surd $10\sqrt {2} + 3\sqrt {5}$

We know that the when sum of two terms and the difference of the same two terms are multiplied, the product is always a rational number.

Let us apply this concept to a binomial surd

$(10\sqrt { 2 } +3\sqrt { 5 })$ .

When we multiply this with the difference of the same two terms, that is, with

$(10\sqrt { 2 } -3\sqrt { 5 })$ , the product is:

$(10\sqrt { 2 } +3\sqrt { 5 } )(10\sqrt { 2 } -3\sqrt { 5 } )=\left( 10\sqrt { 2 } \right) ^{ 2 }-\left( 3\sqrt { 5 } \right) ^{ 2 }\\=(10\times 2)-(3\times 5)=20-15=5\quad (∵a^{ 2 }-b^{ 2 }=(a+b)(a-b))$

Since

$5$ is a rational number.

Hence,

$(10\sqrt { 2 } -3\sqrt { 5 })$ is the conjugate of

$(10\sqrt { 2 } +3\sqrt { 5 })$ .

Write the conjugates of the binomial surds $x + 3\sqrt {y}$

We know that the when sum of two terms and the difference of the same two terms are multiplied, the product is always a rational number.

Let us apply this concept to a binomial surd

$(x +3\sqrt { y })$ .

When we multiply this with the difference of the same two terms, that is, with

$(x -3\sqrt { y })$ , the product is:

$(x+3\sqrt { y } )(x-3\sqrt { y } )=\left( x \right) ^{ 2 }-\left( 3\sqrt { y } \right) ^{ 2 }=x^{ 2 }-9y\quad \quad (∵a^{ 2 }-b^{ 2 }=(a+b)(a-b))$

Since

$x^2-9y$ is a rational number.

Hence,

$(x -3\sqrt { y })$ is the conjugate of

$(x +3\sqrt { y })$ .

Write the conjugates of the binomial surd $\sqrt {8} - 5$

We know that the when sum of two terms and the difference of the same two terms are multiplied, the product is always a rational number.

Let us apply this concept to a binomial surd

$(\sqrt { 8 }-5)$ .

When we multiply this with the sum of the same two terms, that is, with

$(\sqrt { 8 }+5)$ , the product is:

$(\sqrt { 8 } -5)(\sqrt { 8 } +5)=\left( \sqrt { 8 } \right) ^{ 2 }-\left( 5 \right) ^{ 2 }=8-25=-17\quad \quad (∵a^{ 2 }-b^{ 2 }=(a+b)(a-b))$

Since

$-17$ is a rational number.

Hence,

$(\sqrt { 8 }+5)$ is the conjugate of

$(\sqrt { 8 }-5)$ .

Write the conjugates of the binomial surds given as $\sqrt {x} - 2\sqrt {y}$

We know that the when sum of two terms and the difference of the same two terms are multiplied, the product is always a rational number.

Let us apply this concept to a binomial surd

$(\sqrt { x } -2\sqrt { y })$ .

When we multiply this with the sum of the same two terms, that is, with

$(\sqrt { x } +2\sqrt { y })$ , the product is:

$(\sqrt { x } -2\sqrt { y } )(\sqrt { x } +2\sqrt { y } )=\left( \sqrt { x } \right) ^{ 2 }-\left( 2\sqrt { y } \right) ^{ 2 }=x-4y\quad \quad (∵a^{ 2 }-b^{ 2 }=(a+b)(a-b))$

Since

$x-4y$ is a rational number.

Hence,

$(\sqrt { x } +2\sqrt { y })$ is the conjugate of

$(\sqrt { x } -2\sqrt { y })$ .

Write the conjugate of the binomial surd $x\sqrt {a} + y\sqrt {b}$

We know that the when sum of two terms and the difference of the same two terms are multiplied, the product is always a rational number.

Let us apply this concept to a binomial surd

$(x\sqrt { a } +y\sqrt { b })$ .

When we multiply this with the difference of the same two terms, that is, with

$(x\sqrt { a } -y\sqrt { b })$ , the product is:

$\left( x\sqrt { a } +y\sqrt { b } \right) \left( x\sqrt { a } -y\sqrt { b } \right) =\left( x\sqrt { a } \right) ^{ 2 }-\left( y\sqrt { b } \right) ^{ 2 }=ax^{ 2 }-by^{ 2 }\quad \quad (∵a^{ 2 }-b^{ 2 }=(a+b)(a-b))$

Since

$ax^2-by^2$ is a rational number.

Hence,

$(x\sqrt { a } -y\sqrt { b })$ is the conjugate of

$(x\sqrt { a } +y\sqrt { b })$ .

${\dfrac{\sqrt 3 - 1} {\sqrt 3 + 1}} = a + b\sqrt 3$ find $a$ and $b$

$\cfrac { \sqrt { 3 } -1 }{ \sqrt { 3 } +1 } \times\cfrac { \sqrt { 3 } -1 }{ \sqrt { 3 } -1 } \\ =\cfrac { 3+1-2\sqrt { 3 } }{ 3-1 } \\ =\cfrac { 4-2\sqrt { 3 } }{ 2 } =2-\sqrt { 3 } \\ a=2,\\ b=-1$

If $x=\sqrt{2}+1$ . Find the value of $x+\large{\frac{1}{x}}$ .

Given

$x=\sqrt{2}+1$

$\therefore \large{\frac{1}{x}}=\dfrac{1}{\sqrt{2}+1}$

$=\large{\dfrac{\sqrt{2}-1}{(\sqrt{2})^2-1^2}}$ {Using method of Rationalisation).

$=\large{\dfrac{\sqrt{2}-1}{2-1}}=$

$\sqrt{2}-1$

$\therefore x+\large{\dfrac{1}{x}}$

$=(\sqrt{2}+1+\sqrt{2}-1)=2\sqrt{2}$

Write the conjugates of the binomial surd $\dfrac {1}{2}x + \dfrac {1}{2}\sqrt {y}$

We know that the when sum of two terms and the difference of the same two terms are multiplied, the product is always a rational number.

Let us apply this concept to a binomial surd

$\left( \frac { 1 }{ 2 } x+\frac { 1 }{ 2 } \sqrt { y } \right)$ .

When we multiply this with the difference of the same two terms, that is, with

$\left( \frac { 1 }{ 2 } x-\frac { 1 }{ 2 } \sqrt { y } \right)$ , the product is:

$\left( \frac { 1 }{ 2 } x+\frac { 1 }{ 2 } \sqrt { y } \right) \left( \frac { 1 }{ 2 } x-\frac { 1 }{ 2 } \sqrt { y } \right) =\left( \frac { 1 }{ 2 } x \right) ^{ 2 }-\left( \frac { 1 }{ 2 } \sqrt { y } \right) ^{ 2 }=\frac { 1 }{ 4 } x^{ 2 }-\frac { 1 }{ 4 } y=\frac { 1 }{ 4 } (x^{ 2 }-y)\quad \quad (∵a^{ 2 }-b^{ 2 }=(a+b)(a-b))$

Since

$\frac {1}{4}(x^2-y)$ is a rational number.

Hence,

$\left( \frac { 1 }{ 2 } x-\frac { 1 }{ 2 } \sqrt { y } \right)$ is the conjugate of

$\left( \frac { 1 }{ 2 } x+\frac { 1 }{ 2 } \sqrt { y } \right)$ .

Locate $\sqrt {10}$ on number line

We can do it by using Pythagoras Theorem.

We can write

$\sqrt { 10 }$ as follows:

$\sqrt { 10 } =\sqrt { 9+1 } \\ \Rightarrow \sqrt { 10 } =\sqrt { { 3 }^{ 2 }+1^2 }$

The construction steps are as follows:

1. Take a line segment

$AO = 3$ unit on the

$x$ -axis. (consider

$1$ unit =

$2$ cm)

2. Draw a perpendicular on

$O$ and draw a line

$OC = 1$ unit

3. Now join

$AC$ with

$\sqrt { 10 }$ .

4. Take

$A$ as center and

$AC$ as radius, draw an arc which cuts the

$x$ -axis at point

$E$ .

5. The line segment

$AC$ in the below diagram represents

$\sqrt { 10 }$ units.

Find a point corresponding to $3+\sqrt{2}$ on the number line.

First plotting

$\sqrt{2}$ on number line and then shift the origin

$3$ units to the left will, get the point

$(3+\sqrt{2})$ on number ilne.

Steps:-

i) Draw AB of

$1$ - unit,

ii) draw BC of length

$1$ unit perpendicular to AB from B,

ii) take AC as radius and draw and arc which cuts horizontal line in D,

iv) thus distance AD=

$\sqrt{2}$ ,

v) distance

$OD=3+\sqrt{2}$ units.

Write the conjugate of the binomial surd $xy\sqrt {z} + yz\sqrt {x}$

We know that the when sum of two terms and the difference of the same two terms are multiplied, the product is always a rational number.

Let us apply this concept to a binomial surd

$(xy\sqrt { z } +yz\sqrt { x })$ .

When we multiply this with the difference of the same two terms, that is, with

$(xy\sqrt { z } -yz\sqrt { x })$ , the product is:

$\left( xy\sqrt { z } +yz\sqrt { x } \right) \left( xy\sqrt { z } -yz\sqrt { x } \right) =\left( xy\sqrt { z } \right) ^{ 2 }-\left( yz\sqrt { x } \right) ^{ 2 }=zx^{ 2 }y^{ 2 }-xz^{ 2 }y^{ 2 }\quad \quad (∵a^{ 2 }-b^{ 2 }=(a+b)(a-b))$

Since

$zx^2y^2-xz^2y^2$ is a rational number.

Hence,

$(xy\sqrt { z } -yz\sqrt { x })$ is the conjugate of

$(xy\sqrt { z } +yz\sqrt { x })$ .

Write the conjugates of the binomial surd $\dfrac {1}{2} + \sqrt {2}$

We know that the when sum of two terms and the difference of the same two terms are multiplied, the product is always a rational number.

Let us apply this concept to a binomial surd

$\left( \frac { 1 }{ 2 } +\sqrt { 2 } \right)$ .

When we multiply this with the difference of the same two terms, that is, with

$\left( \frac { 1 }{ 2 } -\sqrt { 2 } \right)$ , the product is:

$\left( \frac { 1 }{ 2 } +\sqrt { 2 } \right) \left( \frac { 1 }{ 2 } -\sqrt { 2 } \right) =\left( \frac { 1 }{ 2 } \right) ^{ 2 }-\left( \sqrt { 2 } \right) ^{ 2 }=\frac { 1 }{ 4 } -2=\frac { 1-8 }{ 4 } =-\frac { 7 }{ 4 } \quad \quad (∵a^{ 2 }-b^{ 2 }=(a+b)(a-b))$

Since

$-\frac {7}{4}$ is a rational number.

Hence,

$\left( \frac { 1 }{ 2 } -\sqrt { 2 } \right)$ is the conjugate of

$\left( \frac { 1 }{ 2 } +\sqrt { 2 } \right)$ .

Explain with an example how irrational numbers differ from rational numbers?

A rational number is any number that can be expressed as the quotient or

fraction

$\dfrac { p }{ q }$ of two integers, a numerator

$p$ and a non-

zero denominator

$q$ . Some of the examples of rational numbers are:

$-\dfrac { 3 }{ 11 } ,\dfrac { 1 }{ 2 } ,\dfrac { 5 }{ 3 }$

On the other hand, an irrational number is a number that can not be

expressed as the quotient or fraction

$\dfrac { p }{ q }$ of two integers, a

numerator

$p$ and a non-zero denominator

$q$ . Some of the examples of irrational numbers are as follows:

$\dfrac { 4 }{ 0 }$ as the denominator is not non-zero.

$\pi=3.141592....$ and

$\sqrt { 2 } =1.414....$

Simplify
(i) $10\sqrt{2}-2\sqrt{2}+4\sqrt{32}$
(ii) $\sqrt{48}-3\sqrt{72}-\sqrt{27}+5\sqrt{18}$
(iii) $\sqrt[3]{16}+8\sqrt[3]{54}-\sqrt[3]{128}$ .

(i)

$10\sqrt{2}-2\sqrt{2}+4\sqrt{32}$

$=10\sqrt{2}-2\sqrt{2}+4\sqrt{16\times 2}$

$=10\sqrt{2}-2\sqrt{2}+4\times 4\times \sqrt{2}$

$=(10-2+16)\sqrt{2}=24\sqrt{2}$
(ii)

$\sqrt{48}-3\sqrt{72}-\sqrt{27}+5\sqrt{18}$

$=\sqrt{16\times 3}-3\sqrt{36\times 2}-\sqrt{9\times 3}+5\sqrt{9\times 2}$

$=\sqrt{16}\sqrt{3}-3\sqrt{36}\sqrt{2}=\sqrt{9}\sqrt{3}+5\sqrt{9}\sqrt{2}$

$=4\sqrt{3}-18\sqrt{2}-3\sqrt{3}+15\sqrt{2}$

$=(-18+15)\sqrt{2}+(4-3)\sqrt{3}=-3\sqrt{2}+\sqrt{3}$
(iii)

$\sqrt[3]{16}+8\sqrt[3]{54}-\sqrt[3]{128}$

$=\sqrt[3]{8\times 2}+8\sqrt[3]{27\times 2}-\sqrt[3]{64\times 2}$

$=\sqrt[3]{8}\sqrt[3]{2}+8\sqrt[3]{27}\sqrt[3]{2}-\sqrt[3]{64}\sqrt[3]{2}$

$=2\sqrt[3]{2}+8\times 3\times \sqrt[3]{2}-4\sqrt[3]{2}$

$=2\sqrt[3]{2}+24\sqrt[3]{2}-4\sqrt[3]{2}$

$=(2+24-4)\sqrt[3]{2}=22\sqrt[3]{2}$

Find $a^{3}$ if ${ \left( \dfrac { 2 }{ 7 } \right) }^{ a }={ \left( \dfrac { 16 }{ 21 } \right) }^{ -5 }\times { \left( \dfrac { 3 }{ 8 } \right) }^{ -5 }$

$\begin{array}{l} { \left( { \dfrac { 2 }{ 7 } } \right) ^{ a } }={ \left( { \dfrac { { 16 } }{ { 21 } } } \right) ^{ -5 } }{ \left( { \dfrac { 3 }{ 8 } } \right) ^{ -5 } } \\ { \left( { \dfrac { 2 }{ 7 } } \right) ^{ a } }={ \left( { \dfrac { 2 }{ 7 } } \right) ^{ -5 } } \\ { \left( { \dfrac { 2 }{ 7 } } \right) ^{ a+5 } }={ \left( { \dfrac { 2 }{ 7 } } \right) ^{ 0 } } \\ \therefore a+5=0 \\ a=-5 \\ Hence,\, \, { a^{ 3 } }={ \left( { -5 } \right) ^{ 3 } }=-125. \end{array}$

Solve:
$\dfrac{(-3)^{3}\times (-2)^{5}\times 7^{2}}{(-1)^{5}\times (-7)^{2}\times 2^{5}}$

$\dfrac { { (-3) }^{ 3 }\ast { (-2) }^{ 5 }\ast { 7 }^{ 2 } }{ { (-1) }^{ 5 }\ast { (-7) }^{ 2 }\ast { 2 }^{ 5 } } =\dfrac { (−3\ast -3\ast -3)×(−2\ast -2\ast -2\ast -2\ast -2)×(7\ast 7) }{ (−1\ast -1\ast -1\ast -1\ast -1)×(−7\ast -7)×(2\ast 2\ast 2\ast 2\ast 2) }$

$=\dfrac { (-27)\ast (-32)\ast (49) }{ (-1)\ast 49\ast 32 } =-27$

Find the value of $8\sqrt{60}\times 5\sqrt{3}$

1310799_1034727_ans_2490626eba194353981e0c85f3dd0383.jpeg

Find the value of $9\sqrt{5}\times 20\sqrt{45}$

1310803_1034729_ans_42b3f61cde0045f2958d2d55ea4262a4.jpeg

Solve:
$x^{11}\div x^{15}$

$\begin{array}{l} { x^{ 11 } }\div { x^{ 15 } } \\ \Rightarrow { x^{ 11-15 } }={ x^{ -4 } }=\frac { 1 }{ { { x^{ 4 } } } } \, \, \, \, \, \, \left[ { { a^{ -1 } }=\frac { 1 }{ a } } \right] \end{array}$

$\cot { \left( \dfrac { 15 }{ 2 } \right) } =\sqrt { 2 } +\sqrt { 3 } +\sqrt { 4 } +\sqrt { 6 }$

$\begin{array}{l} \cot \left( { \frac { { 15 } }{ 2 } } \right) =\frac { { 2\times \cos \frac { { 15 } }{ 2 } \times \left( { \cos \frac { { 15 } }{ 2 } } \right) } }{ { 2\times \sin \frac { { 15 } }{ 2 } \times \left( { \cos \frac { { 15 } }{ 2 } } \right) } } \\ =\frac { { 2{ { \cos }^{ 2 } }\left( { \frac { { 15 } }{ 2 } } \right) } }{ { \sin 15 } } \\ =\frac { { 1+\cos \left( { 45-30 } \right) } }{ { \sin \left( { 45-30 } \right) } } \\ =\frac { { 1+\cos 45.\cos 30+\sin 45.\sin 30 } }{ { \sin 45.\cos 30-\cos 45\sin 30 } } \\ =\frac { { 1+\frac { 1 }{ { \sqrt { 2 } } } \times \frac { { \sqrt { 3 } } }{ 2 } +\frac { 1 }{ { \sqrt { 2 } } } \times \frac { 1 }{ 2 } } }{ { \frac { 1 }{ { \sqrt { 2 } } } \times \frac { { \sqrt { 3 } } }{ 2 } -\frac { 1 }{ { \sqrt { 2 } } } \times \frac { 1 }{ 2 } } } \\ =\frac { { 2\sqrt { 2 } +\sqrt { 3 } +1 } }{ { \sqrt { 3 } -1 } } \\ =\frac { { \left( { 1+\sqrt { 3 } +2\sqrt { 2 } } \right) \left( { \sqrt { 3 } +1 } \right) } }{ { \left( { \sqrt { 3 } -1 } \right) \left( { \sqrt { 3 } +1 } \right) } } \\ =\frac { { \sqrt { 3 } +3+2\sqrt { 6 } +1+\sqrt { 3 } +2\sqrt { 2 } } }{ { 3-1 } } \\ =\frac { { 4+2\sqrt { 3 } +2\sqrt { 6 } +2\sqrt { 2 } } }{ 2 } \\ =\sqrt { 2 } +\sqrt { 3 } +\sqrt { 4 } +\sqrt { 6 } \end{array}$

Carry out the following additions of rational numbers.
$1\ \dfrac {2}{3}+2\ \dfrac {4}{5}$

$= 4\dfrac{7}{{15}}$

Simplifying $1\dfrac{2}{3} + 2\dfrac{4}{5}$ .

$1\dfrac{2}{3} + 2\dfrac{4}{5} = \dfrac{5}{3} + \dfrac{{14}}{5}$

Taking the LCM,

$\dfrac{5}{3} + \dfrac{{14}}{5} = \dfrac{{5\left( 5 \right) + 14\left( 3 \right)}}{{15}}$

$= \dfrac{{67}}{{15}}$

$= 4\dfrac{7}{{15}}$

Find the number of digits in the numeral for $(875)^{16}$

$875=25\times 35$

$=5^{2}\times 5\times 7$

$=5^{3}\times 7$

$(875)^{16}=(5^{3}\times 7)^{16}$

$=(5^{3})^{16}\times 7^{16}$

$=5^{48}\times 7^{16}$

Taking

$log_{10}$

$log(875)^{16}=log_{10}5^{48}+log_{10}7^{16}$

$\therefore log (ab)= log_{10}a+log_{10}b$

$= 48 log_{10}5+16log_{10}7$

$= 48log_{10}(\frac{10}{2})+16log_{10}7$

$(\because log_{10}10=1)$

$=48[1-log2]+16 log 7$

$=16[3(1-0.3010)+0.845]$

$16\times [3\times 0.699+0.845]$

$=47.072$

$\simeq 48$ digits.

1226854_1298290_ans_0eb6a9eee9834fbcae6f4e844e1b2d36.jpg

Simplify and express each of the following as a rational number: $\dfrac {3^{5}\times 25\times 10^{5}}{5^{7}\times 6^{5}}$ .

2039273_1145118_ans_f4cbb1b561264cea86bae8e1490af9f9.jpg

Find the smallest and greatest among $\sqrt{7}-\sqrt{5}$ and $\sqrt{8}-\sqrt{6}$

R.F of

$\sqrt{7}-\sqrt{5}$ is

$\sqrt{7}+\sqrt{5}$

$\therefore \sqrt{7}-\sqrt{5}=\sqrt{7}-\sqrt{5}\times\dfrac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}$

$=\dfrac{7-5}{\sqrt{7}+\sqrt{5}}$

$=\dfrac{2}{\sqrt{7}+\sqrt{5}}$

R.F of

$\sqrt{8}-\sqrt{6}$ is

$\sqrt{8}+\sqrt{6}$

$\therefore \sqrt{8}-\sqrt{6}=\sqrt{8}-\sqrt{6}\times\dfrac{\sqrt{8}+\sqrt{6}}{\sqrt{8}+\sqrt{6}}$

$=\dfrac{8-6}{\sqrt{8}+\sqrt{6}}$

$=\dfrac{2}{\sqrt{8}+\sqrt{6}}$

$\sqrt{7}+\sqrt{5}<\sqrt{8}+\sqrt{6}$

$\Rightarrow \dfrac{1}{\sqrt{7}+\sqrt{5}}>\dfrac{1}{\sqrt{8}+\sqrt{6}}$

$\Rightarrow \dfrac{2}{\sqrt{7}+\sqrt{5}}>\dfrac{2}{\sqrt{8}+\sqrt{6}}$

$\Rightarrow \sqrt{7}-\sqrt{5}>\sqrt{8}-\sqrt{6}$

$\therefore \sqrt{8}-\sqrt{6}$ is smaller than

$\sqrt{7}-\sqrt{5}$

Find the Rationalising factor of $\sqrt[3]{3}+\sqrt[3]{2}$

$\sqrt[3]{3}+\sqrt[3]{2}$

We know that R.F of

$\sqrt[3]{a}+\sqrt[3]{b}=\sqrt[3]{{a}^{2}}-\sqrt[3]{ab}+\sqrt[3]{{b}^{2}}$

$=\sqrt[3]{{3}^{2}}-\sqrt[3]{3\times 2}+\sqrt[3]{{2}^{2}}$

$=\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}$

Simplify:
$2b^{6}.b^{3}.5b^{4}$

$\begin{array}{l} 2{ b^{ 6 } }\cdot { b^{ 3 } }\cdot 5{ b^{ 4 } }\, \, \, \, \, \left[ { { b^{ x } }\cdot { b^{ y } }={ b^{ x+y } } } \right] \\ 10{ b^{ 13 } } \end{array}$

What can you say about the product of a non-zero rational and irrational number?

The product of a non zero rational and irrational number is always an irratonal number.

For example:-

$2\times\sqrt2=2\sqrt2$

Here,

$2$ is a non-zero rational number and

$\sqrt2$ is an irrational number but the product is an irrational number.

Express $\cfrac {2157}{625}$ in decimal form and state what it is terminating or not?

2044104_1562200_ans_1f5cd043b5f84bc792a3c3c5f41fdaa7.jpg

$\sqrt{6}.\sqrt{3}$ =_______

$\sqrt{6}.\sqrt{3}$ =

$\sqrt{18}$ =

$3\sqrt{2}$

Identify the following as rational or irrational number. Give the decimal representation rational number.
$\sqrt {\dfrac {9}{27}}$ .

1961079_1602709_ans_11a43477e9ad4cc78c5e01bc8cd720fe.jpg

The longest $(\sqrt{5}-\sqrt{2})$ $(\sqrt{5}+\sqrt{2})$

1324765_1587034_ans_f93a991cb43f4c05b0d81fd015e2b9e5.jpg

Find the value of 'a' and 'b' $\dfrac { 7+\sqrt { 5 } }{ 7-\sqrt { 5 } } -\dfrac { 7-\sqrt { 5 } }{ 7+\sqrt { 5 } } =a+\dfrac { 7 }{ 11 } \sqrt { 5 } b$

1869494_1529115_ans_07e1f86c4d1d476d88dc66b830e2b76f.jpg

Find which of the variables $x, y, z$ and $u$ represent rational numbers and which irrational numbers:
$y^{2} = 9$ .

$y^2 = 9$

$y = \sqrt{9}$

$y = \pm 3$

$\rightarrow$ The number is integer

$\rightarrow y$ is rational number

1412289_1668345_ans_bf421e0a172f4f34bda702d03fd25a10.jpg

Rationalise the denominator of the following.
$\dfrac{4}{2+\sqrt{3}+\sqrt{7}}$ .

1677299_1715540_ans_ed2f9192014943f29b63ad22ee739f62.jpg

Represent $\sqrt{9.3}$ on the number line.

1960823_1868705_ans_9f5a2abb8010453197e11b42a0861ffd.JPG

It being given that $\sqrt{2}=1.414, \sqrt{3}=1.732, \sqrt{5}=2.236$ and $\sqrt{10}=3.162$ , find the value to three places of decimals, of the following.
$\dfrac{\sqrt{10}-\sqrt{5}}{\sqrt{2}}$ .

1673970_1715436_ans_c8665ea0ef554d098d34a466b847cdf0.jpg

Prove that $\sqrt{3}+\sqrt{5}$ is an irrational number.

1870470_1668389_ans_a2a58d06fcf14e379425b150c3ac782e.jpg

Find three rational number lying between $\dfrac{3}{5}$ and $\dfrac{7}{8}$ . How many rational numbers can be determined between these two numbers?

There will be infinite rational numbers lying between given two rational numbers
1961108_1714912_ans_eebebc3223fd4433916a5d0c28d1e1fa.jpg

1961108_1714912_ans_eebebc3223fd4433916a5d0c28d1e1fa.jpg

Rationalise the denominator of the following.
$\dfrac{3}{\sqrt{3}+\sqrt{5}-\sqrt{2}}$ .

1676872_1715538_ans_67cc3821da5e4f4d963fe8e1388d93f4.jpg

If $p=\dfrac{3-\sqrt{5}}{3+\sqrt{5}}$ and $q=\dfrac{3+\sqrt{5}}{3-\sqrt{5}}$ , find the value of $p^2+q^2$ .

1676831_1715535_ans_f6bf1760b9e445c2ace7f29923148d56.jpg

Number Systems - Class 9 Maths - Extra Questions

Following are the five rational numbers which are smaller than 22 ⇒\Rightarrow 1,12,0,−1,−121,\dfrac{1}{2},0,-1,\dfrac{-1}{2}.If true then enter 11 and if false then enter 00.

y2=9y^{2}=9

Simplify and express the result in power notation with positive exponent.(−3)4×(53)4(-3)^4 \times \left ( \dfrac{5}{3} \right )^4 is (m)4(m)^4mm is

Find the value of (31+41+51)0(3^{1}+4^{1}+5^{1})^{0}.

Find the value of.(30+4−1)×22(3^0+4^{-1}) \times 2^2

Some rational numbers are integers, if true then enter 11 and if false then enter 00

5√3+2√3\sqrt{3}+2\sqrt{3} = 7√6\sqrt{6}enter 11 for true and 00 for false

Is zero a rational number?

Zero can be written in the form of pq\displaystyle\ \frac{p}{q}, where p and q are integers and q≠\neq 0?If above statement is true then enter 11 and if false then enter 0.0.

5√3−2√3\sqrt{3} -2\sqrt{3} = 3√3\sqrt{3}enter 11 for true and 00 for false

Rationalise the denominator and simplify:If 2√5+√3\displaystyle \frac{2}{\sqrt{5}+\sqrt{3}} is √m−√3\displaystyle \sqrt{m}-\sqrt{3}, then mm is ?

Rationalise the denominator and simplify:If \displaystyle \frac{3}{4-\sqrt{3}}\displaystyle \frac{3}{4-\sqrt{3}} is \displaystyle \frac{3}{13}(4+\sqrt{m})\displaystyle \frac{3}{13}(4+\sqrt{m}), then mm is ?

Rationalise the denominator and simplify:If \displaystyle \frac{1}{2+\sqrt{3}}\displaystyle \frac{1}{2+\sqrt{3}} is \displaystyle 2-\sqrt{m}\displaystyle 2-\sqrt{m}, then mm is ?

Find the product of 3-\sqrt{7}3-\sqrt{7} and its conjugate.

Find the value-of (3^0 - 4^0)\times5^2(3^0 - 4^0)\times5^2:

Express as rational number \displaystyle \left ( \frac{-5}{7} \right )^{-3}\displaystyle \left ( \frac{-5}{7} \right )^{-3} is \dfrac{-m}{125}\dfrac{-m}{125} Value of mm is

Simplify: \displaystyle \left (\frac{1}{2} \right )^{-2} + \left (\frac{1}{3} \right )^{-2}+ \left (\frac{1}{4} \right )^{-2}\displaystyle \left (\frac{1}{2} \right )^{-2} + \left (\frac{1}{3} \right )^{-2}+ \left (\frac{1}{4} \right )^{-2}

Express as a power of a rational number with positive exponent \displaystyle \left [ \left ( \frac{4}{5} \right )^{-2} \right ]^4\displaystyle \left [ \left ( \frac{4}{5} \right )^{-2} \right ]^4Answer is (\dfrac{5}{4})^m(\dfrac{5}{4})^mValue of mm is

Express as rational number \displaystyle \left ( \frac{-5}{7} \right )^{-1}\displaystyle \left ( \frac{-5}{7} \right )^{-1} is =\displaystyle \frac{-m}{5} =\displaystyle \frac{-m}{5} Value of mm is

Evaluate \displaystyle \left ( \frac{1}{2} \right )^{-5}\displaystyle \left ( \frac{1}{2} \right )^{-5}

If \displaystyle \left \{ \left (\frac{3}{4} \right )^{-1} - \left (\frac{1}{4} \right )^{-1} \right \}=\displaystyle \left \{ \left (\frac{3}{4} \right )^{-1} - \left (\frac{1}{4} \right )^{-1} \right \}= - \dfrac{3}{m}- \dfrac{3}{m}, the the value of mm is

If 4^{-3}4^{-3} is \dfrac{1}{m}\dfrac{1}{m}, then the value of mm is:

Express with positive index \displaystyle x^{ \frac{-1}{2}}\displaystyle x^{ \frac{-1}{2}}

Simplify \displaystyle \left ( - \frac{2}{5} \right )^{-3} \times \left ( - \frac{2}{5} \right )^{4} \displaystyle \left ( - \frac{2}{5} \right )^{-3} \times \left ( - \frac{2}{5} \right )^{4}

Find the value of (3^0 + 4^{-1}) \times 2^2(3^0 + 4^{-1}) \times 2^2

Find the value of \displaystyle \left ( \left ( \frac{4}{7} \right )^{8} \right )^0\displaystyle \left ( \left ( \frac{4}{7} \right )^{8} \right )^0.

Find the value of mm:(2^{-1} \times 4^{-1}) + 2^{-2}=\dfrac{3}{m}(2^{-1} \times 4^{-1}) + 2^{-2}=\dfrac{3}{m}

Simplify \displaystyle \left ( \frac{-3}{4} \right )^{-5} + \left ( \frac{-3}{4} \right )^{-3}\displaystyle \left ( \frac{-3}{4} \right )^{-5} + \left ( \frac{-3}{4} \right )^{-3}

Find the value of mm: \displaystyle \left ( \frac{5}{2} \right )^3 \times \left ( \frac{5}{2} \right )^7 =\left( \dfrac { 5 }{ 2 } \right) ^{ m}\displaystyle \left ( \frac{5}{2} \right )^3 \times \left ( \frac{5}{2} \right )^7 =\left( \dfrac { 5 }{ 2 } \right) ^{ m}

Find (0.04)^{-1.5} = ?(0.04)^{-1.5} = ?

The conjugate of 3 - \sqrt {5 + x}3 - \sqrt {5 + x} is 3 + \sqrt {5 + x}3 + \sqrt {5 + x}. Verify

Verify that the conjugate of \sqrt {3} + \sqrt {2}\sqrt {3} + \sqrt {2} is \sqrt {3} - \sqrt {2}\sqrt {3} - \sqrt {2}

Write five rational numbers which are smaller than 22.

Write five rational numbers which are smaller than \dfrac{5}{6}\dfrac{5}{6}.

Write.(i) The rational number that does not have a reciprocal.(ii) The rational numbers that are equal to their reciprocals.(iii) The rational number that is equal to its negative.

Express the following with positive exponents:{4}^{-7}{4}^{-7}

Find the value of 'xx' such that{(-2)}^{x+1}\times {(-2)}^{7}={(-2)}^{12}{(-2)}^{x+1}\times {(-2)}^{7}={(-2)}^{12}

Write 0, 7, 10, -40, 7, 10, -4 in \dfrac{p}{q}\dfrac{p}{q} form

Can you guess the value of 'xx' when {2}^{x}=1{2}^{x}=1

Express the following with positive exponents:{ \left( \cfrac { 4 }{ 7 } \right) }^{ -3 }{ \left( \cfrac { 4 }{ 7 } \right) }^{ -3 }

Express the following with positive exponents:\dfrac{1}{{(5)}^{-4}}\dfrac{1}{{(5)}^{-4}}

Give one example for the following statement:A number which is rational but not an integer.

Classify the following number as rational or irrational.5 - \sqrt {3}5 - \sqrt {3}

Give one example each to the following statements.A number which is natural number, whole number, integer and rational number

Give one example each to the following statements.An integer which is not a whole number

Give one example each to the following statements.A whole number which is not a natural number

Rationalise the denominator and simplify:\dfrac {\sqrt {3} + \sqrt {2}}{\sqrt {3} - \sqrt {2}}\dfrac {\sqrt {3} + \sqrt {2}}{\sqrt {3} - \sqrt {2}}

Number \dfrac {3}{625}\dfrac {3}{625} is a terminating decimal or a non-terminating repeating decimal? Write it in decimal form.

Rationalise the denominator and simplify:\cfrac { \sqrt { 6 } +\sqrt { 3 } }{ \sqrt { 6 } -\sqrt { 3 } } \cfrac { \sqrt { 6 } +\sqrt { 3 } }{ \sqrt { 6 } -\sqrt { 3 } }

Is zero a rational number? Give reasons for your answer.

Rationalising factor of \left( \sqrt { x+y } \right) \left( \sqrt { x+y } \right) is ____

Find the value of the following:\left (\dfrac {7}{4}\right )^{0}\times 3\left (\dfrac {7}{4}\right )^{0}\times 3

Find x2^{2x \, + \, 2} \, - \, 6^x \, - \, 2 \times 3 ^{2x \, + \, 2} \, = \, 0 2^{2x \, + \, 2} \, - \, 6^x \, - \, 2 \times 3 ^{2x \, + \, 2} \, = \, 0

Solve the following equation:\displaystyle\, \frac{0.2^{x + 0.5}}{\sqrt{5}} = \frac{(0.04)^x}{25}\displaystyle\, \frac{0.2^{x + 0.5}}{\sqrt{5}} = \frac{(0.04)^x}{25}

Solve the following equations.\displaystyle\, \left ( \frac{5}{12} \right )^x \cdot\left ( \frac{6}{5} \right )^{x -1} = (0.3)^{-1}\displaystyle\, \left ( \frac{5}{12} \right )^x \cdot\left ( \frac{6}{5} \right )^{x -1} = (0.3)^{-1}

Is 0.90.9 a rational number?

Solve 17^2 \cdot 17^{-5}17^2 \cdot 17^{-5}

Solve:\dfrac {\sqrt{24}}{8}+\dfrac {\sqrt{54}}{9}\dfrac {\sqrt{24}}{8}+\dfrac {\sqrt{54}}{9}.

Simplify: \dfrac { 25 }{ \sqrt { 5 } }\dfrac { 25 }{ \sqrt { 5 } }

Find a rational number between \dfrac{2}{3}\dfrac{2}{3} and \dfrac{3}{4}\dfrac{3}{4}.

Find four rational numbers between \dfrac{2}{3}\dfrac{2}{3} and \dfrac{3}{5}\dfrac{3}{5}.

Write five rational numbers which are greater than \dfrac{-3}{2}\dfrac{-3}{2}.

Prove that \dfrac{{{{({a^{p + q}})}^2}{{({a^{q + r}})}^2}{{({a^{r + p}})}^2}}}{{{{({a^p}.{a^q}.{a^r})}^4}}} = 1\dfrac{{{{({a^{p + q}})}^2}{{({a^{q + r}})}^2}{{({a^{r + p}})}^2}}}{{{{({a^p}.{a^q}.{a^r})}^4}}} = 1

How many of the following numbers are rational numbers?(i) \dfrac{5}{-8}\dfrac{5}{-8} (ii) \dfrac{-6}{11}\dfrac{-6}{11} (iii) \dfrac{7}{15}\dfrac{7}{15} (iv) -3-3 (v) 00 (vi) \dfrac{1}{0}\dfrac{1}{0}

If \frac{{{{\left( { - 2} \right)}^x} \times {{\left( { - 2} \right)}^7}}}{{3 \times {4^6}}} = \frac{1}{{12}}\frac{{{{\left( { - 2} \right)}^x} \times {{\left( { - 2} \right)}^7}}}{{3 \times {4^6}}} = \frac{1}{{12}}, then the value of x-3 is

Simplify(16)^{\dfrac {-1}{4}}x {^{4}\sqrt {16}}(16)^{\dfrac {-1}{4}}x {^{4}\sqrt {16}}

Evaluate:-{3^3} \times \left( {243} \right)\,{\,^{ - \cfrac{2}{3}}}\,\, \times {9^{\,\, - \cfrac{1}{3}}}{3^3} \times \left( {243} \right)\,{\,^{ - \cfrac{2}{3}}}\,\, \times {9^{\,\, - \cfrac{1}{3}}}

Add 7\sqrt 2 + 5\sqrt 3 7\sqrt 2 + 5\sqrt 3 and \sqrt 2 - 7\sqrt 3 \sqrt 2 - 7\sqrt 3

Use the laws of exponents and simplify {3}^{-7}+{3}^{-4}{3}^{-7}+{3}^{-4}

Solve {2^0} + {1^0} - {4^0} - {7^0}{2^0} + {1^0} - {4^0} - {7^0}

Represent \sqrt {4.5}\sqrt {4.5} on the number line.

Simplify \dfrac {(6.7\times 10^{-11})(6 \times 10^{24})(7.4 \times 10^{22})}{(3.84 \times 10^{8})^{2}}\dfrac {(6.7\times 10^{-11})(6 \times 10^{24})(7.4 \times 10^{22})}{(3.84 \times 10^{8})^{2}}

Find the value of {\left( \cfrac{6}{5} \right)}^{\dfrac{1}{2}}{\left( \cfrac{6}{5} \right)}^{\dfrac{1}{2}}

Simplify: \cfrac { { 3 }^{ 5 }\times { 10 }^{ 5 }\times 2^5 }{ { 5 }^{ 7 }\times { 6 }^{ 5 } } \cfrac { { 3 }^{ 5 }\times { 10 }^{ 5 }\times 2^5 }{ { 5 }^{ 7 }\times { 6 }^{ 5 } }

Find the co-efficient of a^0a^0 in the expression a^2+3a+2a^2+3a+2.

Find the value to three places of decimals of each of the following. It is given that \sqrt {2}=1.414, \sqrt {3}=1.732, \sqrt {5}=2.236\sqrt {2}=1.414, \sqrt {3}=1.732, \sqrt {5}=2.236 and \sqrt {10}=3.162\sqrt {10}=3.162\dfrac {3}{\sqrt {10}}\dfrac {3}{\sqrt {10}}

Simplify:2+\dfrac{\sqrt{3}}{3}2+\dfrac{\sqrt{3}}{3}

Find the value to three places of decimals of each of the following. It is given that \sqrt {2}=1.414, \sqrt {3}=1.732, \sqrt {5}=2.236\sqrt {2}=1.414, \sqrt {3}=1.732, \sqrt {5}=2.236 and \sqrt {10}=3.162\sqrt {10}=3.162\dfrac{2+\sqrt {3}}{3}\dfrac{2+\sqrt {3}}{3}

Find the value to three places of decimals of each of the following. It is given that \sqrt {2}=1.414, \sqrt {3}=1.732, \sqrt {5}=2.236\sqrt {2}=1.414, \sqrt {3}=1.732, \sqrt {5}=2.236 and \sqrt {10}=3.162\sqrt {10}=3.162\dfrac {\sqrt {2}-1}{\sqrt {5}}\dfrac {\sqrt {2}-1}{\sqrt {5}}

Following are the five rational numbers which are smaller than $2$ $\Rightarrow$ $1,\dfrac{1}{2},0,-1,\dfrac{-1}{2}$ .
If true then enter $1$ and if false then enter $0$ .

$y^{2}=9$

Simplify and express the result in power notation with positive exponent.

$(-3)^4 \times \left ( \dfrac{5}{3} \right )^4$ is $(m)^4$
$m$ is

Find the value of $(3^{1}+4^{1}+5^{1})^{0}$ .

Find the value of.

$(3^0+4^{-1}) \times 2^2$

Some rational numbers are integers, if true then enter $1$ and if false then enter $0$

5 $\sqrt{3}+2\sqrt{3}$ = 7 $\sqrt{6}$
enter $1$ for true and $0$ for false

Zero can be written in the form of $\displaystyle\ \frac{p}{q}$ , where p and q are integers and q $\neq$ 0?
If above statement is true then enter $1$ and if false then enter $0.$

5 $\sqrt{3} -2\sqrt{3}$ = 3 $\sqrt{3}$
enter $1$ for true and $0$ for false

Rationalise the denominator and simplify:
If $\displaystyle \frac{2}{\sqrt{5}+\sqrt{3}}$ is $\displaystyle \sqrt{m}-\sqrt{3}$ , then $m$ is ?

Rationalise the denominator and simplify:
If $\displaystyle \frac{3}{4-\sqrt{3}}$ is $\displaystyle \frac{3}{13}(4+\sqrt{m})$ , then $m$ is ?

Rationalise the denominator and simplify:
If $\displaystyle \frac{1}{2+\sqrt{3}}$ is $\displaystyle 2-\sqrt{m}$ , then $m$ is ?

Find the product of $3-\sqrt{7}$ and its conjugate.

Find the value-of $(3^0 - 4^0)\times5^2$ :

Express as rational number $\displaystyle \left ( \frac{-5}{7} \right )^{-3}$ is

Value of $m$ is

Simplify: $\displaystyle \left (\frac{1}{2} \right )^{-2} + \left (\frac{1}{3} \right )^{-2}+ \left (\frac{1}{4} \right )^{-2}$

Express as a power of a rational number with positive exponent $\displaystyle \left [ \left ( \frac{4}{5} \right )^{-2} \right ]^4$
Answer is $(\dfrac{5}{4})^m$
Value of $m$ is

Express as rational number $\displaystyle \left ( \frac{-5}{7} \right )^{-1}$ is $=\displaystyle \frac{-m}{5}$
Value of $m$ is

Evaluate $\displaystyle \left ( \frac{1}{2} \right )^{-5}$

If $\displaystyle \left \{ \left (\frac{3}{4} \right )^{-1} - \left (\frac{1}{4} \right )^{-1} \right \}=$ $- \dfrac{3}{m}$ , the the value of $m$ is

If $4^{-3}$ is $\dfrac{1}{m}$ , then the value of $m$ is:

$\displaystyle x^{ \frac{-1}{2}}$

Simplify $\displaystyle \left ( - \frac{2}{5} \right )^{-3} \times \left ( - \frac{2}{5} \right )^{4}$

Find the value of $(3^0 + 4^{-1}) \times 2^2$

Find the value of $\displaystyle \left ( \left ( \frac{4}{7} \right )^{8} \right )^0$ .

Find the value of $m$ :
$(2^{-1} \times 4^{-1}) + 2^{-2}=\dfrac{3}{m}$

Simplify $\displaystyle \left ( \frac{-3}{4} \right )^{-5} + \left ( \frac{-3}{4} \right )^{-3}$

Find the value of $m$ : $\displaystyle \left ( \frac{5}{2} \right )^3 \times \left ( \frac{5}{2} \right )^7 =\left( \dfrac { 5 }{ 2 } \right) ^{ m}$

Find $(0.04)^{-1.5} = ?$

The conjugate of $3 - \sqrt {5 + x}$ is $3 + \sqrt {5 + x}$ . Verify

Verify that the conjugate of $\sqrt {3} + \sqrt {2}$ is $\sqrt {3} - \sqrt {2}$

Write five rational numbers which are smaller than $2$ .

Write five rational numbers which are smaller than $\dfrac{5}{6}$ .

Write.
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) The rational number that is equal to its negative.

Express the following with positive exponents:
${4}^{-7}$

Find the value of ' $x$ ' such that
${(-2)}^{x+1}\times {(-2)}^{7}={(-2)}^{12}$

Write $0, 7, 10, -4$ in $\dfrac{p}{q}$ form

Can you guess the value of ' $x$ ' when ${2}^{x}=1$

Express the following with positive exponents:
${ \left( \cfrac { 4 }{ 7 } \right) }^{ -3 }$

Express the following with positive exponents:
$\dfrac{1}{{(5)}^{-4}}$

Give one example for the following statement:
A number which is rational but not an integer.

Classify the following number as rational or irrational.
$5 - \sqrt {3}$

Give one example each to the following statements.
A number which is natural number, whole number, integer and rational number

Give one example each to the following statements.
An integer which is not a whole number

Give one example each to the following statements.
A whole number which is not a natural number

Rationalise the denominator and simplify:
$\dfrac {\sqrt {3} + \sqrt {2}}{\sqrt {3} - \sqrt {2}}$

Number $\dfrac {3}{625}$ is a terminating decimal or a non-terminating repeating decimal? Write it in decimal form.

Rationalise the denominator and simplify:
$\cfrac { \sqrt { 6 } +\sqrt { 3 } }{ \sqrt { 6 } -\sqrt { 3 } }$

Rationalising factor of $\left( \sqrt { x+y } \right)$ is ____

Find the value of the following:
$\left (\dfrac {7}{4}\right )^{0}\times 3$

Find x
$2^{2x \, + \, 2} \, - \, 6^x \, - \, 2 \times 3 ^{2x \, + \, 2} \, = \, 0$

Solve the following equation:
$\displaystyle\, \frac{0.2^{x + 0.5}}{\sqrt{5}} = \frac{(0.04)^x}{25}$

Solve the following equations.
$\displaystyle\, \left ( \frac{5}{12} \right )^x \cdot\left ( \frac{6}{5} \right )^{x -1} = (0.3)^{-1}$

Is $0.9$ a rational number?

Solve $17^2 \cdot 17^{-5}$

Solve:
$\dfrac {\sqrt{24}}{8}+\dfrac {\sqrt{54}}{9}$ .

Simplify: $\dfrac { 25 }{ \sqrt { 5 } }$

Find a rational number between $\dfrac{2}{3}$ and $\dfrac{3}{4}$ .

Find four rational numbers between $\dfrac{2}{3}$ and $\dfrac{3}{5}$ .

Write five rational numbers which are greater than $\dfrac{-3}{2}$ .

Prove that $\dfrac{{{{({a^{p + q}})}^2}{{({a^{q + r}})}^2}{{({a^{r + p}})}^2}}}{{{{({a^p}.{a^q}.{a^r})}^4}}} = 1$

How many of the following numbers are rational numbers?
(i) $\dfrac{5}{-8}$ (ii) $\dfrac{-6}{11}$ (iii) $\dfrac{7}{15}$ (iv) $-3$ (v) $0$ (vi) $\dfrac{1}{0}$

If $\frac{{{{\left( { - 2} \right)}^x} \times {{\left( { - 2} \right)}^7}}}{{3 \times {4^6}}} = \frac{1}{{12}}$ , then the value of x-3 is

Simplify
$(16)^{\dfrac {-1}{4}}x {^{4}\sqrt {16}}$

Evaluate:-
${3^3} \times \left( {243} \right)\,{\,^{ - \cfrac{2}{3}}}\,\, \times {9^{\,\, - \cfrac{1}{3}}}$

Add $7\sqrt 2 + 5\sqrt 3$ and $\sqrt 2 - 7\sqrt 3$

Use the laws of exponents and simplify ${3}^{-7}+{3}^{-4}$

Solve ${2^0} + {1^0} - {4^0} - {7^0}$

Represent $\sqrt {4.5}$ on the number line.

Simplify $\dfrac {(6.7\times 10^{-11})(6 \times 10^{24})(7.4 \times 10^{22})}{(3.84 \times 10^{8})^{2}}$

Find the value of ${\left( \cfrac{6}{5} \right)}^{\dfrac{1}{2}}$

Simplify: $\cfrac { { 3 }^{ 5 }\times { 10 }^{ 5 }\times 2^5 }{ { 5 }^{ 7 }\times { 6 }^{ 5 } }$

Find the co-efficient of $a^0$ in the expression $a^2+3a+2$ .

Find the value to three places of decimals of each of the following. It is given that $\sqrt {2}=1.414, \sqrt {3}=1.732, \sqrt {5}=2.236$ and $\sqrt {10}=3.162$
$\dfrac {3}{\sqrt {10}}$

Simplify: $2+\dfrac{\sqrt{3}}{3}$

Find the value to three places of decimals of each of the following. It is given that $\sqrt {2}=1.414, \sqrt {3}=1.732, \sqrt {5}=2.236$ and $\sqrt {10}=3.162$
$\dfrac{2+\sqrt {3}}{3}$

Find the value to three places of decimals of each of the following. It is given that $\sqrt {2}=1.414, \sqrt {3}=1.732, \sqrt {5}=2.236$ and $\sqrt {10}=3.162$
$\dfrac {\sqrt {2}-1}{\sqrt {5}}$

$\left( {{3^ \circ } + {4^{ - 1}}} \right) \times {2^2}$

${\left( {\frac{1}{2}} \right)^{ - 2}}\, + \,{\left( {\frac{1}{3}} \right)^{ - 2}}\, + \,{\left( {\frac{1}{4}} \right)^{ - 2}}$

Write any two irrational numbers between $0.23$ and $0.3$ .