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CBSE Questions for Class 9 Maths Number Systems Quiz 2 - MCQExams.com
CBSE
Class 9 Maths
Number Systems
Quiz 2
Two rational numbers between $$\dfrac{2}{3}$$ and $$\dfrac{5}{3}$$ are :
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$$\dfrac{1}{6}$$ and $$\dfrac{2}{6}$$
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$$\dfrac{1}{2}$$ and $$\dfrac{2}{1}$$
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$$\dfrac{5}{6}$$ and $$\dfrac{7}{6}$$
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$$\dfrac{2}{3}$$ and $$\dfrac{4}{3}$$
Explanation
Changing the denominators of both numbers to 6, we get
$$\dfrac { 2 }{ 3 } =\dfrac { 4 }{ 6 } \quad \& \quad \dfrac { 5 }{ 3 } =\dfrac { 10 }{ 6 }$$
Numbers between the given rational numbers from the options are
$$\dfrac { 5 }{ 6 } \quad \& \quad \dfrac { 7 }{ 6 } $$
So, correct answer is option C.
State true or false:
There are numbers which cannot be written in the form $$\frac{p}{q}$$, where $$q\neq 0$$ and
both
p, q are integers.
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True
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False
Explanation
The statement is true as there are Irrational numbers which dont satisfy the condition of rational numbers i.e irrational number cannot be written in the form of $$ \frac { p }{ q } $$ $$ q\neq 0\quad , where \quad p,q\quad $$ are integers.
Example,
$$ \sqrt { 3 } ,\sqrt { 99 } \quad $$
The value of $$2\sqrt {3} + \sqrt {3}$$ is equal to
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$$2\sqrt {6}$$
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$$3\sqrt {3}$$
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$$4\sqrt {6}$$
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$$6$$
Explanation
$$2\sqrt { 3 } +\sqrt { 3 } =\left( 2+1 \right) \sqrt { 3 } =3\sqrt { 3 }$$
Between any two rational numbers,
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there is no rational number
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there is exactly one rational number
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there are infinitely many rational numbers
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there are only rational numbers and no irrational numbers
Explanation
Re
call that to find a rational number between $$r$$ and $$s,$$ you can add
$$r$$ and $$s$$ and divide the sum by $$2,$$ that is $$\displaystyle \frac { r+s }{ 2 }$$ lies between r and s.
For example, $$\displaystyle \frac { 5 }{ 2 }$$ is a number between $$2$$ and
$$3.$$ We can proceed in this manner to find many more rational numbers between $$2$$ and $$3.$$
Hence, we can conclude that there are infinitely many rational numbers between any two given rational numbers.
Every rational number is
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A natural number
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An integer
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A real number
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A whole number
Explanation
$$\textbf{Step-1: Explain property of rational numbers}$$
$$\text{A real number is a number which can be expressed in the form}$$ $$\dfrac{p}{q},$$ $$\text{where}$$ $$q\neq{0.}$$
$$\textbf{Step-2: Proving that every rational number is a real number}$$
$$\text{Real numbers are numbers that include bothe rational and irrational numbers.}$$
$$\text{Hence, every rational number is a real number.}$$
$$\textbf{Final Answer: Every rational number is a real number. The correct option is (C).}$$
The product of a non - zero rational number with an irrational number is always :
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Irrational number
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Rational number
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Whole number
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Natural number
Explanation
By definition, an irrational number in decimal form goes on forever without repeating (a non-repeating, non-terminating decimal). By definition, a rational number in decimal form either terminates or repeats.
By multiplying a non repeating non terminating number to repeating or terminating/repeating number, the result will always be a non terminating non repeating number.
So, option A is correct.
$$\sqrt { 2 } ,\sqrt { 3 }$$ are
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Whole numbers
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Rational numbers
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Irrational numbers
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Integers
Explanation
$$\sqrt{2}, \sqrt{3}$$ are irrational numbers.
Because these two values cannot be written in the form of p/q where p and q both are integers and q should not be equal to zero .
State whether the given statement is True or False.
After rationalising the denominator of $$\dfrac{5}{3\sqrt{2}-2\sqrt{3}}$$, we get its denominator as $$7.$$
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True
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False
Explanation
The denominator of $$\dfrac{5}{3\sqrt2-2\sqrt3}$$ is $$3\sqrt2-2\sqrt3$$
Rationalising the denominator, we get
$$(3\sqrt2-2\sqrt3)(3\sqrt2+2\sqrt3)$$
$$=\left( 3\sqrt { 2 } \right) ^{ 2 }-\left( 2\sqrt { 3 } \right) ^{ 2 }$$
$$=18-12$$
$$=6$$
So, the denominator is $$6$$.
Thus, answer is
false.
Find the product.
$$(a^2) (2a^{22}) (4a^{26})$$
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$$8a^{40}$$
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$$8a^{50}$$
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$$8a^{30}$$
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$$8^{20a}$$
Explanation
$$(a^2) (2a^{22}) (4a^{26})$$
$$=(a^2)\times (2a^{22})\times (4a^{26})$$
$$=8a^{2+22+26}$$
$$=8a^{50}$$
The value of $$(6 + \sqrt{27}) - (3 + \sqrt{3}) + (1 - 2\sqrt{3})$$ when simplified is :
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positive and irrational
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negative and rational
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positive and rational
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negative and irrational
Explanation
$$6+\sqrt { 27 } -(3+\sqrt { 3 } )+(1-2\sqrt { 3 } )=6+3\sqrt { 3 } -3-\sqrt { 3 } +1-2\sqrt { 3 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =4$$
$$4$$ is a positive rational number
Hence, correct answer is option C.
The value of $$\cfrac{3^0+7^0}{5^0}$$ is:
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$$2$$
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$$0$$
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$$\dfrac{9}{5}$$
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$$\dfrac{1}{5}$$
Explanation
$$\Rightarrow$$ $$\dfrac{3^0+7^0}{5^0}$$
$$\Rightarrow$$ $$\dfrac{1+1}{1}$$ $$[\,\because\,x^0=1]$$
$$\Rightarrow$$ $$\dfrac{2}{1}$$
$$\Rightarrow$$ $$2$$
$$\therefore$$
$$\dfrac{3^0+7^0}{5^0}=2$$
$$\sqrt{5}$$ is an irrational number.
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True
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False
Explanation
An irrational number is any real number that cannot be expressed as a ratio $$\dfrac ab,$$ where $$a$$ and $$b$$ are integers and $$b$$ is non-zero.
$$\sqrt5$$ is irrational as it can never be expressed in the form $$\dfrac ab.$$
Find the five rational numbers between
$$\displaystyle \frac{1}{2}$$ and $$\displaystyle \frac{3}{2}$$
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$$0.5 < 0.6 < 0.7 < 0.8 ... < 1.1 < ... < 1.15 < 1.50$$
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$$0.5 < 0.6 < 1.7 < 3.8 ... < 1.8< ... < 1.15 < 1.50$$
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$$0.5 < 0.6 < 0.7 < 2.8 ... < 1.1 < ... < 1.15 < 1.50$$
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$$0.5 < 0.6 < 0.7 < 0.8 ... < 3.1 < ... < 1.15 < 1.50$$
Which of the following numbers are rational ?
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$$1$$
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$$ -6$$
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$$3\dfrac{1}{2}$$
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All above are rational
Explanation
$$\Rightarrow$$ A rational number is a type of real numbers which
can be expressed in the form of $$\dfrac pq, $$ where $$ q \neq0.$$
$$\Rightarrow$$
All the numbers are rational as they are in the form of
$$\dfrac pq, $$ where $$ q \neq0.$$
i.e, $$\dfrac{1}{1}\ , \dfrac{-6}{1}\ ,\dfrac{7}{2}$$
The rationalizing factor of $$(a+\sqrt{b})$$ is
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$$a-\sqrt{b}$$
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$$\sqrt{a}-b$$
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$$\sqrt{a}-\sqrt{b}$$
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None of these
Explanation
The rationalizing factor of a+$$\sqrt { b }$$ is a-$$\sqrt { b }$$ as the product of these two expressions give a rational number.
Simplify:
$$3\sqrt{3} + 10\sqrt{3}$$
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$$13\sqrt{3}$$
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$$10\sqrt{3}$$
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$$12\sqrt{3}$$
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$$11\sqrt{3}$$
Explanation
$$3 \sqrt{3}+ 10 \sqrt{3} =13 \sqrt{3}$$
The value of $$5\sqrt{3} - 3\sqrt{12} + 2\sqrt{75} $$ on simplifying is :
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$$5\sqrt{3}$$
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$$6\sqrt{3}$$
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$$\sqrt{3}$$
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$$9\sqrt{3}$$
Explanation
The given expression is
$$5\sqrt { 3 } -3\sqrt { 12 } +2\sqrt { 75 } \\ =5\sqrt { 3 } -6\sqrt { 3 } +10\sqrt { 3 } \\ =9\sqrt { 3 }$$.
Therefore, option $$D$$ is correct.
State True or False.
A rational number can always be written in a fraction $$\dfrac{a}{b}$$, where a and b are not integers $$(b \neq 0)$$.
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True
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False
Explanation
A number that can always be written in the form of $$p/q$$, where p is any integer and q is a non-zero integer, is a rational number.
The given statement is false.
Find conjugate of:
$$\sqrt{3}+\sqrt{2}$$
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$$\sqrt{3}-\sqrt{2}$$
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$$\sqrt{3}+\sqrt{2}$$
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$$\sqrt{3}\sqrt{2}$$
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None of these
Explanation
$$\sqrt{3}+\sqrt{2}$$
The procedure of multiplying a surd by another surd to get a rational number is called RationalisationThe operands are called rationalizing factor (RF) of the other
$$(\sqrt{3}+\sqrt{2}) \times (\sqrt{3}-\sqrt{2})=\sqrt{3}^2-\sqrt{2}^2=9-4=5$$
The Rationalizing factor of $$\sqrt{3}+\sqrt{2}$$ is $$\sqrt{3}-\sqrt{2}$$
Rationalise the denominator of :
$$\displaystyle\ \frac{\sqrt{6}}{\sqrt{12}}$$
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$$\displaystyle\ \frac{\sqrt{2}}{6}$$
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$$\displaystyle\ \frac{\sqrt{2}}{5}$$
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$$\displaystyle\ \frac{\sqrt{2}}{2}$$
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$$\displaystyle\ \frac{\sqrt{2}}{3}$$
Explanation
$$\cfrac{\sqrt6}{\sqrt{12}} $$=
$$\cfrac {\sqrt6\times\sqrt{12}}{\sqrt{12}\times\sqrt{12}}$$=$$\cfrac{6\sqrt2}{12}$$=$$\cfrac{\sqrt2}{2}$$
Find conjugate of:
3$$\sqrt{2}$$ -1
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3$$\sqrt{2}$$ +1
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3$$\sqrt{2}$$ -1
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3$$\sqrt{2}$$
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None of these
Explanation
$$3\sqrt{2}+1$$
The procedure of multiplying a surd by another surd to get a rational number is called Rationalisation
The operands are called rationalizing factor (RF) of the other
$$(3\sqrt{2}+1) \times (3\sqrt{2}-1)=(3\sqrt{2})^2-1^2=18-1=17$$
The Rationalizing factor of $$3\sqrt{2}+1$$ is $$3\sqrt{2}-1$$
Rationalise the denominator of :
$$\displaystyle\ \frac{2\sqrt{2}}{\sqrt{3}}$$
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$$\displaystyle\ \frac{2\sqrt{7}}{3}$$
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$$\displaystyle\ \frac{2\sqrt{6}}{3}$$
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$$\displaystyle\ \frac{2\sqrt{2}}{3}$$
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$$\displaystyle\ \frac{2\sqrt{6}}{10}$$
Explanation
$$\cfrac{2\sqrt{2}}{\sqrt{3}}$$
$$=\cfrac{2\sqrt{2}\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}$$
$$=\cfrac{2\sqrt{6}}{3}$$
Rationalise the denominator of :
$$\displaystyle\ \frac{6}{\sqrt{10}-2}$$
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$$\sqrt{10}-2$$
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$$\sqrt{10}+2$$
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$$2\sqrt{10}$$
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None of these
Explanation
$$\quad \\ Multiplying\quad and\quad dividing\quad by\quad \sqrt { 10 } +2:\\ \cfrac{6(\sqrt { 10 } +2)}{(10-4)}=\sqrt { 10 } +2\\ $$
Rationalise the denominator of:
$$\displaystyle\ \frac{2}{\sqrt{5}+\sqrt{3}}$$
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$$\sqrt{5}-\sqrt{3}$$
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$$\sqrt{4}-\sqrt{3}$$
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$$\sqrt{2}-\sqrt{3}$$
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$$\sqrt{6}-\sqrt{3}$$
Explanation
$$\quad \\ Multiplying\quad and\quad dividing\quad by\quad \sqrt { 5 } -\sqrt { 3 } :\\\cfrac{ 2(\sqrt { 5 } -\sqrt { 3 } )}{(5-3)}=\sqrt { 5 } -\sqrt { 3 } \\ $$
Rationalise the denominator
(i) $$\displaystyle\ \frac{22}{2\sqrt{3}+1}$$
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$$(2\sqrt{3}-1)$$
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$$3(2\sqrt{3}-1)$$
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$$2(2\sqrt{3}-1)$$
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$$4(2\sqrt{3}-1)$$
Explanation
$$\\ Multiplying\quad and\quad dividing\quad by\quad 2\sqrt { 3 } -1:\quad \\ \cfrac{22(2\sqrt { 3 } -1)}{(12-1)}=2(2\sqrt { 3 } -1)\\ $$
Write the simplest rationalisation factor of the following surds:
$$\sqrt{32}$$
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$$\sqrt{2}$$
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$$\sqrt{7}$$
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$$\sqrt{5}$$
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$$\sqrt{3}$$
Explanation
$$\sqrt{32}$$
$$=\sqrt{2\times2\times2\times2\times2}$$
$$=4\sqrt2$$
The procedure of multiplying a surd by another surd to get a rational number is called Rationalisation
The operands are called Rationalizing factor (RF) of the other.
Here, $$\sqrt 2$$ is the Rationalisation Factor.
State whether true or false.
$$\displaystyle \frac{5}{11}$$ is a rational number.
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True
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False
Explanation
True.
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,$$ \dfrac {5}{11} $$, is of the form $$ \dfrac {p}{q} $$, where p and q are integers and $$ q \neq 0 $$
Hence, it is a rational number.
Rationalise the denominator of :
$$\displaystyle\ \frac{5}{\sqrt{7}+\sqrt{2}}$$
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$$\sqrt{7}+\sqrt{2}$$
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$$\sqrt{7}-\sqrt{2}$$
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$$\sqrt{7}\sqrt{2}$$
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None of these
Explanation
$$\quad \\ Multiplying\quad and\quad dividing\quad by\quad \sqrt { 7 } -\sqrt { 2 } :\\ \cfrac{5(\sqrt { 7 } -\sqrt { 2 } )}{(7-2)}=\sqrt { 7 } -\sqrt { 2 } \\ $$
State true or false:
There can be a pair of irrational numbers whose sum is irrational s
uch as
$$\displaystyle \sqrt{3}+2$$ and $$\displaystyle 5+\sqrt{2}$$.
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True
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False
Explanation
To get the sum as irrational, both the numbers need to have an irrational part which are different from each other.
Example, the given pair of numbers $$ \sqrt{3} + 2 $$ and $$ 5 + \sqrt {2} $$ have the sum $$ \sqrt{3} + 2 + 5 + \sqrt {2} = 7 + \sqrt {2} + \sqrt {3} $$ which is an irrational number too.
Hence, the given statement is true and option $$A$$ is correct.
State true or false:
$$\sqrt3$$ is an irrational number
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True
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False
Explanation
$$\sqrt{3}$$ is irrational number because it will have non terminating and non recurring decimal
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