Class 8 Maths - Rational Numbers

Write the additive inverse of each of the following rational numbers:
$\dfrac{3}{-11}$ .

If it is $\dfrac{3}{11}$ enter $1$ else enter $0$

Since

$\dfrac{3}{-11} = \dfrac{-3}{11}$ and

$\dfrac{-3}{11} + \dfrac{3}{11} = \dfrac{-3+3}{11} = \dfrac{0}{11} = 0$

$\Rightarrow \dfrac{3}{11}$ is additive inverse for

$\dfrac{-3}{11}$

Represent $\dfrac{7}{4}$ on the number line.

Let's draw number line from

$0$ to

$2$

Since in

$\dfrac{7}{4}$ , denominator is

$4$

We divide line between

$0$ and

$1$ into

$4$ equal parts.

1332618_1403797_ans_a9290dc9215c4e9c8fd883a0ee3455f4.png

Write the additive inverse of each of the following rational numbers:
$\dfrac{-2}{17}$

Any number when added to given number sums to

$0$ is called additive inverse

$\dfrac{-2}{17} + \dfrac{2}{17} = \dfrac{-2+2}{17} = \dfrac{0}{17} = 0$

$\Rrightarrow \dfrac{2}{17}$ is additive inverse for

$\dfrac{-2}{17}$ .

By what number should $\left( 5\right) ^{ -5 }$ be multiplied so that the product is 1?

Let the number be

$x$ .

$\therefore$

${ 5 }^{ -5 }x=1$

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

$x={ \left( 5 \right) }^{ 5 }$ , which is the required number.

Write the negative (additive inverse) of each of the following:
$\dfrac{-2}{5}$

The given number is

$\dfrac{-2}{5}$

Since,

$\dfrac{-2}{5} + \dfrac{2}{5} = \dfrac{-2+2}{5} = \dfrac{0}{5} = 0$

Hence,

$\dfrac{2}{5}$ is the additive inverse of

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

Write the additive inverse of $\dfrac{-11}{-25}$

We know that

$\dfrac{-11}{-25} = \dfrac{11}{25}$

and

$\dfrac{11}{25} - \dfrac{11}{25} = \dfrac{11-11}{25}$

$= \dfrac{0}{25} =0$

Thus

$\dfrac{-11}{25}$ is additive inverse for

$\dfrac{-11}{-25}$

Represent the following rational numbers on the number line.
$4\dfrac {3}{5}$ .

To represent

$4\dfrac {3}{5}$ on number line:

divide the line between

$4$ and

$5$ in five equal spaces by drawing four marking. The

$3^{rd}$ marking is

$4\dfrac {3}{5}$ as shown in figure.

1710804_1806133_ans_16d3f57e9f7547dbb70baec24ceef864.jpg

Represent the following rational numbers on the number line.
$-9/7$ .

$-9/7 = -1\dfrac {2}{7}$
The given rational number on the number line is shown as below.
1710810_1806136_ans_b4e29aeacaa24500bec576c1491be408.jpg

Write the additive inverse (negative ) of :
$- \dfrac{1}{3}$

The additive inverse of

$-\dfrac{1}{3} ~is = \dfrac{1}{3}$

Fill in the blanks :
$\dfrac{5}{8}$ and $\dfrac{-3}{10}$ are on the ............ sides of zero.

$\dfrac{5}{8}$ and

$\dfrac{-3}{10}$ are on the opposite sides of zero.

Since both have opposite sign.

Write the negative (additive inverse) of $\dfrac{7}{-9}$ .

If the answer is $\dfrac ab$ , then the value of $(a+b)$ is:-

Since,

$\dfrac{7}{-9} = \dfrac{-7}{9}$ and

$\dfrac{-7}{9} + \dfrac{7}{9} = \dfrac{-7+7}{9} = \dfrac{0}{9} = 0$

$\Rightarrow \dfrac{7}{9}$ is additive inverse for

$\dfrac{7}{-9}$

$\therefore$ the value of

$(a+b)$ is

$(7+9)=16$

Fill in the blank:
$(-5) \times (......) = 0$

Using the closure property of multiplication,

$[a \times b = c]$

Let,

a = -5, missing number b = x, c = 0

$(-5) \times (x) = 0$

( By sending -5 from Left hand side to the denominator of the right hand side in the multiplication,

= x =

$(\frac{0}{5})$

= x = 0

(

$\because$ Anything is divided by zero the answer is zero itself)

Find the multiplicative inverse (i.e. reciprocal of :
$-1$ .

The multiplicative inverse of a no.

$\dfrac ab$ is

$\dfrac ba$

$\because\dfrac ab \times \dfrac ba=1$

The multiplicative inverse of

$-1$ is

$-1$ .

Represents each of the following rational numbers on the number line:
$(-1/3)$

We know that

$-1/3$ is greater than

$0$ and less than

$-1$

$\therefore \$ It lies between

$4$ and

$5$ . It can \can be represented on number line as,
1626559_1779780_ans_98e75fc16f894967bc9a4aa470303865.PNG

Find the additive inverse of:
$\cfrac3{14}$

Additive inverse of a number is a number when added to original number yields result

$0$ .

So, let's assume additive inverse of

$\cfrac{3}{14}$ is

$x$

So, according to definition of additive inverse

$\cfrac{3}{14}+x=0$

$\therefore x=-\cfrac{3}{14}$

$\therefore$ Additive inverse of

$\cfrac{3}{14}$ is

$-\cfrac{3}{14}$

Represents each of the following rational numbers on the number line:
$\cfrac{-43}9$

Now above question can be written as,

$\cfrac{-43}{9}$

$=\left[-4-\cfrac{7}{9}\right]$

So, we know that

$\cfrac{-43}{9}$ is greater than

$-5$ and less than

$-4$

$\therefore \$ It lies between

$-5$ and

$-4$ .

1624051_1779785_ans_0dd695c710394272b644966802fc04c5.png

Represents each of the following rational numbers on the number line:
$(37/8)$

Now above question can be written as,

$=(37/8)=[4+(5/8)]$
We know that

$227/8$ is greater than

$4$ and less than

$5$

$\therefore \$ It lies between

$4$ and

$5$ . It can be represented on number line as,
1626553_1779779_ans_4948ac4e5ed24b1abac7dc1cd59b25b9.PNG

Find the additive inverse of:
$\cfrac{-11}{15}$

Additive inverse of a number is a number when added to original number yields result

$0$ .

So, let's assume additive inverse of

$\cfrac{-11}{15}$ is

$x$

So, according to definition of additive inverse

$\cfrac{-11}{15}+x=0$

$\therefore x=\cfrac{11}{15}$

$\therefore$ Additive inverse of

$\cfrac{-11}{15}$ is

$\cfrac{11}{15}$

Find the additive inverse of:
$\cfrac{15}{-4}$

Additive inverse of a number is a number when added to original number yields result

$0$ .

$\cfrac{15}{-4}$ can also be written as

$\cfrac{-15}4$

So, let's assume additive inverse of

$\cfrac{-15}{4}$ is

$x$

So, according to definition of additive inverse

$\cfrac{-15}{4}+x=0$

$\therefore x=\cfrac{15}{4}$

$\therefore$ Additive inverse of

$\cfrac{15}{-4}$ is

$\cfrac{15}{4}$

Find the additive inverse of:
$\cfrac{-18}{-13}$

Additive inverse of a number is a number when added to original number yields result

$0$ .

$\cfrac{-18}{-13}$ can also be written as

$\cfrac{18}{13}$

So, let's assume additive inverse of

$\cfrac{18}{13}$ is

$x$

So, according to definition of additive inverse

$\cfrac{18}{13}+x=0$

$\therefore x=-\cfrac{18}{13}$

$\therefore$ Additive inverse of

$\cfrac{-18}{-13}$ is

$-\cfrac{18}{13}$

Fill the additive inverse of:
$-9$

Additive inverse of a number is a number when added to original number yields result

$0$ .

So, let's assume additive inverse of

$-9$ is

$x$

So, according to definition of additive inverse

$-9+x=0$

$\therefore x=9$

$\therefore$ Additive inverse of

$-9$ is

$9$

Represents each of the following rational numbers on the number line:
$\cfrac{36}{-5}$

Now above question can be written as,

$\cfrac{36}{-5}$

$=\left[-7-\cfrac{1}{5}\right]$

So, we know that

$\cfrac{36}{-5}$ is greater than

$-8$ and less than

$-7$

$\therefore \$ It lies between

$-8$ and

$-7$ .

1624046_1779784_ans_9a8d3e30e8a943f59c311686be19da6d.png

Find the additive inverse of:
$5$

Additive inverse of a number is a number when added to original number yields result

$0$ .

So, let's assume additive inverse of

$5$ is

$x$

So, according to definition of additive inverse

$5+x=0$

$\therefore x=-5$

$\therefore$ Additive inverse of

$5$ is

$-5$

Find the additive inverse of:
$\cfrac{1}{-6}$

Additive inverse of a number is a number when added to original number yields result

$0$ .

$\cfrac{1}{-6}$ can also be written as

$\cfrac{-1}{6}$

So, let's assume additive inverse of

$\cfrac{-1}{6}$ is

$x$

So, according to definition of additive inverse

$\cfrac{-1}{6}+x=0$

$\therefore x=\cfrac{1}{6}$

$\therefore$ Additive inverse of

$\cfrac{1}{-6}$ is

$\cfrac{1}{6}$

Fill in the blanks to make the statements true.
Additive inverse of $\dfrac{2}{3}$ is ________.

Negative of a number is additive inverse.
Hence,

$\dfrac{-2}{3}$ additive inverse of

$\dfrac{2}{3}$

Fill in the blanks to make the statements true.
On a number line, $\dfrac{4}{3}$ is to the ________ of zero $(0)$ .

$\dfrac{4}{3}$ is a positive number. Hence, it is right side of zero

$(0)$ .

Fill in the blanks to make the statements true.
On a number line, $\dfrac{-1}{2}$ is to the ________ of zero $(0).$

$\dfrac{-1}{2}$ is a negative rational number.
On a number line, Negative numbers are to the left side of zero

$(0).$

Find a rational number exactly halfway between:
$\dfrac{-1}{3}$ and $\dfrac{1}{3}$

As we know that exactly halfway between two rational numbers is

$\dfrac{a+b}{2}$
Rational numbers are

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

$\dfrac{1}{3}$
Here,

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

$b=\dfrac{1}{3}$

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

$=\dfrac{0}{2}$

$=0$
Therefore, the exactly halfway between

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

$\dfrac{1}{3}$ is zero.

Find the multiplicative inverse or reciprocal of each of the following:
$0$

when we multiply a number by its "Multiplicative Inverse" we get 1. But not when number is 0 as 1/0 is undifined.

The reciprocal of

$0$ is not exist.

How many rational number p/q are there between 0 and 1 for which q<P ?

There are no rational number between $0$ and $1$ such that $q<p$ .

Let $\dfrac{p}{q}$ where $q<p$

$p=5,q=3$

Therefore $\dfrac{5}{3}=1.666>1.$

Which integers have a multiplication inverse?

1972271_1871522_ans_5c4621e7feec458b96d13471bcd2bb43.gif

Fill n the blanks :
$- \dfrac{-5}{-12} +$ its Additive inverse = _____

$- \dfrac{-5}{-12} +$ its Additive inverse =

$- \dfrac{-5}{-12} + \dfrac{5}{12} = 0$ (The additive inverse of

$a$ is

$-a$ )

Write the additive inverse (negative ) of :
$\dfrac{-3}{1}$

The additive inverse of

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

Write the additive inverse (negative ) of :
$\dfrac{4}{13}$

Let the additive inverse of

$\dfrac{4}{13}$ is

$x$

$\dfrac{4}{13}+x=0$

$x=- \dfrac{4}{13}$

Write the additive inverse (negative ) of :
$\dfrac{-7}{5}$

The additive inverse of

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

$\dfrac{7}{5}$

Fill n the blanks :
If $\dfrac{a}{b}$ is additive inverse of $\dfrac{-c}{d'}$ then $\dfrac{-c}{d}$ is additive inverese of ________ ?

The additive inverse of 1=-1

Find the additive inverse of each of the following:
$-18$ .

Additive inverse of any number say,

$\dfrac {a}{b}$ is

$-\dfrac {a}{b}$ such that,

$\dfrac {a}{b} + \left (\dfrac {-a}{b}\right ) = 0$

$\therefore$ By using the above logic

The additive inverse of

$-18$ is

$18$ .

Find the additive inverse of the following:
$\dfrac {1}{3}$

Additive inverse of any number say,

$\dfrac {a}{b}$ is

$-\dfrac {a}{b}$ such that,

$\dfrac {a}{b} + \left (-\dfrac {a}{b}\right ) = 0$

$\therefore$ By using the above logic
The additive inverse of

$\dfrac {1}{3}$ is

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

Fill in the blank:
$\{(-5) \times 3\} (-6)= (......) \times \{3 \times (-6)\}$

Using the Associative law of Multiplication

$[(a \times b) \times c = a \times (b \times c)]$

Let,

a = -5, b = 3, c = -6

$\{(-5) \times 3\} \times (-6) = (-5) \times \{3 \times (-6)\}$

Find the multiplicative inverse of $\dfrac02.$

The multiplicative inverse of

$\dfrac02$ is

$\dfrac{2}{0}$ . Since

$\dfrac{2}{0}$ is undefined. The multiplicative inverse of

$\dfrac02$ is undefined.

Write the additive inverse (negative) of:
0

We know that, additive inverse of a number

$a$ , is another number

$b$ such that

$a+b=0$ .

Let

$a=0$

Hence,

$0+b=0$

$\Rightarrow b=0$

Hence, additive inverse of

$0$ is

$0$ .

Enter 1 if true else enter 0.
The numbers $1$ and $-1$ are their own reciprocals.

Let the number

$x$ be its own reciprocal.

Then by definition,

$x\times x=1$

$\Rightarrow x^2=1$

$\Rightarrow x=+1,-1$

Hence, the numbers

$1$ and

$-1$ are their own reciprocals

Can you find a rational number whose multiplicative inverse is $-1$ ?

Let the number be

$x$

Given, multiplicative inverse of

$x$ is

$-1$

Then,

$x\times(-1)=1$

$\Rightarrow x=-1$

Hence, the number is

$-1$ itself.

Find the multiplicative inverse (i.e. reciprocal of :
$-1/8$ .

The multiplicative inverse of a no.

$\dfrac ab$ is

$\dfrac ba$

$\because\dfrac ab \times \dfrac ba=1$

The multiplicative inverse of

$(-1/8)$ is

$(8/-1)$

We put it in the standard form i.e.

$(8/-1)\times -1$ we get,

$-8/1 = -8$ .

Fill n the blanks :
Additive inverse of $\dfrac{-5}{-12}$ = ________

Additive inverse of

$\dfrac{-5}{-12}$ =

$\dfrac{5}{12}$

$\displaystyle\frac{2}{3}\times\displaystyle\frac{6}{21}=\displaystyle\frac{6}{21}\times$ $\displaystyle\frac{2}{3}$

If true then enter $1$ and if false then enter $0$

Commutative property is true for multiplication.

$a\times b=b \times a$

$\displaystyle\frac{2}{3}\times\displaystyle\frac{6}{21}=\displaystyle\frac{6}{21}\times\displaystyle\frac{2}{3}=\dfrac 4{21}$

The given statement is true.

The addition of rational numbers is commutative.
If true then enter $1$ and if false then enter $0$ .

True

Let say

$a,b,c$ be three rational number then their addition would be commutative if,

$a+(b+c)=(a+b)+c$

For example,

$\frac{1}{2}+(\frac{2}{5}+\frac{3}{4})=(\frac{1}{2}+\frac{2}{5})+\frac{3}{4}$

$LHS=RHS$

$\displaystyle\frac{7}{8}\times\begin{pmatrix}\displaystyle\frac{17}{13}+\displaystyle\frac{14}{13}\end{pmatrix}=\begin{pmatrix}\displaystyle\frac{7}{8}\times\displaystyle\frac{14}{13}\end{pmatrix}+\begin{pmatrix}\displaystyle\frac{7}{8}\times\displaystyle\frac{17}{13}\end{pmatrix}$
If true then enter $1$ and if false then enter $0$

$\displaystyle\frac{7}{8}\times\begin{pmatrix}\displaystyle\frac{17}{13}+\displaystyle\frac{14}{13}\end{pmatrix}=\begin{pmatrix}\displaystyle\frac{7}{8}\times\displaystyle\frac{14}{13}\end{pmatrix}+\begin{pmatrix}\displaystyle\frac{7}{8}\times\displaystyle\frac{17}{13}\end{pmatrix}$
Multiplication is distributive over addition of two rational number.

If $\cfrac{-5}{9}$ is additive inverse of $\cfrac{5}{p}$ , then $p$ is equal to

Additive inverse of a number X is -X.
For example, additive inverse of:-

$\dfrac{-5}{9}$ is

$\dfrac{5}{9}$ .

Comparing this with

$\dfrac{5}{p}$ , we have

$p=9$ .

The additive inverse of $\dfrac{2}{8}$ is $\dfrac{-2}{q}$ , $q$ is

Additive inverse of a number

$x$ is

$-x.$
For example, additive inverse of

$\dfrac{1}{2}$ is

$-\dfrac{1}{2}$

Here, for

$\dfrac{2}{8}$ additive inverse is

$-\dfrac{2}{8}.$

The rational number that is equal to its negative.

Let

$x$ be the rational number

$x=-x$

$=> x+x=0$

$=> 2x=0$

$=> x=0$

Represent the number on the number line: $\dfrac{7}{4}$

Since, the denominator of each given fraction is

$4$ ; divide the space between every pairs of two consecutive integers (on the number line) in

$4$ equal parts. Each part so obtained will represent the fraction

$\dfrac{1}{4}$ and the number line obtained will be of the form:

$Fractions and Number Line$

Verify that: $-(-x)=x$ for $x=\displaystyle\frac{11}{15}$ .

$x=\cfrac{11}{15}$

$\therefore-(-x)=-\begin{pmatrix}-\displaystyle\frac{11}{15}\end{pmatrix}=\displaystyle\frac{11}{15}=x$
Hence, proved.

Substraction of rational numbers is associative.
If true then enter $1$ and if false then enter $0$

False

$a-(b-c)\neq (a-b)-c$

$\frac{1}{2}-(\frac{2}{5}-\frac{3}{4})\neq(\frac{1}{2}-\frac{2}{5})-\frac{3}{4}$

$\displaystyle\frac{18}{29}\times\displaystyle\frac{44}{35}=\displaystyle\frac{-44}{35}\times\displaystyle\frac{18}{29}$
If true then enter $1$ and if false then enter $0$

$\frac { 18 }{ 29 } \times \frac { 44 }{ 35 } =\frac { -44 }{ 35 } \times \frac { 18 }{ 29 } \\ If\frac { 18 }{ 29 } \times \frac { 44 }{ 35 } \quad \quad =x\quad \\ Then\quad \quad x\quad =\quad -x\quad is\quad false..$

$\displaystyle\frac{9}{-13}+\displaystyle\frac{4}{-16}=\displaystyle\frac{4}{-16}+$ $\displaystyle\frac{9}{-13}$ .
If true then enter $1$ and if false then enter $0$

$\displaystyle\frac{9}{-13}+\displaystyle\frac{4}{-16}=\displaystyle\frac{4}{-16}+\displaystyle\frac{9}{-13}$
Commutative property is true for addition.

Multiplication distributes over substraction of rational numbers.
If true then enter $1$ and if false then enter $0$

True
Multiplication distributes over substraction of rational numbers.

$a\times (b-c)=ab-ac$

If $\begin{pmatrix}\displaystyle\frac{-11}{14}\times\displaystyle\frac{6}{13}\end{pmatrix}+\begin{pmatrix}\displaystyle\frac{-11}{14}\times\displaystyle\frac{7}{13}\end{pmatrix}=\dfrac{m}{14}$ .Then find $-m$

$\begin{pmatrix}\displaystyle\frac{-11}{14}\times\displaystyle\frac{6}{13}\end{pmatrix}+\begin{pmatrix}\displaystyle\frac{-11}{14}\times\displaystyle\frac{7}{13}\end{pmatrix}$
Using distributive property of multiplication,

$=\dfrac{-11}{14}(\dfrac{6}{13}+\frac{7}{13})$

$=\dfrac{-11}{14}(\dfrac{6+7}{13})$

$=\dfrac{-11}{14}(\dfrac{13}{13})$

$=\dfrac{-11}{14}$

$m=-11$

$\displaystyle\frac{-17}{18}\times\displaystyle\frac{18}{-17}=...........$ .

$\displaystyle\frac{-17}{18}\times\displaystyle\frac{18}{-17}=1$
The number multiplied by its multiplicative inverse results

$1$

$\begin{pmatrix}\displaystyle\frac{-9}{13}\times\displaystyle\frac{5}{6}\end{pmatrix}-\begin{pmatrix}\displaystyle\frac{-9}{13}\times\displaystyle\frac{4}{6}\end{pmatrix}= \displaystyle\frac{-9}{13} \times\begin{pmatrix}\displaystyle\frac{5}{6}-\displaystyle\frac{4}{6}\end{pmatrix}$ .
If true then enter $1$ and if false then enter $0$

Multiplication is distributive over substraction of two rational number.

$A ( B+ C) = AB + AC$

$\begin{pmatrix}\displaystyle\frac{-9}{13}\times\displaystyle\frac{5}{6}\end{pmatrix}-\begin{pmatrix}\displaystyle\frac{-9}{13}\times\displaystyle\frac{4}{6}\end{pmatrix}=\displaystyle\frac{-9}{13}\times\begin{pmatrix}\displaystyle\frac{5}{6}-\displaystyle\frac{4}{6}\end{pmatrix}$

Check whether the following statements are true or false. If true, then state the property illustrated by the statement.
$\displaystyle\frac{-2}{3}+0=0+\begin{pmatrix}\displaystyle\frac{-2}{3}\end{pmatrix}=\displaystyle\frac{-2}{3}$
$\displaystyle\frac{-4}{5}+\displaystyle\frac{7}{8}=\displaystyle\frac{7}{8}+\begin{pmatrix}\displaystyle\frac{-4}{5}\end{pmatrix}$
If true then enter $1$ and if false then enter $0$

1) True by Additive Identity

Additive Identity is

$0$ , because adding

$0$ to a number does not change it:

$a + 0 = 0 + a = a$
2) False, since subtraction of numbers is not commutative.

The reciprocal of a positive rational number is negative.
If true then enter $1$ and if false then enter $0$

The reciprocal of a positive rational number is positive

If additive inverse of $\dfrac{4}{18}$ is added to it then the result will be ___.

$\displaystyle\frac{-4}{18}+\displaystyle\frac{4}{18}=0$
A number when added to its additive inverse results

$0$

Find the multiplicative inverse of the following.
(i) $-13$

(ii) $\dfrac{1}{5}$

(iii) $\dfrac{-5}{8}\times\dfrac{-3}{7}$

(v) $-1$

Multiplicative inverse of a number

$P$ is

$\dfrac1P$ as

$P\times\dfrac1P = 1$

i) Multiplicative inverse of

$-13$ is

$-\dfrac{1}{13}$

ii) Multiplicative inverse of

$\dfrac 15$ is

$5$

iii)

$\dfrac { -5 }{ 8 } \times \dfrac { -3 }{ 7 } =\dfrac { 15 }{ 56 }$

Multiplicative inverse =

$\dfrac { 56 }{ 15 }$

iv)

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

Multiplicative inverse

$=\dfrac {5 }{ 2 }$

v) Multiplicative inverse of

$-1$ is

$-1$ .

State what property allows you to compute $\dfrac{1}{3}\times\begin{pmatrix}6\times\dfrac{4}{3}\end{pmatrix}$ as $\begin{pmatrix}\dfrac{1}{3}\times6\end{pmatrix}\times\dfrac{4}{3}$ .

Associative property.

According to the above property, the placement of the terms can change but the end product remains the same.

$a\times (b \times c)=(a\times b) \times c=a\times b \times c$

Is $0.3$ the multiplicative inverse of $3\dfrac{1}{3}$ ? Why or why not?

$3\dfrac{1}{3}=\dfrac{10}{3}$

If it is a multiplicative inverse of

$3\dfrac13$ then the product should be

$1$

$0.3\times\dfrac{10}{3}=\dfrac{3}{10}\times\dfrac{10}{3}=1$

Product

$=1$

$\therefore\;0.3$ is a Multiplicative Inverse of

$3\dfrac{1}{3}$ .

Name the property indicated in the following:
$\dfrac {8}{17} + \dfrac {-8}{17} = 0$

Fact: If sum of two real numbers are zero, they numbers are called additive inverse of each other. We can see that in our example

$\dfrac{8}{17}$ and

$\dfrac{-8}{17}$ are additive inverse of each other as,

$\dfrac{8}{17}+\dfrac{-8}{17}=0$

Hence

$\dfrac{8}{17}+\dfrac{-8}{17}=0$ is an application of additive inverse property.

Identify the property in the following statements:
$2 \cdot 8 = 8\cdot 2$

We know that commutative law for multiplication

$a\cdot b=b\cdot a$

Let us take

$a = 2 \ and \ b = 8$ , we can see the application of the above law

Hence

$2\cdot 8=8\cdot 2$ is commutative law for multiplication.

Identify the property in the following statements:
$2 + (3 + 4) = (2 + 3) + 4$

We know that:

$a+(b+c)=(a+b)+c$ is called associative law for addition

In the given question if we take

$a =2 , b= 3 , c= 4$ then we can see the application of the above law.

Hence above property is associative law for addition.

Represent $\dfrac{-2}{11}$ , $\dfrac{-5}{11}$ , $\dfrac{-9}{11}$ on the number line.

Check the commutative property of multiplication for $\dfrac {-7}{13}, \dfrac {25}{27}$

Commutative property of multiplication for any element

$a,b$ is given by

$a \times b = b \times a$
To check,

$\dfrac{-7}{13} \times \dfrac{25}{27} = \dfrac{25}{27} \times \dfrac{-7}{13}$

LHS

$=\dfrac{-7}{13} \times \dfrac{25}{27} = \dfrac{(-7) \times 25}{13 \times 27} = \dfrac{-175}{351}$

RHS

$=\dfrac{25}{17} \times \dfrac{-7}{13} = \dfrac{(-7) \times 25}{13 \times 27} = \dfrac{-175}{351}$

$\therefore$ RHS = LHS

Hence,

$\dfrac{-7}{13}, \dfrac{25}{27}$ are commutative with respect to multiplication..

Prove that $(-1)\times (-1)=1$

Since 1 is the multiplicative identity, we know that

$1\times (-1) = -1$ . Using distributive property,

$0=(1-1)\times (-1) = 1\times (-1)+(-1)\times (-1)$ .
Thus

$(-1)\times (-1)+(-1)=0$ . Adding 1 both sides, we get

$(-1)\times (-1) = 1$

Write the rational number that does not have any reciprocal.

${\textbf{Step - 1: Defining rational numbers}}$

${\text{Rational numbers are those which can be written in the form of }}\dfrac{{\text{p}}}{{\text{q}}}{\text{, where q}} \ne {\text{ 0}}$

${\textbf{Step - 2: Writing rational number which does not have reciprocal}}$

${\text{The rational number which does not has reciprocal is 0,}}$

${\text{as 0 can be written as }}\dfrac{{\text{0}}}{{\text{1}}}{\text{, so its reciprocal will be }}\dfrac{{\text{1}}}{{\text{0}}}{\text{ which is not defined}}{\text{.}}$

${\textbf{Thus, the rational number which does not has reciprocal is 0}}$

Check the commutative property of multiplication for the following pairs:
$\dfrac {-8}{9}, \dfrac {-17}{19}$

Commutative property of multiplication is given by

$a \times b = b \times a$
To check,

$\dfrac{-8}{9} \times \dfrac{-17}{19} = \dfrac{-17}{19} \times \dfrac{-8}{9}$
Consider LHS,

$=\dfrac{-8}{9} \times \dfrac{-17}{19}$

$=\dfrac{(-8) \times -17}{9 \times 19}$ =

$\dfrac{136}{171}$

RHS

$=\dfrac{-17}{19} \times \dfrac{-8}{9}$

$= \dfrac{(-17) \times -8}{19 \times 9}$ =

$\dfrac{136}{171}$

$\therefore$ RHS = LHS

Hence,

$\dfrac{-8}{9}, \dfrac{-17}{19}$ are commutative with respect to multiplication.

Check the commutative property of addition for $\dfrac {-7}{9}, \dfrac {-18}{19}$

Commutative property of addition for any element

$a,b$ is given by

$a + b = b + a$
To check,

$\dfrac{-7}{9} + \dfrac{-18}{19} = \dfrac{-18}{19} + \dfrac{-7}{9 }$
Consider LHS,

$=\dfrac{-7}{9} + \dfrac{-18}{19}$ =

$\dfrac{[(-7) \times 19+(-18) \times 9)]}{9 \times 19}$ =

$\dfrac{-133 -162}{171}$ =

$\dfrac{-295}{171}$
RHS,

$=\dfrac{-18}{19} + \dfrac{-7}{9 } = \dfrac{[(-18) \times 9+(-7) \times 19)]}{19 \times 9}$ =

$\dfrac{-162-133}{171} = \dfrac{-295}{171}$

$\therefore$ RHS = LHS

Hence,

$\dfrac{-7}{9}, \dfrac{-18}{19}$ are commutative.

Find the additive inverse of each of the following numbers:
$\dfrac {8}{5}, \dfrac {6}{10}, \dfrac {-3}{8}, \dfrac {-16}{3}, \dfrac {-4}{1}$

Two numbers are called additive inverse of each other if their sum is

$0$

Hence additive inverse a number

$a$ is

$-a$

Then the additive inverse of

$\dfrac{8}{5},\dfrac{6}{10},-\dfrac{3}{8},-\dfrac{16}{3}, -\dfrac{4}{1}$

are

$-\dfrac{8}{5},-\dfrac{6}{10},\dfrac{3}{8},\dfrac{16}{3}, \dfrac{4}{1}$ respectively

Check the commutative property of multiplication for $\dfrac {22}{4}, \dfrac {3}{4}$

Commutative property of multiplication for any element

$a,b$ is given by

$a \times b = b \times a$
To check,

$\dfrac{22}{4} \times \dfrac{3}{4} = \dfrac{3}{4} \times \dfrac{22}{4}$
Consider,

LHS

$=\dfrac{22}{4} \times \dfrac{3}{4} = \dfrac{22 \times 3}{4 \times 4} = \dfrac{66}{16}$

RHS

$=\dfrac{3}{4} \times \dfrac{22}{4 }$ =

$\dfrac{3 \times 22}{4 \times 4} = \dfrac{66}{16}$

$\therefore$ RHS = LHS

Hence,

$\dfrac{22}{4}, \dfrac{3}{4}$ are commutative with respect to multiplication.

Write the additive inverse of the given rational number:
$\dfrac{-2}{-3}$

The additive inverse of a number is that number, when added to the given number, gives a sum of zero.

$\displaystyle \frac{-2}{-3} = \frac{-1 \times 2}{-1 \times 3} = \frac{2}{3}$

So, the additive inverse of

$\displaystyle \frac{2}{3}$ is

$\displaystyle -\frac{2}{3}$ as

$\displaystyle \frac{2}{3}\, +\, \displaystyle (-\frac{2}{3})\, =\, 0$

$\dfrac{7}{-13}$ .

The additive inverse of a number is a number that is when added to the given number, gives a sum of zero.

So, the additive inverse of

$\displaystyle \frac{7}{-13}$ is

$\displaystyle \frac{7}{13}$ because

$\displaystyle -\frac{7}{13}\, +\, \displaystyle \frac{7}{13}\, =\, 0.$

Mark the following rational numbers on the number line.
$\cfrac { 10 }{ 3 }$

Draw 10/3 on a number line.

As the number 10/3 a positive number so it will be on right of zero. So the rational number lies between 3 and 4.

That is represented on above diagram.
1047325_704026_ans_f2edac5e04b34edcb4552adfb7a54394.png

Mark the following rational numbers on the number line.
$\cfrac { 3 }{ 2 }$

Since the given rational number is greater than one. So, it will always be represented on the right side of one on the number line. So, first of all we need to divide the number line between zero and two into 4 equal parts and the third part of the four parts will be the representation of $\dfrac{3}{2}$ on number line. It can be represented as shown in the number line.

1025517_704024_ans_0f99c77501864f8e8728602862ba340c.jpg

Name the property used in each of the following:
$\dfrac {-1}{2}+0=\dfrac {-1}{2}=0+\dfrac {-1}{2}$

Since

$0$ is the identity element of the set of rationals under the addition.

$a+0=a=0+a$ .

This property exists for every

$a\in \mathbb{Q}$ ,

So, the given property

$\dfrac {-1}{2}+0=\dfrac {-1}{2}=0+\dfrac {-1}{2}$ , is the identity property on set of rationals.

Mark the following rational numbers on the number line.
$\cfrac { 3 }{ 4 }$

Since the given rational number is greater than zero. So, it will always be represented on the right side of zero on the number line. So, first of all we need to divide the number line between zero and one into 4 equal parts and the third part of the four parts will be representation of $\dfrac{3}{4}$ on number line. It can be represented as.

1025507_704021_ans_aee02e8a3b0344fb89a8e29bddfaefa5.jpg

Write the multiplicate inverse of
$\dfrac { 3 }{ 2 }$

We say that the number

$1$ the multiplicative identity because if we multiply the any number with

$1$ , we get the same number as our answer.

For exam we take a number

$a$ such that if we multiply it with

$b$ we get

$1$ i.e.

$b$ is multiplicative inverse of

$a$

$a \times b=1$

Now, Multiplicative inverse of

$\frac{3}{2}$ is

$\frac{2}{3}$

$\frac{3}{2} \times \frac{2}{3}=1$

Are rational number always closed under subtraction?

Yes rational numbers are closed under subtraction.

As for

$a,b$ being rational number,

$a-b$ or

$b-a$ is always going to be a rational number.

Write the additive inverse of $\left( \dfrac { -3 }{ 10 } +\dfrac { 6 }{ 10 } \right)$

Additive inverse of

$\dfrac{{ - 3}}{{10}} + \dfrac{6}{{10}}$ is

$\dfrac{3}{{10}} + \dfrac{{\left( { - 6} \right)}}{{10}}$

Write the multiplicate inverse of:
$-5\dfrac { 6 }{ 7 }$

$- 5\frac{6}{7} = - \frac{{35 + 6}}{7} = \frac{{ - 41}}{7}$

Multiplicative inverse of

$- 5\frac{6}{7} = \frac{{ - 7}}{{41}}$

Tell what property allows you to compute $\frac{1}{3} \times \left( {6 \times \frac{4}{3}} \right)$ as $\left( {\frac{1}{3} \times 6} \right) \times \frac{1}{4}$

It is form of :

$a(b\times c) =(a\times b) \times c$ , which we call this property as associative property

Therefore given expression is computed by associative property

Verify the following:
$\left( \dfrac { -9 }{ 5 } \times \dfrac { -10 }{ 3 } \right) \times \dfrac { 21 }{ -4 } =\dfrac { -9 }{ 5 } \times \left( \dfrac { -10 }{ 3 } \times \dfrac { 21 }{ -4 } \right)$

$L.H.S = \left( {\dfrac{{ - 9}}{5} \times \dfrac{{ - 10}}{3}} \right) \times \frac{{21}}{{ - 4}}$ d

$= {\left( 6 \right)^3} \times \dfrac{{21}}{{ - 4}}$

$= \dfrac{{ - 63}}{2}$

$R.H.S = \dfrac{{ - 9}}{5} \times \left( {\dfrac{{ - 10}}{3} \times \dfrac{{21}}{{ - 4}}} \right)$

$= \dfrac{{ - 9}}{5} \times \left( {\dfrac{{35}}{2}} \right)$

$= \dfrac{{ - 63}}{2}$

$L.H.S = R.H.S$

Find six rational numbers between $- \dfrac{1}{4}$ and $- \dfrac{1}{2}$ .

Consider the given equation is,

we know that, if

$a$ and

$b$ are two rational number such that

$a<b$ then

$\dfrac{1}{2}(a+b)$ is a rational number between us.

A rational number between

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

$\dfrac{-1}{2}$

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

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

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

A rational number between

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

$\dfrac{-3}{8}$

$=\dfrac{1}{2}\left ( \dfrac{-1}{4}+\dfrac{-3}{8} \right )$

$=\dfrac{1}{2}\left ( \dfrac{-5}{8}\right )$

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

A rational number between

$\dfrac{-3}{8}$ and

$\dfrac{-5}{16}$

$=\dfrac{1}{2}\left ( \dfrac{-3}{16}+\dfrac{-5}{8} \right )$

$=\dfrac{1}{2}\left ( \dfrac{-11}{16} \right )$

$=\dfrac{-11}{32}$

A rational number between

$\dfrac{-3}{8}$ and

$\dfrac{-1}{2}$

$=\dfrac{1}{2}\left ( \dfrac{-3}{8}+\dfrac{-1}{2} \right )$

$=\dfrac{1}{2}\left ( \dfrac{-7}{8}\right )$

$=\dfrac{-7}{16}$

A rational number between

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

$\dfrac{-7}{16}$

$=\dfrac{1}{2}\left ( \dfrac{-1}{2} +\dfrac{-7}{16}\right )$

$=\dfrac{1}{2}\left ( \dfrac{-15}{16} \right )$

$=\dfrac{-15}{32}$

A rational number between

$\dfrac{-11}{32}$ and

$\dfrac{-15}{32}$

$=\dfrac{1}{2}\left ( \dfrac{-11}{32} +\dfrac{-15}{32}\right )$

$=\dfrac{1}{2}\left ( \dfrac{-26}{32} \right )$

$=\dfrac{-26}{64}$

Hence, six rational number are

$\dfrac{-3}{8},\dfrac{-5}{16},\dfrac{-11}{32},\dfrac{-7}{16},\dfrac{-15}{32},\dfrac{-26}{64}$ .

Mark the following rational number on the number line.
$\dfrac { 5 }{ 2 }$

1074754_1053548_ans_5f916d3b369e4f2a86b4d1628bd2c130.png

Write the multiplicate inverse of $-5-\dfrac { 1 }{ -5 }$

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

$= - 5 + \dfrac{1}{5}$

$= \dfrac{{1 - 25}}{5}$

$= \dfrac{{ - 24}}{5}$

So , multiplicative inverse of $- 5 - \dfrac{1}{{ - 5}}$ $= \dfrac{{ - 5}}{{24}}$

Verify the following:
$\left( \dfrac { 5 }{ 7 } \times \dfrac { 12 }{ 13 } \right) \times \dfrac { 7 }{ 18 } =\dfrac { 5 }{ 7 } \left( \dfrac { 12 }{ 13 } \times \dfrac { 7 }{ 18 } \right)$

$L.H.S = \left( {\dfrac{5}{7} \times \dfrac{{12}}{{13}}} \right) \times \dfrac{7}{{18}}$

$= \dfrac{{60}}{{11}} \times \dfrac{7}{{18}}$

$= \dfrac{{420}}{{1638}}$

$R.H.S = \dfrac{5}{7}\left( {\dfrac{{12}}{{13}} \times \dfrac{7}{{18}}} \right) = \dfrac{5}{7}\left( {\dfrac{{84}}{{234}}} \right) = \dfrac{{420}}{{1638}}$

$Since\,L.H.S = R.H.S\,proved$

Insert two rational numbers between $1$ and $2$

First rational number

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

Second rational number

$=\dfrac{1+\dfrac{3}{2}}{2}=\dfrac{3}{2}=\dfrac{5}{4}$

Hence, this is the answer.

Find $12$ rational number between $-1$ and $2$ ?

Find 12 Rational Numbers between -1 and 2

as we have to find 12 Rational Numbers

we multiple the Numbers by

$\dfrac{13}{13}$

$-1\times \dfrac{13}{13}=\dfrac{-13}{13}$

$2\times \dfrac{13}{13}= \dfrac{26}{13}$

thus, 12 Rational Numbers between -1 and 2

are

$\dfrac{-12}{13},\dfrac{-11}{13},\dfrac{-10}{13},\dfrac{-9}{13},\dfrac{-8}{13},\dfrac{-7}{13},\dfrac{-6}{13}$

$\dfrac{-5}{13},\dfrac{-4}{13},\dfrac{-3}{13},\dfrac{-2}{13},\dfrac{-1}{13}....$

1159113_1144014_ans_047d4b194a244ec58f200e4dc032f113.jpg

Find the multiplicative inverse of $3$ and $-7$ .

The multiplicative identity for rational numbers with out

$0$ is

$1$ . [ Since the rational numbers forms a group with respect to multiplication of rational numbers]

Now the multiplicative inverse of

$3$ and

$-7$ are

$\dfrac{1}{3}$ and

$-\dfrac{1}{7}$ as

$3.\dfrac{1}{3}=\dfrac{1}{3}.3=1$ and

$(-7).-\dfrac{1}{7}=-\dfrac{1}{7}.(-7)=1$ .

Represents these number on number line
$\dfrac {-5}{6}$

Consider the following number.

$\dfrac{-5}{6}$

To mark $\dfrac{-5}{6}$ ; move five parts on the left-side of zero.

Hence, this is the correct answer.

1053355_1085551_ans_c3d101e670bf4f829560719845370a3a.png

Write the additive inverse of the $\dfrac{3}{8}$

$={\dfrac 38}$

${\text{let}}\;{\mkern 1mu} x\;{\mkern 1mu} {\text{be}}{\mkern 1mu} \;{\text{additive}}\;{\mkern 1mu} {\text{inverse,}}\;{\mkern 1mu} {\text{than}}\;{\mkern 1mu} \dfrac{3}{8} + x = 0$

$x = - {\dfrac 3 8}\left( {additive\,inverse} \right)$

Find six rational numbers between $\dfrac {3}{7}$ and $\dfrac {4}{8}$

The given numbers can be written as

$\dfrac{6}{14},\dfrac{7}{14}$

So, six rational number between these are,

$\dfrac{7}{15},\dfrac{8}{5},\dfrac{7}{16},\dfrac{9}{20},\dfrac{13}{28},\dfrac{33}{70}$

Write three rational numbers that lie between the two given numbers.
$\dfrac{2}{7}, \dfrac{6}{7}$

Solve by taking LCM of denominator,

Three rational no. between given numbers are,

$\dfrac{3}{7},\dfrac{4}{7},\dfrac{5}{7}$

Find a rational number between $\dfrac{1}{5}$ and $\dfrac{3}{4}$

1114188_1162448_ans_d448646e351348dcbae430b2b30d4bb0.jpeg

Insert the rational number between $7$ and $8$ .

A rational number between

$7$ and

$8$ is

$\cfrac { 7+8 }{ 2 } =\cfrac { 15 }{ 2 }$

Fill in the blanks
Multiplicative inverse of $2^{-16}$ is _____ .

When a number is multiplied by its multiplicative inverse, it should be equal to $1$ .

$“n”$ is a number, then the multiplicative inverse or reciprocal of a number is

$\dfrac{1}{n}.$

Multiplicative inverse of

$2^{-16}$ is

$\dfrac{1}{2^{-16}}=2^{16}$

Find a rational number between $\dfrac {1}{4}$ and $\dfrac {1}{3}$

Let us consider a rational number between

$\dfrac{1}{4}$ and

$\dfrac{1}{3}$ be

$x$
We can find the rational number using the formula

$x = \dfrac {1}{2} (\dfrac{1}{4} + \dfrac{1}{3} )$

So,

$x = \dfrac {1}{2} (\dfrac{1}{4} + \dfrac{1}{3} )$

$= \dfrac {1}{2} (\dfrac{3+4}{12})$

$= \dfrac {1}{2} \times \dfrac{7}{12}$

$= \dfrac{7}{24}$

$\therefore$ a rational number between

$\dfrac {1}{4}$ and

$\dfrac{1}{3}$ is

$\dfrac{7}{24}$ .

Write a rational number which does not lie between the rational numbers $-\dfrac{2}{3}$ and $-\dfrac{1}{5}$ .

Given: Two rational number

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

$\dfrac{-1}{5}$

Rational number which does not lies between the above rational numbers can be found as follows:

$\dfrac{-2\times5}{3\times5}$ &

$\dfrac{-1\times3}{5\times3}$

$\dfrac{-10}{15}$ &

$\dfrac{-3}{15}$ By taking LCM

so number which does not lies betwewn

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

$\dfrac{-1}{5}$ are

$\dfrac{-2}{15}$ ,

$\dfrac{-1}{15}$ and continue if you want to move but don't choose between

$\dfrac{-10}{15}$ &

$\dfrac{-3}{15}$

Fill in the blank:
The quotient when a rational number is divided by its additive inverse is ________

The additive inverse of a number

$x$ is

$-x$ .

Therefore, quotient when

$x$ is divide by

$-x$ is

$-1$ .

Verify the following $\dfrac { 1 } { 2 } - \left( \dfrac { 3 } { 4 } - \dfrac { 5 } { 4 } \right)= \left(\dfrac { 1 } { 2 } - \dfrac { 3 } { 4 } \right)- \left(\dfrac { 5 } { 4 } \right)$

To verify

$\dfrac{1}{2}-\left [ \dfrac{3}{4}-\dfrac{5}{4} \right ] =\left [ \dfrac{1}{2}-\dfrac{3}{4} \right ]-\left [ \dfrac{5}{4} \right ]$

L H S

$=\dfrac{1}{2}-\left [ \dfrac{3}{4}-\dfrac{5}{4} \right ] =\dfrac{1}{2}-\left [ \dfrac{-2}{4} \right ]$

$=\dfrac{1}{2}+\dfrac{1}{2}=1$

R H S

$=\left [ \dfrac{1}{2}-\dfrac{3}{4} \right ]-\left [ \dfrac{5}{4} \right ] =\dfrac{-1}{4}- \dfrac{5}{4}=\dfrac{-6}{4}$

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

$\therefore LHS\neq RHS \Rightarrow$ They are not equal.

Find 1 rational numbers between 2.1 and 2.2

The rational number between

$a$ and

$b$ is given as

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

Here

$a=2.1$ and

$b=2.2$

So the rational number is given as

$\dfrac{2.1+2.2}2=\dfrac{4.3}2=2.15$

How many rational numbers are there in between two given rational numbers?

There are countless rational numbers(by density property of rational numbers)

Find the additive inverse of the number $\dfrac{2}{8}$ .

The additive inverse of

$\dfrac{2}{8}$ is

$-\dfrac{2}{8}$ .

Since,

$\dfrac{2}{8}+\left(-\dfrac{2}{8}\right)=0$

Represent each of the following number on the number line:
$1\dfrac {3}{4}$ .

We know that

$1\dfrac {3}{4} = ((1 \times 4) + 3)/4 = 7/4$ is greater than

$1$ and lesser than

$2$ .

$\therefore$ it lies between

$1$ and

$2$ . It can be represented on number line as.
1602774_1366067_ans_7d812e3949a54c55adf5b162909e2017.jpg

Name the property used in each of the following:
$\dfrac {-3}{2}+\left(\dfrac {1}{3}+\dfrac {4}{7}\right)=\left(\dfrac {-3}{5}+\dfrac {1}{3}\right)+\dfrac {4}{7}$

Given,

$\dfrac {-3}{2}+\left(\dfrac {1}{3}+\dfrac {4}{7}\right)=\left(\dfrac {-3}{5}+\dfrac {1}{3}\right)+\dfrac {4}{7}$ .

This property is called the associative property of rational numbers.

Since for all

$a,b,c\in \mathbb{Q}$ we've,

$a+(b+c)=(a+b)+c$ .

Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number:
$\dfrac{2}{5}+\dfrac{8}{3}+\dfrac{-11}{15}+\dfrac{4}{5}+\dfrac{-2}{3}$

$\dfrac{2}{5}+\dfrac{8}{3}+\dfrac{-11}{15}+\dfrac{4}{5}+\dfrac{-2}{3}$

$= \dfrac{2}{5} +\dfrac{4}{5}+\dfrac{8}{3} + \dfrac{-2}{3} + \dfrac{-11}{3}$

$= \dfrac{2+4}{5} + \dfrac{8-2}{3} + \dfrac{-11}{15}$

$= \dfrac{6}{5} + \dfrac{6}{3} + \dfrac{-11}{15}$

$= \dfrac{6}{5} + 2 + \dfrac{-11}{15}$

$= \dfrac{6+2\times 5}{5} + \dfrac{-11}{15}$

$= \dfrac{16}{5} + \dfrac{-11}{15}$

$= \dfrac{16\times 3 + (-11)}{15}$

$=\dfrac{48 -11}{15}$

$= \dfrac{37}{15}$

Write the negative (additive inverse) of $1$

We know that

$1 + (-1) = 1 - 1 = 0$

$\therefore\ -1$ is the additive inverse of

$1$

Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number:
$\dfrac{3}{7}+\dfrac{-4}{9}+\dfrac{-11}{7}+\dfrac{7}{9}$

$\dfrac{3}{7} + \dfrac{-4}{9} + \dfrac{-11}{7} + \dfrac{7}{9}$

$= \dfrac{3}{7} + \dfrac{-11}{7} + \dfrac{-4}{9} + \dfrac{7}{9}$

$= \dfrac{3-11}{7} + \dfrac{-4+7}{9}$

$= \dfrac{-8}{7} + \dfrac{3}{9}$

$= \dfrac{-8}{7} + \dfrac{1}{3}$

$= \dfrac{-8\times 3 + 1\times 7}{21}$

$= \dfrac{-21+7}{21}$

$= \dfrac{-17}{21}$

Write the negative (additive inverse) of each of the following:
$\dfrac{-16}{13}$

To find the additive inverse of

$\dfrac{-16}{13}$

Since,

$\dfrac{-16}{13} + \dfrac{16}{13} = \dfrac{-16+16}{13} = \dfrac{0}{13} = 0$

Hence, additive inverse of

$\dfrac{-16}{13}$ is

$\dfrac{16}{13}$ .

Use the distributivity of multiplication of rational numbers over their addition to simplify:
$\dfrac{-5}{4}\times\left(\dfrac{8}{5}+\dfrac{16}{5}\right)$

1508879_1671477_ans_59e1274ceeee4ad4b64d1c549d7701fe.jpg

Write the negative (additive inverse) of each of the following:
$\dfrac{-5}{1}$

Since

$\dfrac{-5}{1} + \dfrac{5}{1} = \dfrac{-5+5}{1} = \dfrac{0}{1} = 0$

$\Rightarrow \dfrac{5}{1}$ is additive inverse of

$\dfrac{-5}{1}$ .

Write the negative (additive inverse) of each of the following:
$-1$

Since

$1 + (-1) = 1 - 1 = 0$

$\Rightarrow 1$ is the additive incerse of

$-1$ .

Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number:
$\dfrac{4}{7}+0+\dfrac{-8}{9}+\dfrac{-13}{7}+\dfrac{17}{21}$

$\dfrac{4}{7}+0+\dfrac{-8}{9}+\dfrac{-13}{7}+\dfrac{17}{21}$

$= \dfrac{4}{7} + \dfrac{-13}{7} + \dfrac{-8}{9} + \dfrac{17}{21}$

$=\dfrac{4-13}{7} + \dfrac{-8}{9} + \dfrac{17}{21}$

$= \dfrac{-9}{7} + \dfrac{-8}{9} + \dfrac{17}{21}$

$= \dfrac{-9}{7} + \dfrac{17}{21} + \dfrac{-8}{9}$

$= \dfrac{-9\times 3 + 17 \times 1}{21} + \dfrac{-8}{9}$

$= \dfrac{-27+17}{21} + \dfrac{-8}{9}$

$= \dfrac{-10}{21} + \dfrac{-8}{9}$

$= \dfrac{-10 \times 3-8\times 7}{63}$

$=\dfrac{-30-56}{63}$

$= \dfrac{-86}{63}$

Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number:
$\dfrac{2}{5}+\dfrac{7}{3}+\dfrac{-4}{5}+\dfrac{-1}{3}$

$\dfrac{2}{5} + \dfrac{2}{7} + \dfrac{-4}{5} + \dfrac{-1}{3}$

$= \left(\dfrac{2}{5} + \dfrac{-4}{5}\right) + \left(\dfrac{7}{3} + \dfrac{-1}{3}\right)$

$= \dfrac{2-4}{5} + \dfrac{7-1}{3} = \dfrac{-2}{5} + \dfrac{6}{3}$

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

$= \dfrac{8}{5}$

Write the negative (additive inverse) of each of the following:
$0$

Since,

$0 + 0 = 0$

$\Rightarrow 0$ is its own inverse.

In each of the following cases, write the rational number whose numerator and denominator are respectively as under:
$25+15$ and $81 \div 40$

$25+15$ and

$81÷40$
Let numerator

$p=25+15=40$

Denominator

$q=81\div 40=\dfrac{81}{40}$

Hence,rational number,

$\dfrac{p}{q}=\dfrac{40}{\frac{81}{40}}=\dfrac{40\times 40}{81}=\dfrac{1600}{81}$

Use the distributivity of multiplication of rational numbers over their addition to simplify:
$\dfrac{2}{7}\times\left(\dfrac{7}{16}-\dfrac{21}{4}\right)$

1508880_1671479_ans_b07a628bec794720a0884506d3289828.jpg

Find a rational number exactly halfway between:
$\dfrac{5}{-13}$ and $\dfrac{-7}{9}$

As we know that exactly halfway between two rational numbers is

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

Rational numbers are

$\dfrac{5}{-13}$ and

$\dfrac{-7}{9}$

Here,

$a=\dfrac{5}{-13}$ and

$b=\dfrac{-7}{9}$

$\therefore \dfrac{a+b}{2}=\dfrac{\dfrac{5}{-13}+\dfrac{-7}{9}}{2}$

$=\dfrac{\dfrac{-45-91}{117}}{2}$

$=\dfrac{-136}{117\times2}$

$=\dfrac{-136}{234}$

Therefore, the exactly halfway between

$\dfrac{5}{-13}$ and

$\dfrac{-7}{9}$ is

$\dfrac{-136}{234}.$

Find a rational number exactly halfway between:
$\dfrac{1}{15}$ and $\dfrac{1}{12}$

As we know that exactly halfway between two rational numbers is $\dfrac{a+b}{2}$

Rational numbers are $\dfrac{1}{15}$ and $\dfrac{1}{12}$

Here, $a=\dfrac{1}{15}$ and $b=\dfrac{1}{12}$

$\therefore\dfrac{a+b}{2}=\dfrac{\dfrac{1}{15}+\dfrac{1}{12}}{2}$

$=\dfrac{\dfrac{4+5}{60}}{2}$

$=\dfrac{9}{60\times2}$

$=\dfrac{3}{20\times2}$

$=\dfrac{3}{40}$

Therefore, the exactly halfway between $\dfrac{1}{15}$ and $\dfrac{1}{12}$ is $\dfrac{3}{40}$ .

Represent the following number on the number line:
$-3$ .

The value

$-3$ can be represented on number line as after

$3$ divisions on the left of

$0.$

1602907_1771610_ans_4dbafc2bab2e432aa397e9abe0d0e715.jpg

Use the distributivity of multiplication of rational numbers over their addition to simplify:
$\dfrac{3}{4}\times\left(\dfrac{8}{9}-40\right)$

1508883_1671480_ans_2573c22986f64f708fedf3eca260a925.jpg

Fill in the blanks to make the statement true:
The negative of a negative rational number is always a ............... rational number.

Let

$-x$ be the negative rational number.

then,

$-(-x) = x$

Hence, The negative of a negative rational number is always a

$\text{ positive }$ rational number.

Fill in the blanks to make the statement true:
The rational numbers $\dfrac{1}{3}$ and $\dfrac{-1}{3}$ are on the ........... sides of zero on the number line.

The rational number

$\dfrac13$ is on the right side of zero and

$\dfrac{-1}3$ is on the left side of zero.

Hence, The rational numbers

$\dfrac13$ and

$\dfrac{−1}3$ are on the

$\text{opposite}$ sides of zero on the number line.

Fill in the blanks to make the statement true:
Rational numbers can be added or multiplied in any ..............

Rational numbers can be added or multiplied in any

$\text{ order}$

Tell which property allows you to compute
$\dfrac{1}{5} \times \left[ \dfrac{5}{6} \times \dfrac{7}{9} \right]$ as $\left[\dfrac{1}{5} \times \dfrac{5}{6}\right] \times \dfrac{7}{9}$

Associative property of multiplication.

Find the multiplicative inverse of $-1 \dfrac{1}{8}$

We have,

$-1 \dfrac{1}{8} = \dfrac{-9}{8}$

$\therefore$ multiplicative inverse

$= \dfrac{-8}{9}$

Name the property used in the following-
$\dfrac{1}{3} + \left[\dfrac{4}{9} + \left(\dfrac{-4}{3} \right)\right] = \left[\dfrac{-2}{3} + \dfrac{4}{9} \right] + \left(\dfrac{-4}{3}\right)$

Associative property of addition of rational numbers

Name the property used in the following-
$\dfrac{3}{8} \times 1 = 1 \times \dfrac{3}{8} = \dfrac{3}{8}$

Commutative property of multiplication of rational numbers

Name the property used in the following-
$\dfrac{-2}{3} \times \left[\dfrac{3}{4} + \dfrac{-1}{4} \right] = \left[\dfrac{-2}{3} \times \dfrac{3}{4} \right] + \left[\dfrac{-2}{3} \times \dfrac{-1}{4} \right]$

Distributive property of multiplication over addition

Find the multiplicative inverse of $3\dfrac{1}{3}$

We have,

$3\dfrac{1}{3} = \dfrac{10}{9}$

$\therefore$ multiplicative inverse

$= \dfrac9{10}$

Name the property used in the following-
$\dfrac{-2}{7} + 0 = 0 + \dfrac{2}{7} = \dfrac{-2}{7}$

Commutative property of addition of rational numbers

Tell which property allows you to compare
$\dfrac{2}{3} \times \left[\dfrac{3}{4} \times \dfrac{5}{7}\right]$ as $\left[\dfrac{2}{3} \times \dfrac{5}{7}\right] \times \dfrac{3}{4}$

Associative property of multiplication of rational numbers.

Name the property used in the following-
$-\dfrac{7}{11} \times \dfrac{-3}{5} = \dfrac{-3}{5} \times \dfrac{-7}{11}$

Commutative property of multiplication of rational numbers

Find the sum of additive inverse and multiplicative inverse of $7$ .

$\textbf{Step -1: Finding additive inverse and multiplicative inverse .}$

$\text{Additive inverse of }7=-7$

$\text{Multiplicative inverse of }7=\dfrac17$

$\textbf{Step -2: Adding both the Inverse }$

$\text{Additive inverse + Multiplicative inverse}=-7+\dfrac17$

$\implies \dfrac{-48}7$

$\textbf{Hence, the sum of additive inverse and multiplicative inverse of 7 is }\mathbf{-\dfrac{48}7 .}$

Enter 1 if it is true else enter 0.
$\dfrac{-5}{16}$ .

The correct answer is 1.

1887673_1802391_ans_fcc6b5c94d92448fb408a0a39cdb3da6.jpeg

Verify commutative property of addition for the following pairs of rational numbers.
$9/11$ and $2/13$ .

$9/ 11$ and

$2 / 13$
Adding both the numbers,

$= 9 / 11 + 2 / 13$

$= (117 + 22) / 143$

$= 139 / 143$

And

$2 / 13 + 9 / 11$

$= (22 + 117) / 143$

$= 139 / 143$

Therefore,

$9 / 11 + 2 / 13 = 2 / 13 + 9 / 11$ .

Hence, commutative property is verified.

Enter 1 if true else enter 0.
The rational number $\dfrac{-5}{11}$ lies to the left of zero on the number line

Ans :- 1

Given number is negative.

So, It wil lie left of zero on the number line.

Find the additive inverse of the following rational numbers:
$2/-3$ .

Given

$2/-3$
Additive inverse of

$2/-3 = -(2/-3)$
We get,

$= 2/3$ .

Enter 1 if true else enter 0.
The number $\dfrac{-4}{7}$ lies to the left of zero on the number line.

Ans :- 1

Given number is negative.

So, It will lie left of zero on the number line.

Verify commutative property of addition for the following pairs of rational numbers.
$-4/3$ and $3/7$ .

The commutative property of addition of rational numbers state that:

$a+b=b+a$ , where

$a, b$ are rational numbers.

$a =\dfrac{-4}{3}$ and

$b =\dfrac{3}{7}$
Calculating

$a+b$ :

$a+b= \dfrac{-4}{3}+\dfrac{3}{7}$
Taking L.C.M. we get,

$=\dfrac{-28+9}{21}$

$= \dfrac{- 19}{ 21}$

Calculating

$b+a$ :

$b+a= \dfrac{3}{7}+\dfrac{-4}{3}$
Taking L.C.M. we get,

$=\dfrac{9-28}{21}$

$= \dfrac{- 19}{ 21}=a+b$

Therefore,

$a+b=b+a$ .

Hence, the commutative property of addition is verified.

Using the appropriate properties of operations of rational numbers, evaluate the following:
$2/5\times - 3/7 - 1/14 - 3/7 \times 3/5$ .

$2/5 \times -3/7 - 1/14 - 3/7 \times 3/5$

$= 2/5 \times -3/7 - 3/7 \times 3/5 - 1/14$
Taking common term, we get

$= -3/7 (2/5 + 3/5) - 1/14$

$= -3/7\times (2 + 3) / 5 - 1/14$

$= -3/7 - 1/14$
Taking L.C.M. we get,

$= (-6 - 1)/ 14$

$= -7/14$

$= (-7\div 7)/ (14\div 7)$
We get,

$= -1/2$ .

Verify commutative property of multiplication for the following pairs of rational numbers:
$-7/-20$ and $5/-14$ .

$-7/-20$ and

$5/-14$

$-7/-20 = \left \{-7 \times (-1)\right \}/ \left \{-20\times (-1)\right \}$

$= 7/20$
Now,

$7/20$ and

$5/-14$

$7/20\times 5/-14$

$= (7\times 5) / 20\times (-14)$

$= 35 / -280$
and

$5/-14 \times 7/20$

$= (5\times 7)/ (-14\times 20)$

$= 35/ -280$
Therefore,

$7/20 \times 5 / -14 = 5/ -14 \times 7/20$ .

Find the additive inverse of the following rational number:
$\dfrac{-7}{-12}$

Additive inverse of any real number

$a$ is equal to

$-a$ so, that it can follow

$a+(-a)=0$ .

Hence,

$a=\dfrac{-7}{-12}$

$=\dfrac{7}{12}$

And,

$(-a)=-\dfrac{7}{12}$

Hence additive inverse of

$\dfrac{-7}{-12}$ is

$-\dfrac{7}{12}$ .

Verify commutative property of multiplication for the following pairs of rational numbers:
$13\dfrac {1}{3}$ and $1\dfrac {1}{8}$ .

$13\dfrac {1}{3}$ and

$1\dfrac {1}{8}$
This can be written as,

$40 / 3$ and

$9 / 8$
Now,

$40 / 3 \times 9 / 8$

$= (40 \times 9) / (3 \times 8)$

$= 360 / 24$
We get,

$= 15$
and

$9 / 8 \times 40 / 3$

$= (9 \times 40) / (8 \times 3)$

$= 360 / 24$
We get,

$= 15$
Therefore,

$40 / 3 \times 9 / 8 = 9 / 8 \times 40 / 3$ .

Verify the following and name the property:
$5/9 \times (-3/2 + 7/5) = 5/9 \times -3/2 + 5/9\times 7/5$ .

$5 / 9 \times (- 3 / 2 + 7 / 5) = 5 / 9 \times -3 / 2 + 5 / 9 \times 7 / 5$

$L.H.S = 5 / 9 \times (- 3 / 2 + 7 / 5)$

$= 5 / 9 \times \left \{(- 15 + 14) / 10\right \}$
We get,

$= 5 / 9 \times (- 1 / 10)$

$= - 5 / 90$

$= (- 5 \div 5) / (90 \div 5)$
We get,

$= - 1 / 18$

$R.H.S. = 5 / 9 \times (- 3 / 2) + 5 / 9 \times 7 / 5$
On further calculation, we get,

$= - 15 / 18 + 35 / 45$
Taking L.C.M. we get,

$= (- 75 + 70) / 90$

$= - 5 / 90$

$= (- 5 \div 5) / (90 \div 5)$

$= - 1 / 18$

Hence,

$L.H.S. = R.H.S$ .

The given property is Distributive property

Fill in the following blanks:
$-8/9\times \left \{4/13 + 5/17\right \} = -8/9 \times 4/13 +$ _______.

$-8/9\times \left \{4/13 + 5/17\right \} = -8/9 \times 4/13 +$ _______

$-8/9\times \left \{4/13 + 5/17\right \} = -8/9 \times 4/13 + - 8/9 \times 5/17$ .

Using the appropriate properties of operations of rational numbers, evaluate the following:
$8/9 \times 4/5 + 5/6 - 9/5 \times 8/9$ .

$8 / 9 \times 4 / 5 + 5 / 6 - 9 / 5 \times 8 / 9$

$= 8 / 9 \times 4 / 5 - 9 / 5 \times 8 / 9 + 5 / 6$
Taking common terms, we get,

$= 8 / 9 (4 / 5 - 9 / 5) + 5 / 6$

$= 8 / 9 \left \{(4 - 9) / 5\right \} + 5 / 6$

$= 8 / 9 \times - 5 / 5 + 5 / 6$

$= 8 / 9 \times (- 1) + 5 / 6$
On further calculation, we get

$= - 8 / 9 + 5 / 6$
Taking L.C.M. we get,

$= (- 16 + 15) / 18$

$= - 1 / 18$ .

Find the product of additive inverse and multiplication inverse of $-3/7$ .

The additive inverse of

$-3/7$ is

$3/7$
The multiplicative inverse of

$-3/7$ is

$-7/3$
Therefore,

$3/7 \times (-7/3) = -1$ .

Find the sum of additive inverse and multiplication inverse of $9$ .

The additive inverse of

$9$ is

$- 9$
The multiplicative inverse of

$9$ is

$1 / 9$
Hence,

$- 9 + 1 / 9 = (- 81 + 1) / 9$
We get,

$= - 80 / 9$

$= -8\dfrac {8}{9}$ .

Represent the following rational numbers on the number line.
$11/4$ .

$11/4 = 2\dfrac {3}{4}$
The given rational number on the number line is shown as below:
1710798_1806131_ans_c2e49723343b4a5cb1f23d79aa775dd8.jpg

Fill in the following blanks:
$-6/13 \times \left \{8/9 - 4/7\right \} = -6/13 \times$ ____ $-(-6/13) \times (4/7)$ .

$-6/13 \times \left \{8/9 - 4/7\right \} = -6/13 \times$ ____

$- (-6/13) \times (4/7)$

$-6/13 \times \left \{8/9 - 4/7\right \} = -6/13 \times 8/9- (-6/13) \times (4/7)$ .

Fill in the following blanks:
$3/7\times (-2/8\times$ _____ $) = (3/7 \times -2/8) \times 5/9$ .

$3/7\times (-2/8\times$ _____

$) = (3/7 \times -2/8) \times 5/9$

$3/7\times (-2/8\times 5/9) = (3/7 \times -2/8) \times 5/9$ .

Using distributivity, find
$\left \{9/16 \times 4/12\right \} + \left \{9/16 \times (-3/9)\right \}$ .

$\left \{9/16 \times 4/12\right \} + \left \{9/16 \times (-3/9)\right \}$
Taking common factor, we get

$= 9/16 \times \left \{4/12 + (-3/9)\right \}$

$= 9/16 \times (1/3 - 1/3)$
We get,

$= 9/16 \times 0$

$= 0$ .

Using distributivity, find
$\left \{7/5 \times (-3/12) \right \} + \left \{7/5 + 5/12\right \}$ .

$\left \{7/5 \times (-3/12) \right \} + \left \{7/5 + 5/12\right \}$
Taking common factor, we get

$= 7 / 5 \times (- 3 / 12 + 5 / 12)$

$= 7 / 5 \times \left \{(- 3 + 5) / 12\right \}$

$= 7 / 5 \times 2 / 12$
We get,

$= 7 / 30$ .

For each pair of , verify commutative property of addition of rational numbers.
$3 and \dfrac{-2}{7}$

$3 and \dfrac{-2}{7}$

$\dfrac{3}{1} + \dfrac{-2}{7} = \dfrac{-2}{7} +\dfrac{3}{1}$
LHS

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

$= \dfrac{2 \times 1}{7 \times7}$

$(∵LCM of 1 and 7=7)$

$= \dfrac{21 - 2}{7} + \dfrac{3}{1}$

$= \dfrac{-2 + 21}{7} = \dfrac{19}{7} \therefore RHS = LHS$
Hence, the commutative property for the addition of rational numbers is verified.

Write the additive inverse (negative ) of :
1

The additive inverse of 1=-1

Fill in the blanks :
$\dfrac{-5}{8}$ and $\dfrac{3}{10}$ are on the ............ sides of zero.

$\dfrac{-5}{8}$ and

$\dfrac{3}{10}$ are on the opposite side of zero.

Since both have opposite sign.

Fill in the blanks :
$\dfrac{-5}{8}$ and $\dfrac{-3}{10}$ are on the ............ sides of zero.

$\dfrac{-5}{8}$ and

$\dfrac{-3}{10}$ are on the same sides of zero.

Since both have same sign.

For each pair of , verify commutative property of addition of rational numbers.
$\dfrac{-4}{5} and \dfrac{-13}{-15}$

$\dfrac{-4}{5} + \dfrac{-13}{-15} = \dfrac{-13}{-15}$

$LHS = \dfrac{-4}{5} + \dfrac{13}{15}$
Taking LCM
5 5,15
3 1,3
1,1

$\therefore LCM of 5 and 15=5\times 3 = 15$

$LHS = \dfrac{-4 \times 3}{5 \times 3}+ \dfrac{13 \times 1}{15 \times 1} =\dfrac{-12+13}{15}=\dfrac{1}{15}$

$RHS = \dfrac{13}{15} + \dfrac{-4}{5}$
Hence, the commutative property for the addition of rational numbers is verified.

Fill in the blanks :
$\dfrac{5}{8}$ and $\dfrac{3}{10}$ are on the .............. side of zero.

$\dfrac{5}{8}$ and

$\dfrac{3}{10}$ are on the same side of zero.

Since they are of same sign.

Represent the following rational numbers on the number line.
$\dfrac{2}{5}$ .

The given rational number on the number line is shown above.
1710821_1806140_ans_7fe231b262be494d8f6d9f7ec11810c5.jpg

What is the multiplicative inverse if $1$ ?

We know that the multiplicative inverse or reciprocal for a number $x$ , denoted by $\dfrac{1}{x}$ where $x\ne 0$ , is a number which when multiplied by $x$ yields the multiplicative identity $1.$

Therefore the multiplicative inverse of $1$ is $\dfrac{1}{1}=1.$ Hence, multiplicative inverse of $1$ is $1$ itself.

Write any three rational numbers between the two numbers given below.
1) $0.3$ and $-0.5$
2) $-2.3$ and $-2.33$
3) $5.2$ and $5.3$
4) $-4.5$ and $-4.6$

1) The three rational numbers between

$0.3$ and

$0.5$ are

$-0.4$ ,

$0$ and,

$0.1$
2) The three rational numbers between

$-2.3$ and

$-2.33$ are

$-2.3$ ,

$-2.32$ and

$-2.325$
3) The three rational numbers between

$5.2$ and

$5.3$ are

$5.21$ ,

$5.24$ and

$5.28$
4) The three rational numbers between

$-4.5$ and

$-4.6$ are

$-4.51$ ,

$-4.55$ and

$-4.59$

Name the property under multiplication used in each of the following:
i) $\cfrac{-4}{5}\times1=1\times\cfrac{-4}{5}=-\cfrac{4}{5}$
ii) $-\cfrac{13}{17}\times\cfrac{-2}{7}=\cfrac{-2}{7}\times\cfrac{-13}{17}$
iii) $\cfrac{-19}{29}\times\cfrac{29}{-19}=1$

Existence of multiplicative identity.
Commutative property of multiplication.
Existence of multiplicative inverse.

List three rational numbers between -2 and -1

Let us write -1 and -2 as rational numbers with denominator 5 (why?)
We have -1 =

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

$\displaystyle \frac{-10}{5}$
so

$\displaystyle \frac{-10}{5}$ <

$\displaystyle \frac{-9}{5}$ <

$\displaystyle \frac{-8}{5}$ <

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

$\displaystyle \frac{-6}{5}$ <

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

or -2<

$\displaystyle \frac{-9}{5}$ <

$\displaystyle \frac{-8}{5}$ <

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

$\displaystyle \frac{-6}{5}$ <-1

The three rational numbers between -2 and -1 would be

$\displaystyle \frac{-9}{5}$ ,

$\displaystyle \frac{-8}{5}$ ,

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

(You can take any three of

$\displaystyle \frac{-9}{5},$

$\displaystyle \frac{-8}{5}$ ,

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

$\displaystyle \frac{-6}{5}$ )

Use the method of mean to find $3$ rational numbers between $-1$ and $-2$ .

A rational number between two numbers $a$ and $b = \cfrac{(a + b)}{2}$
So, a rational number between $- 1$ and $-2 = \cfrac {-1 -2}{2} = \cfrac {-3}{2}$
Now, another rational number between $-1$ and $\cfrac {-3}{2} =\cfrac {-1 + \tfrac {-3}{2}}{2} = \cfrac {-5}{4}$
And another rational numbers between $\cfrac {-3}{2}$ and $-2 =\cfrac { \tfrac{-3}{2} - 2 }{2} = \cfrac {-7}{4}$

Using appropriate properties, find
(i) $-\dfrac{2}{3}\times\dfrac{3}{5}+\dfrac{5}{2}-\dfrac{3}{5}\times\dfrac{1}{6}\\$
(ii) $\dfrac25\times\left(-\dfrac37\right) - \dfrac16\times\dfrac32 + \dfrac1{14}\times\dfrac25$

$(i)$

$\dfrac{-2}{3}\times\dfrac{3}{5}-\dfrac{3}{5}\times\dfrac{1}{6}+\dfrac{5}{2}\\$

Using the commutative property, we can write the given expression as:

$\dfrac{-3}{5}\times \dfrac{2}{3}-\dfrac{3}{5}\times\dfrac{1}{6}+\dfrac{5}{2}\\$

Now, using the distributive property of multiplication over addition,

$\begin{pmatrix}\dfrac{-3}{5}\end{pmatrix}\;\begin{pmatrix}\dfrac{2}{3}+\dfrac{1}{6}\end{pmatrix}+\dfrac{5}{2}$

$\Rightarrow \left(\dfrac{-3}{5}\right)\left(\dfrac{5}{6}\right)+\dfrac{5}{2}=\dfrac{-1}{2}+\dfrac{5}{2}$

$\Rightarrow \dfrac{4}{2}=2$

Hence, using appropriate properties,

$\dfrac{-2}{3}\times\dfrac{3}{5}-\dfrac{3}{5}\times\dfrac{1}{6}+\dfrac{5}{2}=2$

$(ii)$

$\dfrac{2}{5}\times\begin{pmatrix}\dfrac{-3}{7}\end{pmatrix}-\dfrac{1}{6}\times\dfrac{3}{2}+\dfrac{1}{14}\times\dfrac{2}{5}$

Using the commutative property, we can write the same expression as:

$\dfrac{2}{5}\times\begin{pmatrix}\dfrac{-3}{7}\end{pmatrix}+\dfrac{1}{14}\times\dfrac{2}{5}-\dfrac{1}{6}\times\dfrac{3}{2}$

Now, using the distributive property of multiplication over addition, we get:

$\dfrac{2}{5}\times\begin{pmatrix}\dfrac{-3}{7}+\dfrac{1}{14}\end{pmatrix}-\dfrac{1}{4}$

$\Rightarrow \dfrac{2}{5}\times\begin{pmatrix}\dfrac{-5}{14}\end{pmatrix}-\dfrac{1}{4}$

$\Rightarrow-\dfrac{1}{7}-\dfrac{1}{4}=\dfrac{-11}{28}$

Hence, using appropriate properties,

$\dfrac{2}{5}\times\begin{pmatrix}\dfrac{-3}{7}\end{pmatrix}-\dfrac{1}{6}\times\dfrac{3}{2}+\dfrac{1}{14}\times\dfrac{2}{5}=\dfrac{-11}{28}$

Write the additive inverse of each of the following.
(i) $\dfrac{2}{8}$ (ii) $\dfrac{-5}{9}$ (iii) $\dfrac{-6}{-5}$ (iv) $\dfrac{2}{-9}$ (v) $\dfrac{19}{-6}$

Additive inverse of

$\dfrac ab$ is

$-\dfrac ab$ , since

$\dfrac ab + \left(-\dfrac ab\right) = 0$

In the same manner:

(i)

Additive inverse of

$\dfrac28$ is

$-\dfrac{2}{8}$

(ii)

Additive inverse of

$\dfrac{-5}9$ is

$\dfrac{5}{9}$

(iii)

Additive inverse of

$\dfrac{-6}{-5}$ is

$\dfrac{-6}{5}$

(iv)

Additive inverse of

$\dfrac2{-9}$ is

$\dfrac{2}{9}$

(v)

Additive inverse of

$\dfrac{19}{-6}$ is

$\dfrac{19}{6}$

Fill in the blanks.
(i) Zero has ___ reciprocal.
(ii) The numbers _ and __ are their own reciprocals.
(iii) The reciprocal of $-5$ is _.
(iv) Reciprocal of $\dfrac{1}{x}$ , where $x\neq0$ is _.
(v) The product of two rational numbers is always a .
(vi) The reciprocal of a positive rational number is _____.

i) Zero has No reciprocal.

(ii) The numbers 1 and -1 are their own reciprocals.

(iii) The reciprocal of

−55 is

$\dfrac{-1}5$ .

(iv) Reciprocal of

$\dfrac1x$ , where

$x≠0$ is

$x$ .

(v) The product of two rational numbers is always a Rational Number.

(vi) The reciprocal of a positive rational number is Positive Rational Number.

Find five rational numbers between $\displaystyle\frac{3}{5}$ and $\displaystyle\frac{4}{5}$ .

2150986_463526_ans_7b1c31d4937b4aa68850b1860b4e0ab7.jpeg

Represent these numbers on the number line.
(i) $\dfrac{7}{4}$ (ii) $\dfrac{-5}{6}$

Find five rational numbers between $3$ and $4$ .

${\textbf{Step -1: State mid point method to find rational numbers}}{\text{.}}$

${\text{Rational numbers always being in form of }}\dfrac{p}{q}.{\text{ where q}} \ne {\text{0}}{\text{.}}$

${\text{Let }}x{\text{ and }}y{\text{ be two numbers, such that }}x{\text{ < }}y{\text{.Then in between there is one rational point}}{\text{.}}$

${\text{That is }}\dfrac{{x + y}}{2}.$

${\textbf{Step -2: Find rational numbers between 3and4}}{\text{.}}$

${\text{So between 3 and 4 the rational number be , }}\dfrac{{3 + 4}}{2} = \dfrac{7}{2}.$

${\textbf{Step -3: Find rational numbers between 3 and }}\mathbf{\dfrac{7}{2}.}$

${\text{So between 3 and }}\dfrac{7}{2}{\text{the rational number will be , }}\dfrac{{3 + \dfrac{7}{2}}}{2} = \dfrac{{\dfrac{{6 + 7}}{2}}}{2} = \dfrac{{13}}{4}.$

${\textbf{Step -4: Find rational numbers between 3 and }}\mathbf{\dfrac{{13}}{4}.}$

${\text{Between 3 and }}\dfrac{{13}}{4}{\text{ the rational number will be , }}\dfrac{{3 + \dfrac{{13}}{4}}}{2} = \dfrac{{25}}{8}.$

${\textbf{Step -5: Find rational numbers between }}\mathbf{\dfrac{7}{2}}{\textbf{ and 4}}{\textbf{.}}$

${\text{Between }}\dfrac{{\text{7}}}{{\text{2}}}{\text{ and 4 the rational number will be , }}\dfrac{{4 + \dfrac{7}{2}}}{2} = \dfrac{{\dfrac{{7 + 8}}{2}}}{2} = \dfrac{{15}}{4}.$

${\textbf{Step -6: Find rational numbers between }}\mathbf{\dfrac{{15}}{4}}{\textbf{ and 4}}{\textbf{.}}$

${\text{Between }}\dfrac{{{\text{15}}}}{{\text{4}}}{\text{ and 4 , the rational number will be , }}\dfrac{{\dfrac{{15}}{4} + 4}}{2} = \dfrac{{31}}{8}.$

${\textbf{Hence , rational numbers between 3 and 4 are }}\mathbf{\dfrac{7}{2},\dfrac{{13}}{4},\dfrac{{15}}{4},\dfrac{{25}}{8},\dfrac{{31}}{{8}}.}$

Is $\dfrac{8}{9}$ the multiplicative inverse of $-1\dfrac{1}{8}$ ? Why or why not?

If it is a multiplicative inverse, then the product

$=1$

$\dfrac{8}{9}\times\begin{pmatrix}-1\dfrac{1}{8}\end{pmatrix}=\dfrac{8}{9}\times\dfrac{-9}{8}=-1$

Product

$=-1$

$\therefore$ It is not a multiplicative inverse.

Name the property under multiplication used in each of the following.
(i) $\dfrac{-4}{5}\times1=1\times\dfrac{-4}{5}=-\dfrac{4}{5}$ (iii) $\dfrac{-19}{29}\times\dfrac{29}{-19}=1$

i) Commutative property

As it follows:

$A\times B = B\times A$

ii) Commutative property

As it follows:

$A\times B = B\times A$

iii) Multiplicative inverse

As it follows:

$A\times\dfrac1A = 1$

Name the property indicated in the following:
$315 + 115 = 430$

We know that for some integers

$a$ and

$b$ , if

$a+b$ is also an integer then the set of integers is said to be closed under the operation of addition.

Here, both

$315$ and

$115$ are integers. Also, the addition of

$315$ and

$115$ is

$430$ which is also an integer.

Hence,

$315+115=430$ satisfies the closure property of addition.

Which property is illustrated by the following statement?
$(16+18)+20=(18+16)+20$

$(16+18)+20=(18+16)+20$
It is associative property of addition

Name the property indicated in the following:
$\dfrac {3}{4} \cdot \dfrac {9}{5} = \dfrac {27}{20}$

We know that for some rational numbers

$\dfrac {a}{b}$ and

$\dfrac {c}{d}$ , if

$\dfrac {a}{b}\times \dfrac {c}{d}$ is also a rational number then the set of rational numbers is said to be closed under the operation of multiplication.

Here, both

$\dfrac {3}{4}$ and

$\dfrac {9}{5}$ are rational numbers. Also, the multiplication of

$\dfrac {3}{4}$ and

$\dfrac {9}{5}$ is

$\dfrac {27}{20}$ which is also a rational number.

Hence,

$\dfrac {3}{4}\times \dfrac {9}{5}=\dfrac {27}{20}$ satisfies the closure property of multiplication.

What are the properties of $R$ used in the following? $a+(\pi + e)=(a+\pi)+e$

We know that the associative property of any binary operation

$*$ states that

$a*(b*c)=(a*b)*c$

Now, since it is given that

$a+(\pi +e)=(a+\pi )+e$ satisfies under addition.

Hence, the associative property of addition is used in

$a+(\pi +e)=(a+\pi )+e$ .

Name the property indicated in the following:
$\dfrac {8}{9}\times 1 = \dfrac {8}{9}$

We know that for the set of rational numbers,

$1$ is said to be multiplicative identity. It means when any rational number is multiplied to

$1$ or

$1$ is multiplied to any rational number, the result is that rational number itself.

Here, a rational number

$\dfrac {8}{9}$ is multiplied to

$1$ , the result is that rational number itself that is

$\dfrac {8}{9}$ .

Hence,

$1$ is the multiplicative identity in the given expression

$\dfrac {8}{9}\times 1=\dfrac {8}{9}$ .

Check the distributive property for the stated triples of rational numbers:
$\dfrac {3}{8}, 0, \dfrac {13}{7}$

We know that the distributive property states that multiplying a sum by a number gives the same result as multiplying each addend by the number and then adding the products together, that is, if

$a,b$ and

$c$ are three numbers, then

$a(b+c)=ab+ac$ .

Let

$a=\dfrac {3}{8},b=0$ and

$c=\dfrac {13}{7}$ .

Let us first find

$a(b+c)$ as follows:

$a(b+c)=\dfrac { 3 }{ 8 } \left( 0+\dfrac { 13 }{ 7 } \right) =\dfrac { 3 }{ 8 } \times \dfrac { 13 }{ 7 } =\dfrac { 39 }{ 56 } ........(1)$

Now, we find

$ab+ac$ as follows:

$ab+ac=\left( \dfrac { 3 }{ 8 } \times 0 \right) +\left( \dfrac { 3 }{ 8 } \times \dfrac { 13 }{ 7 } \right) =0+\dfrac { 39 }{ 56 } =\dfrac { 39 }{ 56 } .......(2)$

Since equation

$(1)$ is equal to equation

$(2)$ ,

$\therefore$

$a(b+c)=ab+ac$ i.e.

$\dfrac{3}{8}\left(0+\dfrac{13}{7}\right)=\left(\dfrac{3}{8}\times 0\right)+\left(\dfrac{3}{8}\times \dfrac{13}{7}\right)$ .

Hence, the given numbers

$\dfrac {3}{8},0$ and

$\dfrac {13}{7}$ satisfies the distributive property.

Check the commutative property of addition for $\dfrac {-8}{13}, \dfrac {23}{27}$

We know that the commutative property of Addition states that changing the order of addends does not change the sum. i.e., if

$a$ and

$b$ are two real numbers, then

$a + b = b + a$ .

Let

$a=\dfrac {-8}{13}$ and

$b=\dfrac {23}{27}$ .

Let us first find

$a+b$ as follows:

$a+b=\dfrac { -8 }{ 13 } +\dfrac { 23 }{ 27 } \\=\dfrac { -8\times 27 }{ 13\times 27 } +\dfrac { 23\times 13 }{ 27\times 13 } \\=\dfrac { -216 }{ 351 } +\dfrac { 299 }{ 351 } \\=\dfrac { 83 }{ 351 } ........(1)$

Now, we find

$b+a$ as follows:

$b+a=\dfrac { 23 }{ 27 } +\left( \dfrac { -8 }{ 13 } \right) \\=\dfrac { 23\times 13 }{ 27\times 13 } -\dfrac { 8\times 27 }{ 13\times 27 } \\=\dfrac { 299 }{ 351 } -\dfrac { 216 }{ 351 } \\=\dfrac { 83 }{ 351 } .......(2)$

Since equation

$(1)$ is equal to equation

$(2)$ ,

$\therefore$

$a + b = b + a$ i.e.

$\dfrac{-8}{13}+\dfrac{23}{27}=\dfrac{23}{27}+\dfrac{-8}{13}$

Hence, the given numbers

$\dfrac {-8}{13}$ and

$\dfrac {23}{27}$ satisfies the commutative property of addition.

What are the properties of $R$ used in the following? $8\times 7 = 7\times 8$

We know that the commutative property of multiplication states that two numbers

$a$ and

$b$ can be multiplied in either order that is

$a\times b=b\times a$

Now, since it is given that

$8\times 7=7\times 8$

Hence, the commutative property of multiplication is used in

$8\times 7=7\times 8$ .

Suppose $\dfrac{m}{n}$ and $\dfrac{p}{q}$ are two positive rational numbers. Where do $\dfrac{m + p}{n + q} lie,$ with respect to $\dfrac{m}{n}$ and $\dfrac{p}{q}$ ?

Let

$\dfrac {m}{n}=\dfrac {1}{2}=0.5$ and

$\dfrac {p}{q}=\dfrac {3}{4}=0.75$

Individually we have,

$m=1,n=2,p=3$ and

$q=4$ .

Now consider,

$\dfrac { m+p }{ n+q } =\dfrac { 1+3 }{ 2+4 } =\dfrac { 4 }{ 6 } =\dfrac { 2 }{ 3 } =0.66$

Hence, the above drawn number line shows that

$\dfrac { m+p }{ n+q }$ lies between

$\dfrac {m}{n}$ and

$\dfrac {p}{q}$ .

Check the distributive property for the stated triples of rational numbers:
$\dfrac {1}{8}, \dfrac {1}{9}, \dfrac {1}{10}$

We know that the distributive property states that multiplying a sum by a number gives the same result as multiplying each addend by the number and then adding the products together, that is, if

$a,b$ and

$c$ are three numbers, then

$a(b+c)=ab+ac$ .

Let

$a=\dfrac {1}{8},b=\dfrac {1}{9}$ and

$c=\dfrac {1}{10}$ .

Let us first find

$a(b+c)$ as follows:

$a(b+c)=\dfrac { 1 }{ 8 } \left( \dfrac { 1 }{ 9 } +\dfrac { 1 }{ 10 } \right) =\dfrac { 1 }{ 8 } \left( \dfrac { 1\times 10 }{ 9\times 10 } +\dfrac { 1\times 9 }{ 10\times 9 } \right) =\dfrac { 1 }{ 8 } \left( \dfrac { 10 }{ 90 } +\dfrac { 9 }{ 90 } \right) =\dfrac { 1 }{ 8 } \times \dfrac { 19 }{ 90 } =\dfrac { 19 }{ 720 }$ ........

$(1)$

Now, we find

$ab+ac$ as follows:

$ab+ac=\left( \dfrac { 1 }{ 8 } \times \dfrac { 1 }{ 9 } \right) +\left( \dfrac { 1 }{ 8 } \times \dfrac { 1 }{ 10 } \right) =\dfrac { 1 }{ 72 } +\dfrac { 1 }{ 80 } =\dfrac { 1\times 80 }{ 72\times 80 } +\dfrac { 1\times 72 }{ 80\times 72 } =\dfrac { 80 }{ 5760 } +\dfrac { 72 }{ 5760 } =\dfrac { 152 }{ 5760 } =\dfrac { 19 }{ 720 }$ .......

$(2)$

Since equation

$(1)$ is equal to equation

$(2)$ ,

$\therefore$

$a(b+c)=ab+ac$ i.e.

$\dfrac{1}{8}\left(\dfrac{1}{9}+\dfrac{1}{10}\right)=\left(\dfrac{1}{8}\times \dfrac{1}{9}\right)+\left(\dfrac{1}{8}\times \dfrac{1}{10}\right)$

Hence, the given numbers

$\dfrac {1}{8},\dfrac {1}{9}$ and

$\dfrac {1}{10}$ satisfies the distributive property.

Check the distributive property for the stated triples of rational numbers:
$\dfrac {-4}{9}, \dfrac {6}{5}, \dfrac {11}{10}$

We know that the distributive property states that multiplying a sum by a number gives the same result as multiplying each addend by the number and then adding the products together, that is, if

$a,b$ and

$c$ are three numbers, then

$a(b+c)=ab+ac$ .

Let

$a=\dfrac {-4}{9},b=\dfrac {6}{5}$ and

$c=\dfrac {11}{10}$ .

Let us first find

$a(b+c)$ as follows:

$a(b+c)=\dfrac { -4 }{ 9 } \left( \dfrac { 6 }{ 5 } +\dfrac { 11 }{ 10 } \right) =\dfrac { -4 }{ 9 } \left( \dfrac { 6\times 2 }{ 5\times 2 } +\dfrac { 11 }{ 10 } \right) =\dfrac { -4 }{ 9 } \left( \dfrac { 12 }{ 10 } +\dfrac { 11 }{ 10 } \right) =\dfrac { -4 }{ 9 } \times \dfrac { 23 }{ 10 } =-\dfrac { 92 }{ 90 } =-\dfrac { 46 }{ 45 } ........(1)$

Now, we find

$ab+ac$ as follows:

$ab+ac=\left( \dfrac { -4 }{ 9 } \times \dfrac { 6 }{ 5 } \right) +\left( \dfrac { -4 }{ 9 } \times \dfrac { 11 }{ 10 } \right) =\dfrac { -24 }{ 45 } -\dfrac { 44 }{ 90 } =\dfrac { -24\times 2 }{ 45\times 2 } -\dfrac { 44 }{ 90 } =\dfrac { -48 }{ 90 } -\dfrac { 44 }{ 90 } =-\dfrac { 92 }{ 90 } =-\dfrac { 46 }{ 45 } .......(2)$

Since equation

$(1)$ is equal to equation

$(2)$ ,

$\therefore$

$a(b+c)=ab+ac$ i.e.

$\dfrac{-4}{9}\left(\dfrac{6}{5}+\dfrac{11}{10}\right)=\left(\dfrac{-4}{9}\times \dfrac{6}{5}\right)+\left(\dfrac{-4}{9}\times \dfrac{11}{10}\right)$ .

Hence, the given numbers

$\dfrac {-4}{9},\dfrac {6}{5}$ and

$\dfrac {11}{10}$ satisfies the distributive property.

Check the commutative property of addition for $\dfrac {102}{201}, \dfrac {3}{4}$

We know that the Commutative Property of Addition states that changing the order of addends does not change the sum. i.e., if

$a$ and

$b$ are two real numbers, then

$a + b = b + a$ .

Let

$a=\dfrac {102}{201}$ and

$b=\dfrac {3}{4}$ .

Let us first find

$a+b$ as follows:

$a+b=\dfrac { 102 }{ 201 } +\dfrac { 3 }{ 4 } =\dfrac { 102\times 4 }{ 201\times 4 } +\dfrac { 3\times 201 }{ 4\times 201 } =\dfrac { 408 }{ 804 } +\dfrac { 603 }{ 804 } =\dfrac { 1011 }{ 804 }......(1)$

Now, we find

$b+a$ as follows:

$b+a=\dfrac { 3 }{ 4 } +\dfrac { 102 }{ 201 } =\dfrac { 3\times 201 }{ 4\times 201 } +\dfrac { 102\times 4 }{ 201\times 4 } =\dfrac { 603 }{ 804 } +\dfrac { 408 }{ 804 } =\dfrac { 1011 }{ 804 }......(2)$

Since equation

$(1)$ is equal to equation

$(2)$ ,

$\therefore a + b = b + a$ i.e.

$\dfrac{102}{201}+\dfrac{3}{4}=\dfrac{3}{4}+\dfrac{102}{201}$

Hence, the given numbers

$\dfrac {102}{201}$ and

$\dfrac {3}{4}$ satisfies the commutative property of addition.

Find the multiplicative inverse of each of the following numbers:
$2, \dfrac {6}{11}, \dfrac {-8}{15}, \dfrac {19}{18}, \dfrac {1}{1000}$

We know that the multiplicative inverse or reciprocal for a number

$x$ , denoted by

$\dfrac {1}{x}$ or

$x^{−1}$ , is a number which when multiplied by

$x$ yields the multiplicative identity

$1$ .

The multiplicative inverse of a fraction

$\dfrac {a}{b}$ is

$\dfrac {b}{a}$

$\left( \because \quad \dfrac { a }{ b } \times \dfrac { b }{ a } =1 \right)$

Similarly,

The multiplicative inverse of a number

$2$ is

$\dfrac {1}{2}$

$\left( \because \quad 2 \times \dfrac { 1 }{ 2 } =1 \right)$

The multiplicative inverse of a fraction

$\dfrac {6}{11}$ is

$\dfrac {11}{6}$

$\left( \because \quad \dfrac { 6 }{ 11 } \times \dfrac { 11 }{ 6 } =1 \right)$

The multiplicative inverse of a fraction

$\dfrac {-8}{15}$ is

$\dfrac {-15}{8}$

$\left( \because \quad \dfrac { -8 }{ 15 } \times \dfrac { -15 }{ 8 } =1 \right)$

The multiplicative inverse of a fraction

$\dfrac {19}{8}$ is

$\dfrac {8}{19}$

$\left( \because \quad \dfrac { 19 }{ 8 } \times \dfrac { 8 }{ 19 } =1 \right)$

The multiplicative inverse of a fraction

$\dfrac {1}{1000}$ is

$1000$

$\left( \because \quad \dfrac { 1 }{ 1000 } \times 1000 =1 \right)$

Represent $-\dfrac {3}{4}$ on number line

The below diagram shows marking of fraction

$-\dfrac { 3 }{ 4 }$ on a number line.

What are the properties of $R$ used in the following? $\pi \times 1 = \pi$

We know that

$1$ is the identity element under multiplication for the real numbers, since if

$a$ is any real number then

$a\times 1 =1\times a = a$

Now, since it is given that

$\pi \times 1=\pi$ , therefore,

$1$ is the identity element.

Hence, the identity element property of multiplication is used in

$\pi \times 1=\pi$ .

Find the additive inverse of: $7+\sqrt[4]{2}$

We know that the additive inverse of a number

$x$ is

$-x$ , which when added to

$x$ yields the additive identity

$0$ that is

$x+(-x)=x-x=0$

Now, since

$(7+\sqrt [ 4 ]{ 2 } )+[-(7+\sqrt [ 4 ]{ 2 } )]=7+\sqrt [ 4 ]{ 2 } -7-\sqrt [ 4 ]{ 2 } =0$

Hence, the additive inverse of

$(7+\sqrt [ 4 ]{ 2 } )$ is

$-(7+\sqrt [ 4 ]{ 2 } )$ .

Find the additive inverse of $\sqrt{5}.$

We know that the additive inverse of a number

$x$ is

$-x$ , which when added to

$x$ yields the additive identity

$0$ that is

$x+(-x)=x-x=0$

Now, since

$\sqrt { 5 } +(-\sqrt { 5 } )=0$

Hence, the additive inverse of

$\sqrt { 5 }$ is

$-\sqrt { 5 }$ .

Find the additive inverse of: $-\frac{3}{\sqrt{\pi}}$

We know that the additive inverse of a number

$x$ is

$-x$ , which when added to

$x$ yields the additive identity

$0$ that is

$x+(-x)=x-x=0$

Now, since

$-\dfrac { 3 }{ \sqrt { \pi } } +\dfrac { 3 }{ \sqrt { \pi } } =0$

Hence, the additive inverse of

$-\dfrac { 3 }{ \sqrt { \pi } }$ is

$\dfrac { 3 }{ \sqrt { \pi } }$ .

Find the additive inverse of: $1+\pi$

We know that the additive inverse of a number

$x$ is

$-x$ , which when added to

$x$ yields the additive identity

$0$ that is

$x+(-x)=x-x=0$

Now, since

$(1+\pi )+[-(1+\pi )]=1+\pi -1-\pi =0$

Hence, the additive inverse of

$(1+\pi)$ is

$-(1+\pi)$ .

Find the additive inverse of each of the following: $(-3+\sqrt{3})^2$

We know that the additive inverse of a number

$x$ is

$-x$ , which when added to

$x$ yields the additive identity

$0$ that is

$x+(-x)=x-x=0$

Now, since

${ (-3+\sqrt { 3 } ) }^{ 2 }+[-{ (-3+\sqrt { 3 } ) }^{ 2 }]={ (-3+\sqrt { 3 } ) }^{ 2 }-{ (-3+\sqrt { 3 } ) }^{ 2 }=0$

Hence, the additive inverse of

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

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

Write the reciprocal of a negative rational number.

1608674_564880_ans_305c5a0d9b5b49588fc44ca7aeadf223.jpg

Find two rational numbers between $\dfrac {-3}{4}$ and $\dfrac {1}{2}$

$a = \dfrac {-3}{5}, b = \dfrac {1}{2}$
Let

$q_{1}$ and

$q_{2}$ be two rational numbers.

$q_{1} = \dfrac {1}{2}(a + b)$

$q_{1} = \dfrac {1}{2} \times \left (\dfrac {-3}{5} + \dfrac {1}{2}\right ) = \dfrac {1}{2}\times \left (\dfrac {-6 + 5}{10}\right ) = \dfrac {1}{2}\times \left (\dfrac {-1}{10}\right ) = \dfrac {-1}{20}$

$q_{2} = \dfrac {1}{2}(a + q_{1}) = = \dfrac {1}{2} \times \left (\dfrac {-3}{5} + \left (\dfrac {-1}{20}\right )\right )$

$= \dfrac {1}{2}\times \left (\dfrac {-12 + (-1)}{20}\right ) = \dfrac {1}{2}\times \left (\dfrac {-12 - 1}{20}\right ) = \dfrac {1}{2} \times \left (\dfrac {-13}{20}\right ) = \dfrac {-13}{40}$
The two rational numbers are

$\dfrac {-1}{20}$ and

$\dfrac {-13}{40}$

Find any two rational numbers $\displaystyle\frac{1}{4}$ and $\displaystyle\frac{3}{4}$ .

A rational number between

$\displaystyle\frac{1}{4}$ and

$\displaystyle\frac{3}{4}=\frac{1}{2}\left(\frac{1}{4}+\frac{3}{4}\right)=\frac{1}{2}(1)=\frac{1}{2}$
Another rational number between

$\displaystyle\frac{1}{2}$ and

$\displaystyle\frac{3}{4}=\frac{1}{2}\left(\frac{1}{2}+\frac{3}{4}\right)=\displaystyle\frac{1}{2}\times \frac{5}{4}=\frac{5}{8}$
The rational numbers

$\displaystyle\frac{1}{2}$ and

$\displaystyle\frac{5}{8}$ lie between

$\displaystyle\frac{1}{4}$ and

$\displaystyle\frac{3}{4}$ .

Represent $\dfrac {8}{5}$ and $\dfrac {-8}{5}$ on a numberline

1130157_569117_ans_f8fd3eccb9d840f0ad02e408dd747a06.png

If a/b is the additive inverse of c/d, then find the value of $\dfrac{a} {b} + \dfrac{c} {d}$

$\cfrac{a}{b}$ is additive inverse of

$\cfrac{c}{d}$

then

$\cfrac{a}{b}+\cfrac{c}{d}=0$

Find a rational number and also an irrational number lying between the numbers $0.401001000100001000001........$ and $0.404004000400004000004..........$ .

Rational numbers between given numbers will be

$0.402$ and

$0.403$ .

A rational number between these numbers

$={\dfrac{0.402+0.403}{2}}$

$=0.4025$ and irrational numbers, given between the numbers are

$0.402002000200002000002........$ and

$0.403003000300003000003..........$

a. Write additive inverse of $\frac{2}{8}and\frac{{ - 6}}{{ - 5}}$
b. Write multiplicative inverse of $- 13{\rm{ and }}\frac{{ - 13}}{{19}}$

$\begin{array}{l}a.{\rm{ }}\frac{2}{8} \to \frac{{ - 2}}{8}\\\frac{{ - 6}}{5} = \frac{6}{5} \to \frac{{ - 6}}{5}\\b.{\rm{ - 13 }} \to {\rm{ }}\frac{1}{{ - 13}}\\\frac{{ - 13}}{{19}} \to \frac{{ - 19}}{{13}}\end{array}$

Write a multiple of $\dfrac{-6}{3}$ rational number.

$\dfrac{-6}{3}\times 2=-1\implies -1$ is a multiple of

$\dfrac{-6}{3}$

Write $5$ rational numbers between $3$ and $\cfrac { 3 }{ 10 }$

$a=3,b=\cfrac { 3 }{ 10 }$

${ x }_{ 1 }=\cfrac { 1 }{ 2 } \left[ \cfrac { 3 }{ 10 } +3 \right]$

$=\cfrac { 1 }{ 2 } \left[ \cfrac { 3+30 }{ 10 } \right]$

${ x }_{ 1 }=\cfrac { 33 }{ 20 }$

${ x }_{ 2 }=\cfrac { 1 }{ 2 } \left[ \cfrac { 3 }{ 10 } +\cfrac { 33 }{ 20 } \right]$

$=\cfrac { 1 }{ 2 } \left[ \cfrac { 6+33 }{ 20 } \right] =\cfrac { 39 }{ 40 }$

${ x }_{ 3 }=\cfrac { 1 }{ 2 } \left[ \cfrac { 3 }{ 10 } +\cfrac { 39 }{ 40 } \right]$

$=\cfrac { 1 }{ 2 } \left[ \cfrac { 12+39 }{ 40 } \right]$

$=\cfrac { 1 }{ 2 } \times \cfrac { 51 }{ 40 } =\cfrac { 51 }{ 80 }$

${ x }_{ 4 }=\cfrac { 1 }{ 2 } \left[ \cfrac { 3 }{ 10 } +\cfrac { 51 }{ 80 } \right] =\cfrac { 1 }{ 2 } \left[ \cfrac { 24+51 }{ 80 } \right]$

$=\cfrac { 1 }{ 2 } \left[ \cfrac { 75 }{ 80 } \right] =\cfrac { 75 }{ 160 }$

${ x }_{ 5 }=\cfrac { 1 }{ 2 } \left[ \cfrac { 3 }{ 10 } +\cfrac { 75 }{ 160 } \right]$

$=\cfrac { 1 }{ 2 } \left[ \cfrac { 48+75 }{ 160 } \right] =\cfrac { 123 }{ 320 }$

Mark the following rational number on number line.
$\dfrac { 4 }{ 5 }$

The value of

$\dfrac{4}{5}$ is between

$\left( {0,\,1} \right)$

As shown in the figure

1163708_1053544_ans_fe5bb9cc6e4f4b9d8533b230d16bf2e2.png

I am a number between $20$ and $30$ . If you divide $47$ and $92$ by me, the remainders are $3$ and $4$ respectively what number am I?

When we divide 47 by the required number then the remainder = 3

∴ 47 - 3 = 44

so the number divides 44 exactly

Between 20 & 30 only 22 divides the number 44 exactly

Similarly,

92 - 4 = 88

Between 20 & 30 only 22 divides the number 88 exactly

so the required number is 22

Represented $\dfrac {-3}{4}$ & $\dfrac {1}{8}$ on the number line

$\dfrac { { -3 } }{ 4 } =\dfrac { { -6 } }{ 8 }$
1380165_1053227_ans_85fb089b94224d7ebc38f1eb1783737b.png

Find five fractions between $\dfrac {2}{3}$ and $\dfrac {4}{5}$ .

Making equal denominators of both the dfractions.

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

$= \dfrac{{10}}{{15}}$

And,

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

$= \dfrac{{12}}{{15}}$

Now, the dfraction between $\dfrac{{10}}{{15}}$ and $\dfrac{{12}}{{15}}$ is,

$\dfrac{{\dfrac{{10}}{{15}} + \dfrac{{12}}{{15}}}}{2} = \dfrac{{22}}{{30}}$

So, $\dfrac{{10}}{{15}} < \dfrac{{22}}{{30}} < \dfrac{{12}}{{15}}$

The dfraction between $\dfrac{{10}}{{15}}$ and $\dfrac{{22}}{{30}}$ is,

$\dfrac{{\dfrac{{10}}{{15}} + \dfrac{{22}}{{30}}}}{2} = \dfrac{{42}}{{30 \times 2}}$

$= \dfrac{{21}}{{30}}$

So, $\dfrac{{10}}{{15}} < \dfrac{{21}}{{30}} < \dfrac{{22}}{{30}} < \dfrac{{12}}{{15}}$

The dfraction between $\dfrac{{22}}{{30}}$ and $\dfrac{{12}}{{15}}$ is,

$\dfrac{{\dfrac{{22}}{{30}} + \dfrac{{12}}{{15}}}}{2} = \dfrac{{46}}{{30 \times 2}}$

$= \dfrac{{23}}{{30}}$

$\dfrac{{10}}{{15}} < \dfrac{{21}}{{30}} < \dfrac{{22}}{{30}} < \dfrac{{23}}{{30}} < \dfrac{{12}}{{15}}$

The dfraction between $\dfrac{{10}}{{15}}$ and $\dfrac{{21}}{{30}}$ is,

$\dfrac{{\dfrac{{10}}{{15}} + \dfrac{{21}}{{30}}}}{2} = \dfrac{{41}}{{30 \times 2}}$

$= \dfrac{{41}}{{60}}$

So, $\dfrac{{10}}{{15}} < \dfrac{{41}}{{60}} < \dfrac{{21}}{{30}} < \dfrac{{22}}{{30}} < \dfrac{{23}}{{30}} < \dfrac{{12}}{{15}}$

The dfraction between $\dfrac{{23}}{{30}}$ and $\dfrac{{12}}{{15}}$ .

$\dfrac{{\dfrac{{23}}{{30}} + \dfrac{{12}}{{15}}}}{2} = \dfrac{{47}}{{30 \times 2}}$

$= \dfrac{{47}}{{60}}$

So, $\dfrac{{10}}{{15}} < \dfrac{{41}}{{60}} < \dfrac{{21}}{{30}} < \dfrac{{22}}{{30}} < \dfrac{{23}}{{30}} < \dfrac{{47}}{{60}} < \dfrac{{12}}{{15}}$

Therefore, the five dfractions between the given numbers are $\dfrac{{41}}{{60}},\dfrac{{21}}{{30}},\dfrac{{22}}{{30}},\dfrac{{23}}{{30}},\dfrac{{47}}{{60}}$ .

Let * be a binary operation on the set $Q$ of rational numbers as follows:
$a*b=a-b$
Find which of the binary operations are commutative and which are associative.

Consider given the given binary operation,

$a*b=a-b$

Commutative –

$a*b=a-b$

$b*a=b-a$

$a-b\ne b-a$

$a*b\ne b*a$

Hence, given operation is not commutative.

Associative-

$a*b=a-b$

$a*\left( b*c \right)=a*\left( b-c \right)$

$=a-\left( b-c \right)=a-b+c$

$\left( a*b \right)*c=\left( a-b \right)*c=\left( a-b \right)-c$

$=a-b-c$

Now,

$a*\left( b*c \right)\ne \left( a*b \right)*c$

$\because a-b+ca\ne a-b-c$

Hence, given operation is not associative.

Verify the following:
$\dfrac { -13 }{ 24 } \times \left( \dfrac { -12 }{ 5 } \times \dfrac { 35 }{ 36 } \right) =\left( \dfrac { -13 }{ 24 } \times \dfrac { -12 }{ 5 } \right) \times \dfrac { 35 }{ 36 }$

$L.H.S = \dfrac{{ - 13}}{{24}} \times \left( {\dfrac{{12}}{5} \times \dfrac{{35}}{{36}}} \right)$

$= \dfrac{{ - 13}}{{24}} \times \left( {\dfrac{{ - 7}}{3}} \right)$

$= \dfrac{{91}}{{72}}$

$R.H.S = \left( {\dfrac{{ - 13}}{{24}} \times \dfrac{{ - 12}}{5}} \right) \times \dfrac{{35}}{{36}}$

$= \dfrac{{13}}{{10}} \times \dfrac{{35}}{{36}}$

$= \dfrac{{91}}{{72}}$

$Since,\,R.H.S = L.H.S$

Write three rational numbers that lie between the two given numbers
$\dfrac {4}{5},\dfrac {2}{3}$

Making the same denominator of the rational numbers $\dfrac{4}{5}$ and $\dfrac{2}{3}$ .

$\dfrac{4}{5} = \dfrac{{4 \times 3}}{{5 \times 3}} = \dfrac{{12}}{{15}}$

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

The rational numbers between $\dfrac{{12}}{{15}}$ and $\dfrac{{10}}{{15}}$ is,

$\dfrac{{\dfrac{{12}}{{15}} + \dfrac{{10}}{{15}}}}{2} = \dfrac{{22}}{{30}}$

So, $\dfrac{{10}}{{15}} < \dfrac{{22}}{{30}} < \dfrac{{12}}{{15}}$

The rational number between $\dfrac{{10}}{{15}}$ and $\dfrac{{22}}{{30}}$ is,

$\dfrac{{\dfrac{{10}}{{15}} + \dfrac{{22}}{{30}}}}{2} = \dfrac{{42}}{{30 \times 2}}$

$= \dfrac{{21}}{{30}}$

The rational number between $\dfrac{{12}}{{15}}$ and $\dfrac{{22}}{{30}}$ is,

$\dfrac{{\dfrac{{12}}{{15}} + \dfrac{{22}}{{30}}}}{2} = \dfrac{{46}}{{30 \times 2}}$

$= \dfrac{{23}}{{30}}$

Thus, $\dfrac{{10}}{{15}} < \dfrac{{21}}{{30}} < \dfrac{{22}}{{30}} < \dfrac{{23}}{{30}} < \dfrac{{12}}{{15}}$ .

Therefore the three rational numbers are $\dfrac{{21}}{{30}},\dfrac{{22}}{{30}},\dfrac{{23}}{{30}}$ .

Represent $-\cfrac { 2 }{ 13 }$ and $\cfrac { -9 }{ 13 }$ on the number line.

$\dfrac{-2}{13}$ and

$\dfrac{-9}{13}$ lie between

$-1$ and

$0$

Now, divide the unit distance between

$-1$ and

$0$ in

$13$ equal parts as shown in the figure.

$\because \dfrac{-2}{13}$ means

$2$ parts out of

$13$ so, among

$13$ equal divisions,

$2^{nd}$ division corresponds to

$\dfrac{-2}{13}$

Similarly

$\dfrac{-9}{13}$ means

$9$ parts out of

$13$ so, among

$13$ equal divisions,

$9^{th}$ division corresponds to

$\dfrac{-9}{13}$

$\therefore$ Representation of

$\dfrac{-2}{13}$ and

$\dfrac{-9}{13}$ is shown in the figure.

1126931_1079177_ans_48b5ad39559b41b2b289f7109943a411.png

Find two rational no between $3$ & $4$ .

A rational number between

$3$ and

$4$

$=1/2(3+4)=7/2$

Then

$3<7/2<4$

A rational number between

$3$ and

$7/2$

$=1/2{3+7/2}=1/2(3/1+72)$

$=\dfrac { 1 }{ 2 } \left( \dfrac { 6+7 }{ 2 } \right) =1/2+13/2=13/4$

$13/4$ ,

$7/2$ are two rational number between

$3$ and

$4$

$3<13/4<7/2<4$ .

Represent these number on number line: $\dfrac {7}{8},\dfrac {-5}{3}$

$(i)\ \cfrac{7}{8}:$

Since the given rational number is greater than zero. So, it will always be represented on the right side of zero on the number line. So, first of all, we need to divide the number line between zero and one into

$8$ equal parts and the seventh part of the eight parts will be representation of

$\cfrac{7}{8}$ on number line. It can be represented as shown in the

$Image \ 1$ of the figure above.

$(ii)\ \cfrac{-5}{3}:$

As we can see that the given fraction is an improper fraction but with a negative sign. So, it will be smaller than zero but greater than -2. Hence, the fraction will lie between zero and negative two. To represent we will divide number line between

$0$ and

$-2$ into

$6$ equal parts and the fifth part of the six parts will be

$\cfrac{-5}{3}$ . This can be represented as shown in the

$Image \ 2$ of the figure above.

1009504_1063361_ans_ce2013cef7b64d428f6147614c410818.png

Represent this numbers on number line:
(1) $\dfrac { 9 }{ 7 }$
(2) $-\dfrac { 7 }{ 5 }$

(i) Representing

$\dfrac{9}{7}$ on the number line (see figure 1)

(ii) Representing

$\dfrac{-7}{5}$ on the number line : (see figure 2)

$\dfrac{-7}{5}=-1\dfrac{2}{5}$

1126929_1079104_ans_32f6f67f19084e2b9278e2af1fdd8e28.png

Find a rational number between the following rational numbers:
(i) $\frac { 3 } { 4 } \text { and } \frac { 4 } { 3 }$
(ii) $5 \text { and } 6$
(iii) $- \frac { 3 } { 4 } \text { and } \frac { 1 } { 3 }$

$(i)\dfrac{\dfrac{3}{4}+\dfrac{4}{3}}{2}=\dfrac{\dfrac{9+16}{12}}{2}=\dfrac{25}{24}$ is a rational number between

$\dfrac{3}{4}$ and

$\dfrac{4}{3}$

$(ii)\dfrac{5+6}{2}=\dfrac{11}{2}$ is a rational number between

$5$ and

$6$

$(iii)\dfrac{\dfrac{-3}{4}+\dfrac{1}{3}}{2}=\dfrac{\dfrac{-9+4}{12}}{2}=\dfrac{-5}{24}$ is a rational number between

$\dfrac{-3}{4}$ and

$\dfrac{1}{3}$

Find 3 rational numbers between $\dfrac{1}{2}\;$ and $\dfrac{1}{3}$

$3$ rational numbers between

$\dfrac{1}{2}$ and

$\dfrac{1}{3}$

$1$ .

$\dfrac{\tfrac{1}{2}+\tfrac{1}{3}}{2}=\dfrac{5}{12}$

$2$ .

$\dfrac{\tfrac{1}{2}+\tfrac{5}{12}}{2}=\dfrac{11}{24}$

$3$ .

$\dfrac{\tfrac{5}{12}+\tfrac{1}{3}}{2}=\dfrac{9}{24}=\dfrac{3}{8}$

The

$3$ three rational numbers are

$\dfrac{5}{12}; \dfrac{11}{24}; \dfrac{3}{8}$ .

Express the each of the following as a rational number of the form $\dfrac {p}{q}$ , where $p$ and $q$ are integers and $q\neq 0$ :
$\left (\dfrac {1}{2}\right)^{-5}$

1514774_1670902_ans_db157de4e6af4132bb8fd3a00395d840.jpg

Prove that the associative property hold true for the addition $\left( \dfrac { 7 } { 8 } + \dfrac { 2 } { 5 } \right) + \dfrac { 8 } { 3 }$

$\begin{array}{l} \text{Here} \\ \left( { \dfrac { 7 }{ 8 } +\dfrac { 2 }{ 5 } } \right) +\dfrac { 8 }{ 3 } =\left( { \dfrac { { 35+16 } }{ { 40 } } } \right)+ \dfrac { 8 }{ 3 } \\ =\dfrac { { 51 } }{ { 40 } } +\dfrac { 8 }{ 3 } \\ =\dfrac { { 103+320 } }{ { 120 } } \\ =\dfrac { { 423 } }{ { 120 } } \\ Now, \\ \dfrac { 7 }{ 8 } +\left( { \dfrac { 2 }{ 5 } +\dfrac { 8 }{ 3 } } \right) \\ =\dfrac { 7 }{ 8 } +\left( { \dfrac { { 6+40 } }{ { 15 } } } \right) \\ =\dfrac { 7 }{ 8 } +\dfrac { { 41 } }{ { 15 } } \\ =\dfrac { { 105+328 } }{ { 120 } } \\ =\dfrac { { 433 } }{ { 120 } } \\ \text{Hence the both value are equal} \\\therefore \text{it holds associative property}. \end{array}$

The multiplicative inverse of
${\left( {\dfrac{1}{2}} \right)^4} \div {\left( {\dfrac{1}{3}} \right)^4} + {\left( { - \dfrac{1}{2}} \right)^{^3}}\;{\text{is}}$

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

$\displaystyle =\frac{1}{16}\div \frac{1}{81}+(\frac{-1}{8})$

$\displaystyle =\frac{1}{16}\times \frac{81}{1}+\frac{-1}{8}$

$\displaystyle =\frac{81}{16}-\frac{1}{8}$

$=\dfrac{81-2}{16}$

$=\dfrac{79}{16}$ Ans

Find the product of additive inverse and multiplicative inverse of $-\dfrac{1}{3}$ .

1979508_1315989_ans_16d22e220989423cb5b812092a2dcd94.jpg

Fill in the blank:
The multiplicative inverse of $0$ _______

The multiplicative inverse of 0 is 1/0.

Any number divided by 0 is undefined.

Find 5 rational numbers between $-4/5\quad and\quad -2/3$

2040784_1327716_ans_b10683c0b7f447108e12b4b124e91d52.jpg

Write any two rational numbers which are between $2.3$ and $2.4$ .

$2.3$ and

$2.4$ can be written as

$2.30$ and

$2.40$

$\Rightarrow 2.31$ and

$2.32$ are two rational numbers in between them.

Find any two rational numbers between $\dfrac{-1}{2}$ and $\dfrac{-2}{3}$

The rational number between

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

$\dfrac{-2}{3}$

$\dfrac{\dfrac{-1}{2}+\dfrac{-2}{3}}{2}=\dfrac{\dfrac{-3-4}{6}}{2}=\dfrac{-7}{12}$

The rational number between

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

$\dfrac{-7}{12}$ is

$\dfrac{\dfrac{-1}{2}+\dfrac{-7}{12}}{2}=\dfrac{\dfrac{-12-14}{24}}{2}=\dfrac{-26}{48}=\dfrac{-13}{24}$

Represent on number line
$\dfrac {2}{3}$

Mark few points on number line :

$-1,0,1,2..$
Divide space between every two integers in three equal parts.
The second part after

$0$ and before

$1$ is

$\dfrac{2}{3}$

1275670_1369103_ans_5c6a1b4065324a5981c8cdff1cc94d57.png

Represent on number line
$\dfrac {1}{6}$

Divide

$1$ cm into

$6$ equal parts.In the same way divide each centimeter in to

$6$ equal parts. and from

$0$ to

$\dfrac{1}{6}$ th part is the point shown in the number line.

1275668_1369099_ans_bb0fc0e1fa0a428f8ec389d4c97476cf.PNG

Represent each of the following numbers on the number line:
$-4\dfrac {3}{5}$ .

1689000_1366163_ans_72f1e9e760f14f378e95bd4b7c9a703e.jpg

Represent $\dfrac {3}{7}$ on a number line.

Divide each division into

$7$ equal parts.

Now each sub division represents

$\dfrac{1}{7}$ .

Thus

$\dfrac{3}{7}$ will be at three sub-division to the right of

$0$ .

Hence, the position of

$\dfrac{3}{7}$ is shown in the figure.

1275669_1369102_ans_76a73fd859fe420a822233d078066698.PNG

Represent each of the following numbers on the number line:
$\dfrac {-3}{4}$

We know that

$-3/4$ is greater than

$-1$ and lesser than

$0$ .

$\therefore$ it lies between

$-1$ and

$0$ . It can be represented on number line as
1602840_1366134_ans_d4510fdff3a44cc8b8f1d7fccd63106f.jpg

Represent on number line
$\dfrac {6}{5}$

Divide

$1$ cm into

$5$ equal parts.In the same way divide each centimeter in to

$5$ equal parts. and from

$0$ to

$\dfrac{6}{5}$ th part is the point shown in the number line.

1275671_1369105_ans_021943be15a046cb82373a7e828de391.PNG

Represent each of the following numbers on the number line:
$2\dfrac {2}{5}$ .

1688964_1366081_ans_127370d44404455b9ee476db15517c3d.jpg

Draw number lines and locate the points on them.
$\displaystyle \frac{1}{2},\frac{1}{4},\frac{3}{4},\frac{4}{4}$

Given numbers are

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

$\dfrac{1}{4}=\dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{4}$

So we have to draw

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

Let's draw number line from

$0$ to

$2.$

Since all numbers are with denominator

$4.$

We divide the line between

$0$ &

$1$ into

$4$ equal parts.

1202330_1416703_ans_b243fc65d2c64b01be18a6e7b8c9f348.png

Express the each of the following as a rational number of the form $\dfrac {p}{q}$ , where $p$ and $q$ are integers and $q\neq 0$ :
$\dfrac {1}{3^{-2}}$

1514773_1670900_ans_b25666878058479aa36291239539573c.jpg

Find three rational numbers between $-3$ and $5$ .

$\textbf{Step 1: Find three rational numbers.}$

$\text{Given numbers are}$

$-3$

$\text{and}$

$5.$

$\text{We know, all the integers are rational numbers.}$

$\text{Three integers between }-3 \text{ and 5 are }-2,1\text{ and 3.}$

$\textbf{Hence, three rational numbers between }$

$\bf{-3}$

$\textbf{and}$

$\bf 5$

$\textbf{are}$

$\bf{-2,1}\text{ and 3.}$

Represent $\dfrac53$ and $\dfrac{-5}3$ on the number line.

1336613_1385120_ans_491810e072ef47b9849874ddbe6fabdb.png

What is the sum of additive inverse and multiplicative inverse of $-7$ .

$\textbf{Step 1: Find the required sum.}$

$\text{The given number is}$

$-7$

$\text{Additive inverse of}$

$-7=7$

$\text{Multiplicative inverse of}$

$-7=-\dfrac{1}{7}$

$\therefore\text{Sum }=7+\left(-\dfrac{1}{7}\right)$

$=7-\dfrac{1}{7}$

$=\dfrac{49-1}{7}$

$=\dfrac{48}{7}$

$\textbf{Hence, the required sum is}$

$\bf{ \dfrac{48}{7} .}$

Represent $\dfrac{-11}{5}$ and $\dfrac{11}{5}$ on the number line

1336610_1385374_ans_82cd5367f65e4712a8bd40703782befe.png

Draw number lines and locate the points on them:
$\displaystyle \frac{2}{5},\frac{3}{5},\frac{8}{5},\frac{4}{5}$

Given numbers are

$\dfrac{2}{5},\dfrac{3}{5},\dfrac{8}{5},\dfrac{4}{5}$

Let's draw number line fro

$0$ to

$2.$

REF.Image

So, the denominator is

$5.$

We divide

$0$ &

$2$ into

$5$ equal parts.

1202327_1416716_ans_48557e9eac3f43898394a4230e61d0e9.png

Write the additive inverse and multiplicative inverse of a rational number.

*The additive inverse of a number a is the number that, when added to a, yields zero.

Additive inverse of

$a$ is

$-a$ .

*The reciprocal of a number obtained is such that when it is multiplied with the original number the value equals identity 1.

$a\times \dfrac{1}{a}=1$

$\dfrac{1}{a}$ is the multiplicative inverse of

$a$ and

$a$ is the multiplicative inverse of

$a$ .

The multiplicative inverse of 5 will be

$\textbf{Step 1: Using definition find inverse}$

$\text{Two numbers are said to be multiplicative inverses if their product is equal to 1}$

$\text{Let inverse of 5 be } x.$

$\therefore 5 \times x = 1$

$\Rightarrow x = \dfrac{1}{5}$

$\textbf{Hence, inverse of 5 will be}$

$\mathbf{\dfrac{1}{5}.}$

Draw the number line and represent the following rational number on it:
$\cfrac{-5}{8}$

1339174_1645508_ans_60db97db61074d198e9e32ddd30f1377.png

Draw the number line and represent the following rational number on it:
$\cfrac{7}{8}$

1338631_1645510_ans_451c8f9dc3be4b90bb5c6c8e437e1a1a.png

By what number should $\left( 2\right) ^{ -7 }$ be multiplied so that the product is $1.$

Let

$x$ be the required number.

Therefore according to question,

${ \left( 2 \right) }^{ -7 }.x=1$

$\therefore$

$x=\dfrac { 1 }{ { 2 }^{ -7 } } ={ 2 }^{ 7 }$

Express the each of the following as a rational number of the form $\dfrac {p}{q}$ , where $p$ and $q$ are integers and $q\neq 0$ :
$\left (\dfrac {2}{3}\right)^{-2}$

1514776_1670906_ans_de326c536a7b47c8b6992931131494cd.jpg

Express the each of the following as a rational number of the form $\dfrac {p}{q}$ , where $p$ and $q$ are integers and $q\neq 0$ :
$(-4)^2$

1514772_1670897_ans_0d66549e8f60490282689c90c7a0fda1.jpg

The points $P,Q,R,S,T,U,A$ and $B$ on the number line are such that $TR=RS=SU$ and $AP=PQ=QB$ . Name the rational numbers represented by $P,Q,R,S$

Each part which is between the two numbers is divided into

$3$ parts
Therefore,

$A=\cfrac{6}{3}, P=\cfrac{7}{3},Q=\cfrac{8}{3}$ and

$B=\cfrac{9}{3}$
Similarly

$T=\cfrac{-3}{3},R=\cfrac{-4}{3},S=\cfrac{-5}{3}$ and

$U=\cfrac{-6}{3}$
Thus the rational numbers represented

$P,Q,R$ and

$S$ are

$\cfrac{7}{3},\cfrac{8}{3},\cfrac{-4}{3}$ and

$\cfrac{-5}{3}$ respectively.

Verify commutativity of addition of rational number for each of the following pairs of rational numbers:
$\dfrac{-11}{5}$ and $\dfrac{4}{7}$

Commutativity of addition of rational number means that if

$\dfrac ab$ and

$\dfrac cd$ are two rational numbers,

Then

$\dfrac ab+\dfrac cd=\dfrac cd+\dfrac ab$

Here we have,

$\dfrac{-11}{5}$ and

$\dfrac47$

$\therefore \dfrac{-11}5+\dfrac45=\dfrac{-11\times7}{5\times 7}+\dfrac{4\times5}{7\times5}=\dfrac{-55}{35}+\dfrac{20}{35}=\dfrac{-77+20}{35}=-\dfrac{57}{35}\\$

And,

$\dfrac45+\dfrac{-11}5=\dfrac{4\times5}{7\times5}+\dfrac{-11\times7}{5\times 7}=\dfrac{20}{35}+\dfrac{-55}{35}=\dfrac{20-77}{35}=-\dfrac{57}{35}$

$\therefore \dfrac{-11}5+\dfrac45=\dfrac45+\dfrac{-11}5\\$

Hence verified.

Verify commutativity of addition of rational number for the following pair of rational numbers:
$-4$ and $\dfrac{4}{-7}$

Commutativity of addition of rational number means that if

$\dfrac ab$ and

$\dfrac cd$ are two rational numbers,

Then

$\dfrac ab+\dfrac cd=\dfrac cd+\dfrac ab$

Here we have,

$-4$ and

$\dfrac{4}{-7}=-\dfrac47$

$\\\therefore -4+\dfrac{4}{-7}$

$=-4-\dfrac47$

$=-\dfrac{4\times7}{1\times 7}-\dfrac{4\times1}{7\times1}$

$=-\dfrac{28}7-\dfrac{4}{7}$

$=\dfrac{-28-4}{7}$

$=-\dfrac{32}{7}\\$

$\text{And, } \dfrac{4}{-7}-4$

$=-\dfrac47-4$

$=-\dfrac{4\times1}{7\times1}-\dfrac{4\times7}{1\times 7}$

$=-\dfrac{4}{7}-\dfrac{28}7$

$=\dfrac{-4-28}{7}$

$=-\dfrac{32}{7}\\$

$\therefore -4+\dfrac{4}{-7}=\dfrac{4}{-7}-4$

Hence verified.

Verify the property: $x\times y=y\times x$ by taking:
$x=-\dfrac{1}{3}, y=\dfrac{2}{7}$

Given,

$x=-\dfrac13\ ,\ y=\dfrac{2}{7}$

To verify:

$x\times y=y\times x$

Therefore,

$LHS=x\times y=-\dfrac13\times \dfrac{2}7=-\dfrac{1\times2}{3\times 7}=-\dfrac2{21}\\$

$RHS= y\times x=\dfrac27\times \left(-\dfrac13\right)=-\dfrac{2\times1}{7\times3}=-\dfrac2{21}$

$\therefore LHS=RHS$

Hence verified.

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

1514779_1670918_ans_6a26b960158d48d798e9c6a9b6a24b46.jpg

Verify commutativity of addition of rational number for each of the following pairs of rational numbers:
$\dfrac{-3}{5}$ and $\dfrac{-2}{-15}$

Commutativity of addition of rational number means that if

$\dfrac ab$ and

$\dfrac cd$ are two rational numbers,

Then

$\dfrac ab+\dfrac cd=\dfrac cd+\dfrac ab$

Here we have,

$\dfrac{-3}{5}=-\dfrac35$ and

$\dfrac{-2}{-15}=\dfrac{2}{15}$

$\\\therefore\dfrac{-3}{5}+\dfrac{-2}{-15}= -\dfrac35+\dfrac{2}{15}=-\dfrac{3\times3}{5\times 3}+\dfrac{2\times1}{15\times1}=-\dfrac{9}{15}+\dfrac{2}{15}=\dfrac{-9+2}{15}=-\dfrac{7}{15}\\$

And,

$\dfrac{-2}{-15}+\dfrac{-3}{5}=\dfrac{2}{15}- \dfrac35=\dfrac{2\times1}{15\times1}-\dfrac{3\times3}{5\times 3}=\dfrac{2}{15}-\dfrac{9}{15}=\dfrac{2-9}{15}=-\dfrac{7}{15}$

$\therefore \dfrac{-3}{5}+\dfrac{-2}{-15}=\dfrac{-2}{-15}+\dfrac{-3}{5}\\$

Hence verified.

Verify commutativity of addition of rational number for each of the following pairs of rational numbers:
$\dfrac{4}{9}$ and $\dfrac{7}{-12}$

Commutativity of addition of rational number means that if

$\dfrac ab$ and

$\dfrac cd$ are two rational numbers,

Then

$\dfrac ab+\dfrac cd=\dfrac cd+\dfrac ab$

Here we have,

$\dfrac{4}{9}$ and

$\dfrac{7}{-12}=-\dfrac{7}{12}.$ So,

$\dfrac{4}{9}+\dfrac{7}{-12}= \dfrac49-\dfrac{7}{12}$

$=\dfrac{4\times4}{9\times 4}-\dfrac{7\times3}{12\times3}$

$=\dfrac{16}{36}-\dfrac{21}{36}$

$=\dfrac{16-21}{36}$

$=-\dfrac{5}{36}$

And,

$\dfrac{7}{-12}+\dfrac{4}{9}=-\dfrac{7}{12}+ \dfrac49$

$=-\dfrac{7\times3}{12\times3}+\dfrac{4\times4}{9\times 4}$

$=-\dfrac{21}{36}+\dfrac{16}{36}$

$=\dfrac{-21+16}{36}$

$=-\dfrac{5}{36}$

$\therefore \dfrac{4}{9}+\dfrac{7}{-12}=\dfrac{7}{-12}+\dfrac{4}{9}\\$

Hence verified.

Find the value of each of the following:
$(5^{-1}\times 2^{-1})\div 6^{-1}$

1514786_1670942_ans_1e56213f346e4c619632840c2a2aca09.jpg

Verify commutativity of addition of rational number for each of the following pairs of rational numbers:
$\dfrac{2}{-7}$ and $\dfrac{12}{-35}$

Commutativity of addition of rational number means that if

$\dfrac ab$ and

$\dfrac cd$ are two rational numbers,

Then

$\dfrac ab+\dfrac cd=\dfrac cd+\dfrac ab$

Here we have,

$\dfrac{2}{-7}=-\dfrac27$ and

$\dfrac{12}{-35}=-\dfrac{12}{35}$

$\dfrac{2}{-7}+\dfrac{12}{-35}=-\dfrac27-\dfrac{12}{35}$

$=-\dfrac{2\times5}{7\times 5}-\dfrac{12}{35}$

$=-\dfrac{10}{35}-\dfrac{12}{35}$

$=\dfrac{-10-12}{35}$

$=-\dfrac{22}{35}$

And,

$\dfrac{12}{-35}+\dfrac{2}{-7}=-\dfrac{12}{35}-\dfrac27$

$=-\dfrac{12}{35}-\dfrac{2\times5}{7\times 5}$

$=-\dfrac{12}{35}-\dfrac{10}{35}$

$=\dfrac{-12-10}{35}$

$=-\dfrac{22}{35}$

$\therefore \dfrac{2}{-7}+\dfrac{12}{-35}=\dfrac{12}{-35}+\dfrac{2}{-7}$

Hence, the commutativity of the addition of rational numbers has been verified.

Write the additive inverse of the following rational number:
$\dfrac{-17}{5}$

If the answer is $\dfrac ab$ , then the value of $(a+b)$ will be what?

Since

$\dfrac{-17}{5} + \dfrac{17}{5} = \dfrac{-17+17}{5} = \dfrac{0}{5} = 0$

$\Rightarrow \dfrac{17}{5}$ is additive inverse for

$\dfrac{-17}{5}$ .

So, by comparing with

$\dfrac ab$ we get

$(17+5)=22$ as the answer.

Find the value of each of the following:
$3^{-1}+4^{-1}$

1514777_1670914_ans_041c732ad4f94bbcbd1ea086650458f0.jpg

Find a rational number between $-3$ and $1$ .

${\textbf{Step - 1: Let us consider the rational number as x.}}$

${\text{The formula for finding rational number is,}}$ $x = \dfrac{1}{2}\left( {\dfrac{a}{b} + \dfrac{c}{d}} \right)$

${\textbf{Step -2: So, using this formula find the rational number between}}$ $\mathbf{- 3}$ ${\textbf{and}}$ $\mathbf{1.}$

$\Rightarrow x = \dfrac{1}{2}\left( { - 3 + 1} \right)$

$\Rightarrow x = \dfrac{1}{2}\left( { - 2} \right)$

$\Rightarrow x = - 1$

${\textbf{Hence, the rational number between }}$ $\mathbf{ - 3}$ ${\textbf{and}}$ $\mathbf{1}$ ${\textbf{is}}$ $\mathbf{- 1.}$

Name the property of multiplication of rational number illustrated by the following statements:
$\dfrac{-3}{2}\times\dfrac{5}{4}+\dfrac{-3}{2}\times \dfrac{-7}{6}=\dfrac{-3}{2}\times\left(\dfrac{5}{4}+\dfrac{-7}{6}\right)$

Given

$\dfrac{-3}{2}\times\dfrac{5}{4}+\dfrac{-3}{2}\times \dfrac{-7}{6}=\dfrac{-3}{2}\times\left(\dfrac{5}{4}+\dfrac{-7}{6}\right)$

This illustrates the distributivity of multiplication over addition of rational numbers, which states that

$\dfrac ab,\ \dfrac cd$ and

$\dfrac ef$ are three rational numbers,

Then

$\dfrac ab\times \dfrac cd+\dfrac ab \times \dfrac ef=\dfrac ab\times \left( \dfrac cd + \dfrac ef\right)$

Name the property of multiplication of rational number illustrated by the following statements:
$\dfrac{-5}{16}\times \dfrac{8}{15}=\dfrac{8}{5}\times \dfrac{-5}{16}$

Given

$\dfrac{-5}{16}\times \dfrac{8}{15}=\dfrac{8}{5}\times \dfrac{-5}{16}$

This illustrates the commutativity of multiplication of rational numbers, which states that

$\dfrac ab$ and

$\dfrac cd$ are two rational numbers,

Then

$\dfrac ab\times\dfrac cd=\dfrac cd\times\dfrac ab$

Name the property of multiplication of rational number illustrated by the following statements:
$\dfrac{-11}{6}\times \dfrac{16}{-11}=1$

Given

$\dfrac{-11}{6}\times \dfrac{16}{-11}=1$

This illustrates the existence of multiplicative inverse, which states that

$\dfrac ab$ is a rational number, there exists a number

$\dfrac ba$ such that,

$\dfrac ab\times \dfrac ba= 1$

Name the property of multiplication of rational number illustrated by the following statements:
$\dfrac{13}{-17}\times 1= \dfrac{13}{-17}=1\times \dfrac{13}{-17}$

Given

$\dfrac{13}{-17}\times 1= \dfrac{13}{-17}=1\times \dfrac{13}{-17}$

This illustrates the existence of multiplicative identity, which states that

$\dfrac ab$ is a rational number,

Then

$\dfrac ab\times 1=1\times \dfrac ab= \dfrac ab$

Use the distributivity of multiplication of rational numbers over their addition to simplify:
$\dfrac{3}{5}\times\left(\dfrac{35}{24}+\dfrac{10}{1}\right)$

1508877_1671476_ans_003eb321939e462f8d1ba671c2a6e7f1.jpg

Draw the number line and represent the following rational number on it: $\dfrac{3}{8}$

Here we have,

$0<\dfrac38<1$

Since the denominator is

$8$ ,

We divide the line between

$0$ and

$1$ into

$8$ equal parts.

Thus, the given number is represented on the number line as shown.

1691956_1676403_ans_8d50b3a0d32e4ecdba30fc0ab1a9347c.png

Find two rational numbers between $\dfrac{-2}{9}$ and $\dfrac{5}{9}$

Given rational numbers are

$\dfrac{-2}{9}$ and

$\dfrac{5}{9}$

Here, the denominators are same which is

$9$

Two rational numbers between the numerators

$-2$ and

$5$ can be taken as

$2$ and

$4$

Therefore, two rational numbers between

$\dfrac{-2}{9}$ and

$\dfrac{5}{9}$ are :

$\dfrac{2}{9}$ and

$\dfrac{4}{9}$

Name the property of multiplication of rational number illustrated by the following statements:
$\dfrac{-5}{9}\times \left(\dfrac{4}{15}\times \dfrac{-9}{8}\right)=\left(\dfrac{-5}{9}\times \dfrac{4}{15}\right)\times \dfrac{-9}{8}$

Given

$\dfrac{-5}{9}\times \left(\dfrac{4}{15}\times \dfrac{-9}{8}\right)=\left(\dfrac{-5}{9}\times \dfrac{4}{15}\right)\times \dfrac{-9}{8}$

This illustrates the associativity of multiplication of rational numbers, which states that

$\dfrac ab,\ \dfrac cd$ and

$\dfrac ef$ are three rational numbers,

Then

$\dfrac ab\times \left( \dfrac cd \times \dfrac ef\right)=\left(\dfrac ab\times \dfrac cd \right) \times \dfrac ef$

Name the property of multiplication of rational number illustrated by the following statements:
$\dfrac{-17}{5}\times 9=9\times \dfrac{-17}{5}$

Given

$\dfrac{-17}{5}\times 9=9\times \dfrac{-17}{5}$

This illustrates the commutativity of multiplication of rational numbers, which states that

$\dfrac ab$ and

$\dfrac cd$ are two rational numbers,

Then

$\dfrac ab\times\dfrac cd=\dfrac cd\times\dfrac ab$

Draw the number line and represent the following rational numbers on it : $-\dfrac{22}{7}$

Since

$-\dfrac{22}{7}<0$

and it lies between

$-4<-\dfrac{22}{7}<-3$

1908150_1676408_ans_b695d7ba89d848b6b212a14e1a30d9c8.png

Draw the number line and represent the following rational numbers on it : $-\dfrac{7}{3}$

Here we have,

$-\dfrac73=-2\dfrac13$

i.e.,

$-3<-\dfrac73<-2$

Since the denominator is

$3$ ,

We divide the line between

$-3$ and

$-2$ into

$3$ equal parts.

Thus, the given number is represented on the number line as shown.

1691974_1676407_ans_db26aa166c274707a503cdf3f0794701.png

Draw the number line and represent the following rational numbers on it:
$\dfrac{-5}{8}$

Since in

$\dfrac {-5} {8}$ , denominator is

$8.$

We divide line between

$-1$ &

$0$ into

$8$ equal parts.

1516880_1676405_ans_afc6d0664350438c80d65d236dd05cb3.png

Draw the number line and represent the following rational numbers on it : $\dfrac{-31}{3}$

Here we have,

$\dfrac{-31}3=-10\dfrac13$

$\therefore -11<-10\dfrac13<-10$

Since the denominator is

$3$ ,

We divide the line between

$-11$ and

$-10$ into

$3$ equal parts.

Thus, the given number is represented on the number line as shown.

1692023_1676409_ans_eed4b750e6304ccdaa42d2c956364923.png

Represent the following rational number on the number line.
$-\dfrac{23}{6}$ .

We can write

$-\dfrac{23}{6}$ as

$-3\dfrac{5}{6}$ .
1563063_1714903_ans_e936ab124a4846abb4981ad06b3156a2.png

Find a rational number between $\dfrac{3}{8}$ and $\dfrac{2}{5}$ .

Consider

$x=\dfrac{3}{8}$ and

$y=\dfrac{2}{5}$

Then

$\dfrac{3}{8} < \dfrac{2}{5}$

Rational number which lies in-between

$x$ and

$y$

$=\dfrac{1}{2}(x+y)$

Substituting the value of

$x$ and

$y$

$=\dfrac{1}{2}\left(\dfrac{3}{8}+\dfrac{2}{5}\right)$

On further calculation

$=\dfrac{1}{2}\left(\dfrac{15+16}{40}\right)$

So we get

$=\dfrac{1}{2}\times \dfrac{31}{40}=\dfrac{31}{80}$

Thus the rational number which lies in-between

$\dfrac{3}{8}$ and

$\dfrac{2}{5}$ is

$\dfrac{31}{80}$ .

Find a rational number between $-\dfrac{3}{4}$ and $-\dfrac{2}{5}$ .

Consider

$x=-\dfrac{3}{4}$ and

$y=-\dfrac{2}{5}$

Then

$-\dfrac{3}{4} < -\dfrac{2}{5}$

Rational number in-between

$x$ and

$y$

$=\dfrac{1}{2}\left(\left(-\dfrac{3}{4}\right)+\left(-\dfrac{2}{5}\right)\right)$

On further calculation

$=\dfrac{1}{2}\left(\dfrac{-15-18}{20}\right)$

So we get

$=\dfrac{1}{2}\times \dfrac{-23}{20}=-\dfrac{23}{40}$

Thus the raitonal number which lies in-between

$-\dfrac{3}{4}$ and

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

$-\dfrac{23}{40}$ .

Represent each of the following numbers on the number line:
$3\dfrac {1}{2}$ .

1961157_1771597_ans_e7cff20bc43e447192c0e96dfb0fe3f8.jpg

Represent each of the following numbers on the number line:
$8$ .

1689026_1771604_ans_d87f30f10ced4d68bb7faae2bde0ad2a.jpg

Represent $\dfrac{5}{7}$ rational number on the number line.

It lies between the numbers

$0$ and

$1$ .

1563058_1714901_ans_13b5d0084a754c74bf478a41f1f86fde.png

Find the additive inverse of each of the following:
$\dfrac {-17}{8}$ .

Additive inverse of any number say,

$\dfrac {a}{b}$ is

$-\dfrac {a}{b}$ such that,

$\dfrac {a}{b} + \left (\dfrac {-a}{b}\right ) = 0$

$\therefore$ By using the above logic

$\dfrac{-17}{8} +x=0$

$x=0-\dfrac{-17}{8}$

$x=\dfrac{17}{8}$

The additive inverse of

$\dfrac {-17}{8}$ is

$\dfrac {17}{8}$ .

Find the additive inverse of each of the following:
$\dfrac {-3}{11}$ .

Additive inverse of any number say,

$\dfrac {a}{b}$ is

$-\dfrac {a}{b}$ such that,

$\dfrac {a}{b} + \left (\dfrac {-a}{b}\right ) = 0$

$\therefore$ By using the above logic

The additive inverse of

$\dfrac {-3}{11}$ is

$\dfrac {3}{11}$ .

Represent each of the following numbers on the number line:
$-2\dfrac56$ .

1689037_1771609_ans_4c0fa6a1462946448bd31e023b9bb29d.jpg

Find the additive inverse of each of the following:
$\dfrac {23}{9}$ .

Additive inverse of any number say,

$\dfrac {a}{b}$ is

$-\dfrac {a}{b}$ such that,

$\dfrac {a}{b} + \left (\dfrac {-a}{b}\right ) = 0$

$\therefore$ By using the above logic
The additive inverse of

$\dfrac {23}{9}$ is

$\dfrac {-23}{9}$ .

Find the additive inverse of each of the following:
$\dfrac {-16}{-5}$ .

Additive inverse of any number say,

$\dfrac {a}{b}$ is

$-\dfrac {a}{b}$ such that,

$\dfrac {a}{b} + \left (\dfrac {-a}{b}\right ) = 0$

$\therefore$ By using the above logic

The additive inverse of

$\dfrac {-16}{-5}$

Firstly we have to change the given number to positive number by multiplying by

$-1$ we get,

$\dfrac {-16\times -1}{-5\times -1} = \dfrac {16}{5}$

$\therefore$ The additive inverse of

$\dfrac {16}{5}$ is

$\dfrac {-16}{5}$ .

Find the additive inverse of each of the following:
$\dfrac {15}{-4}$ .

Additive inverse of any number say,

$\dfrac {a}{b}$ is

$-\dfrac {a}{b}$ such that,

$\dfrac {a}{b} + \left (\dfrac {-a}{b}\right ) = 0$

$\therefore$ By using the above logic

The additive inverse of

$\dfrac {15}{-4}$

Firstly we have to change the given number to positive number by multiplying by

$-1$ we get,

$\dfrac {15\times -1}{-4\times -1} = \dfrac {-15}{4}$

$\therefore$ The additive inverse of

$\dfrac {-15}{4}$ is

$\dfrac {15}{4}$ .

i. Which rational number is its own additive inverse?
ii. Is the difference of two rational numbers are rational number?
iii. Is addition commutative on rational numbers?
iv. Is addition associative on rational numbers?
v. Is subtraction commutative on rational numbers?
vi. Is subtraction associative on rational numbers?
vii. What is the negative of a negative rational number?

i. Let us consider a number

$\dfrac {a}{b}$

The additive inverse of

$\dfrac {a}{b}$ is

$-\dfrac {a}{b}$ such that,

$\dfrac {a}{b} + \dfrac {-a}{b} = 0$

$\therefore 0$ is the number which is additive inverse of its own.

ii. Let us consider two rational numbers i.e.

$\dfrac {a}{b}$ and

$\dfrac {c}{d}$ where,

$b\neq 0$

Now let's subtract

$\dfrac {a}{b} - \dfrac {c}{d}$

By taking LCM of

$b$ and

$d$ i.e.

$bd$

$\dfrac {ad - bc}{bd}$ Where,

$bd \neq 0$

We can say

$\dfrac {ad - bc}{bd}$ is a rational number

Hence, the difference of two rational numbers is a rational number.

iii. Let us consider two rational numbers i.e.

$\dfrac {a}{b}$ and

$\dfrac {c}{d}$ where,

$b\neq 0$ and

$d\neq 0$

By using commutative law

$\dfrac {a}{b} = \dfrac {c}{d} = \dfrac {c}{d} + \dfrac {a}{b}$

$\therefore$ Yes, Addition is commutative on rational numbers

iv. Let us consider three rational numbers i.e.

$\dfrac {a}{b}, \dfrac {c}{d}$ and

$\dfrac {e}{f}$ where,

$b\neq 0, d\neq 0$ and

$f\neq 0$

By using associative law

$\dfrac {a}{b} + \left (\dfrac {c}{d} + \dfrac {e}{f}\right ) = \left (\dfrac {a}{b} + \dfrac {c}{d}\right ) + \dfrac {e}{f}$

$\therefore$ Yes, Addition is associative on rational numbers

v. Let us consider two rational numbers i.e.

$\dfrac {a}{b}$ and

$\dfrac {c}{d}$ where,

$b\neq 0$ and

$d\neq 0$

By using commutative law

$\dfrac {a}{b} + \dfrac {c}{d}\neq \dfrac {c}{d} + \dfrac {a}{b}$

$\therefore$ No, Addition is not commutative on rational numbers

vi. Let us consider three rational numbers i.e.

$\dfrac {a}{b}, \dfrac {c}{d}$ and

$\dfrac {e}{f}$ where,

$b\neq 0, d\neq 0$ and

$f\neq 0$

By using associative law

$\dfrac {a}{b} + \left (\dfrac {c}{d} + \dfrac {e}{f}\right ) \neq \left (\dfrac {a}{b} + \dfrac {c}{d}\right ) + \dfrac {e}{f}$

$\therefore$ No, Addition is not associative on rational numbers

vii. The negative of a negative rational number is the number itself without the negative sign.

Find the multiplicative inverse (i.e. reciprocal of :
$18$ .

The multiplicative inverse of a no.

$\dfrac ab$ is

$\dfrac ba$

$\because\dfrac ab \times \dfrac ba=1$

The multiplicative inverse of

$18$ is

$(1/18)$ .

Find the multiplicative inverse (i.e. reciprocal of :
$(-3) / (-5)$ .

The multiplicative inverse of a no.

$\dfrac ab$ is

$\dfrac ba$

$\because\dfrac ab \times \dfrac ba=1$

The multiplicative inverse of

$\dfrac {-3}{5}$ is

$\dfrac {-5}{-3}$

We put it in the standard form i.e.

$\dfrac {-5}{-3}\times -1$ we get,

$5/3$ .

Find the multiplicative inverse (i.e. reciprocal of :
$-16$ .

The multiplicative inverse of a no.

$\dfrac ab$ is

$\dfrac ba$

$\because\dfrac ab \times \dfrac ba=1$

We put it in the standard form i.e.

$(1/-16)\times -1$ we get,

$-1/16$ .

The multiplicative inverse of

$-16$ is

$(\dfrac {1}{-16})$

Find the multiplicative inverse of $\dfrac{-17}{12}$

The multiplicative inverse of a no.

$\dfrac ab$ is

$\dfrac ba$

$\because\dfrac ab \times \dfrac ba=1$

We put it in the standard form i.e.

$\dfrac{12}{-17}\times -1$ we get,

$\dfrac{-12}{17}$

The multiplicative inverse of

$\dfrac{-17}{12}$ is

$\dfrac{-12}{17}$

Find the additive inverse of each of the following:
$\dfrac {19}{-6}$ .

Additive inverse of any number say,

$\dfrac {a}{b}$ is

$-\dfrac {a}{b}$ such that,

$\dfrac {a}{b} + \left (\dfrac {-a}{b}\right ) = 0$

$\therefore$ By using the above logic
The additive inverse of

$\dfrac {19}{-6}$
Firstly we have to change the given number of positive number by multiplying by

$-1$ we get,

$\dfrac {19\times 1}{-6\times -1} = \dfrac {-19}{6}$

$\therefore$ The additive inverse of

$\dfrac {-19}{6}$ is

$\dfrac {19}{6}$ .

Find the additive inverse of each of the following:
$0$ .

Additive inverse of any number say,

$\dfrac {a}{b}$ is

$-\dfrac {a}{b}$ such that,

$\dfrac {a}{b} + \left (\dfrac {-a}{b}\right ) = 0$

$\therefore$ By using the above logic
The additive inverse of

$0$ is

$0$ .

Find the additive inverse of each of the following:
$\dfrac {-8}{-7}$ .

Additive inverse of any number say,

$\dfrac {a}{b}$ is

$-\dfrac {a}{b}$ such that,

$\dfrac {a}{b} + \left (\dfrac {-a}{b}\right ) = 0$

$\therefore$ By using the above logic
The additive inverse of

$\dfrac {-8}{-7}$
Firstly we have to change the given number of positive number by multiplying by

$-1$ we get,

$\dfrac {-8\times -1}{-7\times -1} = \dfrac {8}{7}$

$\therefore$ The additive inverse of

$\dfrac {8}{7}$ is

$\dfrac {-8}{7}$ .

Find the multiplicative inverse (i.e. reciprocal of:
$13/25$ .

The multiplicative inverse of a no.

$\dfrac ab$ is

$\dfrac ba$

$\because\dfrac ab \times \dfrac ba=1$

So, the multiplicative inverse of

$\dfrac{13}{25}$ is

$\dfrac {25}{13}$

Find the multiplicative inverse (i.e. reciprocal of:
$-7/24$ .

The multiplicative inverse of a no.

$\dfrac ab$ is

$\dfrac ba$

$\because\dfrac ab \times \dfrac ba=1$

We put it in the standard form i.e.

$(24/-7) \times -1$ we get,

$-24/7$ .

The multiplicative inverse of

$(-7/24)$ is

$(24/-7)$

Find the multiplicative inverse (i.e. reciprocal of :
$\dfrac {2}{-5}$ .

The multiplicative inverse of a no.

$\dfrac ab$ is

$\dfrac ba$

$\because\dfrac ab \times \dfrac ba=1$

The multiplicative inverse of

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

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

Fill in the blank:
$(-8) \times (-9) = (-9) \times (.....)$

Using the communicative law of multiplication,

$[a \times b = b \times a]$

Let,

a = -8, b = -9

$(-8) \times (-9) = (-9) \times (-8)$

Fill in the blank:
$7 \times (-3) = (-3) \times (.....)$

Using the communicative law of multiplication

$[a \times b = b \times a]$

Let,

a=7, b = -3

$7 \times (-3) = (-3) \times (7)$

Fill in the blank:
$(-6) \times (.....) = 6$

Using the closure property of multiplication,

$[a \times b = c]$

Let,

a = -6, missing number b = x, c = 6

$(-6) \times (x) = 6$

(By sending -6 from left hand side to the denominator of the right hand side in the multiplication, the sign remain unchanged)

= x =

$(\frac{6}{-6})$

(

$\because \ \div$ Both numerator and denominator by 6)

= x = -1

Fill in the blank:
$(-18) \times (.....) = (-18)$

Using the closure property of multiplication,

$[a \times b = c]$

Let,

a = -18, missing number b = x, c = -18

$(-18) \times (x) = (-18)$

(By sending -18 from left hand side to the denominator of the right hand side in the multiplication, the sign remain unchanged)

= x =

$(\frac{-18}{-18})$

(

$\because \ \div$ Both numerator and denominator by -18)

= x = 1

Find a rational number exactly halfway between:

$\dfrac{1}{6}$ and $\dfrac{1}{9}$

As we know that
exactly halfway between two rational numbers is

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

Rational numbers are

$\dfrac{1}{6}$ and

$\dfrac{1}{9}$

Here,

$a=\dfrac{1}{6}$ and

$b=\dfrac{1}{9}$

$\therefore \dfrac{a+b}{2}=\dfrac{\dfrac{1}{6}+\dfrac{1}{9}}{2}$

$=\dfrac{\dfrac{3+2}{18}}{2}$

$=\dfrac{5}{18\times2}$

$=\dfrac{5}{36}$

Therefore, the exactly halfway between

$\dfrac{1}{6}$ and

$\dfrac{1}{9}$ is

$\dfrac{5}{36}.$

Fill in the blanks to make the statement true:
The two rational numbers lying between $-2$ and $-5$ with denominator as $1$ are .......................... and .................

The two rational numbers lying between

$−2$ and

$−5$ with denominator as

$1$ are

$-3$ and

$-4$ .

1649532_1791844_ans_006d04c675bf421c88bb95991e7444ad.png

Fill in the blanks to make the statement true:
The multiplicative inverse of $\dfrac{4}{3}$ is ..........

Let, The multiplicative inverse of

$\dfrac{4}{3}$ be

$x$

Then

$x\times \dfrac43=1$

$\Rightarrow x=\dfrac34$

$\therefore$ The multiplicative inverse of

$\dfrac{4}{3}$ is

$\dfrac{3}{4}$

Give one example each to show that the rational numbers are closed under addition, subtraction, and multiplication. Are rational numbers closed under division ? Give one example in support of your answer.

Examples of closure property under

(i) Addition-

$\dfrac{1}{2} + \dfrac{2}{3} = \dfrac{3 + 4}{6} = \dfrac{7}{6}$

(ii) Subtraction

$\dfrac{1}{2} - \dfrac{2}{3} = \dfrac{3 - 4}{6} = \dfrac{-1}{6}$

(iii)Multiplication

$\dfrac{1}{2} \times \dfrac{2}{3} = \dfrac{2}{6} = \dfrac{1}{3}$

No, rational numbers are not closed under division.

Let us take two rational numbers

$3$ and

$0$ .

Then their division

$\dfrac30$ is not defined,

Verify $- (-x) = x$ for $x = \dfrac{3}{5}$

Here,

$LHS=-(-x )=-\left( -\dfrac{3}{5}\right) =\dfrac{3}{5} = x=RHS$

Verify the property $x \times y = y \times x$ of rational numbers by using
$x = \dfrac{2}{3}$ and $y = \dfrac{9}{4}$

The given arrangement shows the Commutative property.

$\therefore LHS=\dfrac23\times \dfrac94=\dfrac{18}{12}=\dfrac32 \\RHS= \dfrac94\times\dfrac23=\dfrac{18}{12}=\dfrac32$

Hence,

$LHS=RHS$

Verify the property $x + y = y + x$ of rational numbers by taking
$x = \dfrac{1}{2}, y = \dfrac{1}{2}$

Given

$x = \dfrac{1}{2}, y = \dfrac{1}{2}$

$\therefore LHS=\dfrac{1}{2} + \dfrac{1}{2} = \dfrac{2}{2} =1$

$RHS=\dfrac{1}{2} + \dfrac{1}{2} = \dfrac{2}{2} =1$

Hence,

$RHS=LHS$

Verify the property $x \times y = y \times x$ of rational numbers by using
$x = 7$ and $y = \dfrac{1}{2}$

The given arrangement shows Commutative property-

$\therefore LHS=7 \times \dfrac{1}{2}=\dfrac{7}{2} \\RHS= \dfrac{1}{2} \times 7=\dfrac{7}{2}$

Hence,

$LHS=RHS$

Verify the property $x + y = y + x$ of rational numbers by taking
$x = \dfrac{-3}{7}, y = \dfrac{-20}{21}$

Given

$x =\dfrac{-3}{7} , y =\dfrac{-20}{21}$

$\therefore LHS=\dfrac{-3}{7} +\left (\dfrac{-20}{21}\right) = \dfrac{-9 + (-20)}{21}= \dfrac{-29}{21}$

$RHS=\dfrac{-20}{21} + \left(\dfrac{-3}{7}\right) = \dfrac{-20 + (-9)}{21} = \dfrac{-29}{21}$

Hence,

$RHS=LHS$

Verify the property $x + y = y + x$ of rational numbers by taking
$x = \dfrac{-2}{3}, y = \dfrac{-5}{6}$

Given

$x =\dfrac{-2}{3} , y =\dfrac{-5}{6}$

$\therefore LHS=\dfrac{-2}{3} +\left (\dfrac{-5}{6}\right) = \dfrac{-4 + (-5)}{6}= \dfrac{-9}{6}$

$RHS=\dfrac{-5}{6} + \left(\dfrac{-2}{3}\right) = \dfrac{-5 + (-4)}{6} = \dfrac{-9}{6}$

Hence,

$RHS=LHS$

Verify the property $x \times y = y \times x$ of rational numbers by using
$x = \dfrac{-5}{7}$ and $y = \dfrac{14}{15}$

The given arrangement shows the Commutative property.

$\therefore LHS=\dfrac{-5}{7} \times \dfrac{14}{15}=\dfrac{-70}{105}=\dfrac{-2}{3} \\RHS= \dfrac{14}{15} \times \dfrac{-5}{7}=\dfrac{-70}{105}=\dfrac{-2}{3}$

Hence,

$LHS=RHS$

Simplify the following by using suitable property. Also name the property.
$\left[ \dfrac{1}{2} \times \dfrac{1}{4} \right] + \left[ \dfrac{1}{2} \times 6\right]$

We have

$\left[ \dfrac{1}{2} \times \dfrac{1}{4} \right] + \left[ \dfrac{1}{2} \times 6\right]$

$=\left[ \dfrac{1}{2} \times \left(\dfrac{1}{4} +6\right)\right]\quad\quad\{\text{Using distributive property} \ \ [a\times b+a\times c]=[a\times(b+c)]\}$

$= \dfrac{1}{2} \left[\dfrac{25}{4}\right]$

$= \dfrac{25}{8}$

Simplify the following by using suitable property. Also name the property.
$\dfrac{-3}{5} \times \left \{ \dfrac{3}{7} + \left(\dfrac{-5}{6}\right) \right \}$

We have

$\dfrac{-3}{5} \times \left\{\dfrac{3}{7} +\left(\dfrac{-5}6\right)\right\}$

$=\left\{ \dfrac{-3}5 \times \dfrac37\right\} +\left\{\dfrac{-3}5\times\left(\dfrac{-5}6\right)\right\}\quad\quad\{\text{Using distributive property} \ \ [a\times(b+c)]=[a\times b+a\times c]\}$

$= \left(\dfrac{-9}{35}\right)+\left(\dfrac{15}{30}\right)$

$= \dfrac{-54+105}{210}$

$=\dfrac{51}{210}=\dfrac{17}{70}$

Verify the property $x \times y = y \times x$ of rational numbers by using
$x = \dfrac{-3}{8}$ and $y = \dfrac{-4}{9}$

The given arrangement shows the Commutative property.

$\therefore LHS=\dfrac{-3}{8} \times \dfrac{-4}{9} =\dfrac{12}{72}=\dfrac{1}{6} \\RHS= \dfrac{-4}{9} \times \dfrac{-3}{8}=\dfrac{12}{72}=\dfrac{1}{6}$

Hence,

$LHS=RHS$

Verify the property $x + y = y + x$ of rational numbers by taking
$x = \dfrac{-2}{5}, y = \dfrac{-9}{10}$

Given

$x =\dfrac{-2}{5} , y =\dfrac{-9}{10}$

$\therefore LHS=\dfrac{-2}{5} +\left( \dfrac{-9}{10}\right) = \dfrac{-4 + (-9)}{10}= \dfrac{-13}{10}$

$RHS=\dfrac{-9}{10} + \left(\dfrac{-2}{5}\right) = \dfrac{-9 + (-4)}{10} = \dfrac{-13}{10}$

Hence,

$RHS=LHS$

Verify the property $x \times (y + z) = x \times y + x \times z$ of rational numbers by taking
$x = \dfrac{-1}{2}, y = \dfrac{2}{3}, z = \dfrac{3}{4}$

The given arrangement shows the distributive property of multiplication over addition.

$\therefore LHS=-\dfrac{1}{2} \times \left(\dfrac{2}{3} + \dfrac{3}{4}\right)=\dfrac{-1}{2} \times \dfrac{2 \times 4 + 3 \times 3}{12}=\dfrac{-1}{2} \times \dfrac{8 + 9}{12} =-\dfrac{-17}{24} \\RHS= \left(-\dfrac{1}{2} \times \dfrac{2}{3} \right) + \left(-\dfrac{1}{2} \times \dfrac{3}{4} \right) = \dfrac{-1}{3} + \dfrac{-3}{8}= \dfrac{-8 - 9}{24}= \dfrac{-17}{24}$

Hence,

$LHS=RHS$

Verify the property $x \times (y + z) = x \times y + x \times z$ of rational numbers by taking
$x = \dfrac{-1}{2}, y = \dfrac{3}{4}, z = \dfrac{1}{4}$

The given arrangement shows the distributive property of multiplication over addition.

$\therefore LHS=-\dfrac{1}{2} \times \left(\dfrac{3}{4} + \dfrac{1}{4}\right) =-\dfrac12 \times1=-\dfrac12 \\RHS= \left(-\dfrac{1}{2} \times \dfrac{3}{4} \right) + \left(-\dfrac{1}{2} \times \dfrac{1}{4} \right) = \dfrac{-3}{8} + \dfrac{-1}{8}=- \dfrac12$

Hence,

$LHS=RHS$

Verify the property $x \times (y + z) = x \times y + x \times z$ of rational numbers by taking
$x = \dfrac{-1}{5}, y = \dfrac{2}{15}, z = \dfrac{-3}{10}$

The given arrangement shows the distributive property of multiplication over addition.

$\therefore LHS=\dfrac{-1}{5} \times \left(\dfrac{2}{15} + \dfrac{-3}{10}\right)=\dfrac{-1}{5} \times \dfrac{4 - 9}{30}=\dfrac{-1}{5} \times \dfrac{-5}{30}= \dfrac{1}{30} \\RHS= \left( \dfrac{-1}{5} \times \dfrac{2}{15} \right) + \left(\dfrac{-1}{5} \times \dfrac{-3}{10} \right) = \dfrac{-2}{75} + \dfrac{3}{50}= \dfrac{-4 + 9}{150}=\dfrac{5}{150}=\dfrac{1}{30}$

Hence,

$LHS=RHS$

Verify the property $x \times (y + z) = x \times y + x \times z$ of rational numbers by taking
$x = \dfrac{-2}{3}, y = \dfrac{-4}{6}, z = \dfrac{-7}{9}$

The given arrangement shows the distributive property of multiplication over addition.

$\therefore LHS=\dfrac{-2}{3}\times \left(\dfrac{-4}{6} + \dfrac{-7}{9}\right)\\=\dfrac{-2}{3} \times \left(\dfrac {-12 - 14}{18}\right) \\= \dfrac{-2}{3} \times \dfrac{-26}{18} \\= \dfrac{26}{27} \\RHS= \left( \dfrac{-2}{3} \times \dfrac{-4}{6} \right) + \left(\dfrac{-2}{3} \times \dfrac{-7}{9} \right) \\= \dfrac{4}{9} + \dfrac{14}{27}\\= \dfrac{12 + 14}{27}\\=\dfrac{26}{27}$

Hence,

$LHS=RHS$

Use the distributive of multiplication of rational numbers over addition to simplify
$\dfrac{-5}{4} \times \left[\dfrac{8}{5} + \dfrac{16}{15}\right]$

Using distributive law of multiplication over addition,

$\dfrac{-5}{4} \times \left[\dfrac{8}{5} + \dfrac{16}{15}\right]$

$=\left(-\dfrac{5}{4} \times \dfrac{8}{5}\right) +\left(- \dfrac{5}{4} \times \dfrac{16}{15}\right)$

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

$=\dfrac{10}{3}$

Verify the property $x \times (y \times z) = (x \times y) \times z$ of rational numbers by using
$x = \dfrac{2}{3}, y = \dfrac{-3}{7}, z = \dfrac{1}{2}$

The given arrangement shows the associative property of multiplication.

$\therefore LHS=\dfrac{2}{3} \times \left(\dfrac{-3}{7} \times \dfrac{1}{2}\right) =\dfrac{2}{3} \times \dfrac{-3}{14}=\dfrac{-1}{7} \\RHS= \left(\dfrac{2}{3} \times \dfrac{-3}{7}\right) \times \dfrac{1}{2}= \dfrac{-2}{7} \times \dfrac12=\dfrac{-1}{7}$

Hence,

$LHS=RHS$

Verify the property $x \times (y \times z) = (x \times y) \times z$ of rational numbers by using
$x = 1, y = \dfrac{-1}{2}, z = \dfrac{1}{4}$

The given arrangement shows the associative property of multiplication.

$\therefore LHS=1 \times \left(\dfrac{-1}{2} \times \dfrac{1}{4}\right) =1\times \dfrac{-1}8=\dfrac{-1}{8} \\RHS= \left(1 \times \dfrac{-1}{2}\right) \times \dfrac{1}{4}= \dfrac{-1}{2} \times \dfrac14=\dfrac{1}{8}$

Hence,

$LHS=RHS$

Verify the property $x \times (y \times z) = (x \times y) \times z$ of rational numbers by using
$x = 0, y = \dfrac{1}{2}$

The given arrangement shows the associative property of multiplication.

$\therefore LHS=0\times \left(\dfrac{1}{2} \times z\right) =0 \times \dfrac z2=0\\RHS= \left(0 \times \dfrac{1}{2}\right) \times z= 0 \times z=0$

Hence,

$LHS=RHS$

The multiplicative inverse of $10^{10}$ is

$10^{-10}$
multiplicative inverse is the reciprocal of a number-

$10^{10}=\dfrac{1}{10^{10}}=10^{-10}$

Verify the property $x \times (y \times z) = (x \times y) \times z$ of rational numbers by using
$x = \dfrac{-2}{7}, y = \dfrac{-5}{6}, z = \dfrac{1}{4}$

The given arrangement shows the associative property of multiplication.

$\therefore LHS=\dfrac{-2}{7}\times \left(\dfrac{-5}{6} \times \dfrac{1}{4}\right) =\dfrac{-2}{7} \times \dfrac{-5}{24}=\dfrac{5}{84} \\RHS= \left(\dfrac{-2}{7} \times \dfrac{-5}{6}\right) \times \dfrac{1}{4}= \dfrac{5}{21} \times \dfrac14=\dfrac{5}{84}$

Hence,

$LHS=RHS$

Draw a
number line and represent the following rational numbers on it:

(i) $\dfrac{3}{8}$

(ii) $\dfrac{-3}{4}$

(iii) $\dfrac{-7}{8}$

(iv) $\dfrac{-17}{8}$

Represent the following numbers on the number line:
$7, 7.2, \dfrac {-3}{2}, \dfrac {-12}{5}$ .

1662130_1794029_ans_abbbcecbd0754ab6a46261506a529cd3.png

Find the multiplicative inverse of $(-7)^{-2}\div (90)^{-1}$ .

$a$ is called multiplicative inverse of

$b$ , if

$a\times b=1$ .
We have,

$(-7)^{-2}\div (90)^{-1}$

$=\dfrac{1}{(-7)^2}\div \dfrac{1}{(90)^1}=\dfrac{1}{49}\div \dfrac{1}{90}=\dfrac{1}{49}\times \dfrac{90}{1}=\dfrac{90}{49}$

$\because$

$a^{-m}=\dfrac{1}{a^m}$
Put,

$b=\dfrac{90}{49}$

$\therefore a\times \dfrac{90}{49}=1$

$\Rightarrow a = \dfrac{49}{90}$

Find three rational numbers between:
$0.1$ and $0.11$ .

$0.101, 0.102, 0.103$ (terminating decimals) are three rational numbers which lie between

$0.1$ and

$0.11$ .

Verify commutative property of multiplication for the following pairs of rational numbers:
$\dfrac{4}{5}$ and $\dfrac{-7}{8}$

Commutative property of multiplication states that

$a\times b$ is equal to

$b\times a$ .

So,

$\begin{aligned}{}\frac{4}{5} \times \frac{{ - 7}}{8}& = \frac{{4 \times - 7}}{{5 \times 8}}\\ \\&= \frac{{ - 28}}{{40}}\\\\& = - \frac{7}{{10}}\quad\quad\quad\dots(i)\end{aligned}$

And,

$\begin{aligned}{}\frac{{ - 7}}{8} \times \frac{4}{5} &= \frac{{ - 7 \times 4}}{{8 \times 5}}\\\\ &= \frac{{ - 28}}{{40}}\\\\ &= - \frac{7}{{10}}\quad\quad\quad\dots(ii)\end{aligned}$

From

$(i)$ and

$(ii)$ ,

$\displaystyle \frac{4}{5} \times \frac{{ - 7}}{8}=\frac{{ - 7}}{8} \times \frac{4}{5}$

Hence, verified.

Find the multiplication inverse of the following:
$\left(\dfrac{81}{16}\right)^{-\tfrac{3}{4}}$

$\left(\dfrac{81}{16}\right)^{-\tfrac{3}{4}}$

Let us simplify to find the multiplicative inverse of the given expression,

$\left(\dfrac{81}{16}\right)^{-\tfrac{3}{4}}=\left(\dfrac{16}{81}\right)^{\tfrac{3}{4}}$

$=\left(\dfrac{2^4}{3^4}\right)^{\tfrac{3}{4}}$

$=\left(\dfrac{2}{3}\right)^{4\times \tfrac{3}{4} }$

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

$=\dfrac{8}{27}$

So, the multiplication inverse of

$\dfrac{8}{27}$ is

$\dfrac{27}{8}$ .

Find the multiplication inverse of the following:
$\left(\dfrac 57\right)^{-2}\times \left(\dfrac 57\right)^4 \div \left(\dfrac 57\right)^3$

$\left(\dfrac 57\right)^{-2}\times \left(\dfrac 57\right)^4 \div \left(\dfrac 57\right)^3$

Let us simplify to find the multiplicative inverse of the given expression,

$\left(\dfrac 57\right)^{-2}\times \left(\dfrac 57\right)^4 \div \left(\dfrac 57\right)^3=\left(\dfrac 75\right)^2 \times \left(\dfrac 57\right)^4 \div \left(\dfrac 57\right)^3$

$=\left(\dfrac 75\right)^2 \times \left(\dfrac 57\right)^{4-3}$

$=\left(\dfrac 75\right)^2\times \left(\dfrac 57\right)$

$=\left(\dfrac 75\right)\times \left(\dfrac 75\right)\times \left(\dfrac 57\right)$

$=\dfrac 75$

So, the multiplication inverse of

$\dfrac 75$ is

$\dfrac 57$ .

Show that the set of all non-zero rational numbers is not closed under addition.

Closure property: A set is closed under an operation if performance of that operation on members of the set always produces a member of that set.

Let

$a$ and

$b$ be two non-zero rational numbers

Now,

$A=\dfrac ab \in \{Q-0\}$

$B=\dfrac {-a}{b} \in \{Q-0\}$

again

$A+B=\dfrac{a-a}{b}=0$ which doe not completely belong to set

$\{Q-0\}$

So, the set of all non-zero rational numbers is not closed under addition.

Mark the following pairs of rational numbers on the number line :
$\dfrac{2}{5}$ and $\dfrac{-4}{5}$

$\dfrac{2}{5}$ and

$\dfrac{-4}{5}$
1738490_1825706_ans_a95c68ff5fbe4277a68515ca93ec9254.png

Find the multiplication inverse of the following:
$\left\{(-3/2)^{-4}\right\}^{1/2}$

Let us simplify to find the multiplicative inverse of the given expression,

$\left\{(-3/2)^{-4}\right\}^{1/2}=(-3/2)^{-4\times 1/2}$

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

$=(2/-3)^2$

$=(2/-3)\times (2/-3)$

$=4/9$

So, the multiplication inverse of

$4/9$ is

$9/4$ .

Write the rational numbers for each point labeled with a letter.

The rational numbers for each point labeled with a letter are as follows:

$A = 3/7$

$B = 7/7 = 1$

$C = 8/7 = 1\dfrac {1}{7}$

$D = 12 /7 = 1\dfrac {5}{7}$

$E = 13/7 = 1\dfrac {6}{7}$ .

Mark the following pairs of rational numbers on the number line:
$\dfrac{2}{5}$ and $\dfrac{-3}{5}$

$\dfrac{2}{5}$ and

$\dfrac{-3}{5}$
1738486_1825701_ans_0ba8d064d8244e2e9855f9341a163e4f.png

Mark the following pairs of rational numbers on the number line:
$\dfrac{5}{6}$ and $\dfrac{-2}{6}$

$\dfrac{5}{6}$ and

$\dfrac{-2}{3}$
1738488_1825703_ans_6a308da4138745f990b00ede03e580c6.png

Write the rational numbers for each point labeled with a letter.

The rational numbers for each point labeled with a letter are as follows:

$P -3/8$

$Q = -4/8$ or

$-1/2$

$R = -7/8$

$S = -11/8$

$T = -12/8$ or

$-3/2$ .

For each pair of , verify commutative property of addition of rational numbers.
$\dfrac{5}{9} and \dfrac{5}{-12}$

$\dfrac{5}{9} and \dfrac{5}{-12}$
To prove :

$\dfrac{5}{9} + \dfrac{5}{-12} = \dfrac{5}{-12} + \dfrac{5}{-9}$

$LHS = \dfrac{5}{9} + \dfrac{5}{-12}$ LCM of 9 and

$12 = 2 \times 2 \times 2 \times 3 \times 3 = 36$

$LHS \dfrac{5 \times 4}{5 \times 4} - \dfrac{5 \times3}{12 \times3}$

$= \dfrac{20n- 15}{36} = \dfrac{5}{36}$

$RHS = \dfrac{5}{-12} +\dfrac{5}{9}$

$= \dfrac{5 \times 3}{-12 \times 3} + \dfrac{5 \times 4}{9 \times4}$
(

$\therefore LCM of 9 and 12 = 36$ )

$= \dfrac{-15 + 20}{36} = \dfrac{5}{36} \therefore RHS = LHS$

$i.e \dfrac{5}{9} + \dfrac{5}{-12} = \dfrac{5}{-12} + \dfrac{5}{9}$
Hence, the commutative property for the addition of rational numbers is verified.

Find the multiplicative inverse of the following rational numbers.
$\displaystyle \frac { 8 }{ 13 } ,\frac { -13 }{ 11 } ,\frac { 12 }{ 17 } ,\frac { -101 }{ 100 } ,\frac { 26 }{ 23 }$

We know that the multiplicative inverse or reciprocal for a number $x$ , denoted by $\dfrac{1}{x}$ or $x^{-1},$ is a number which when multiplied by $x$ yields the multiplicative identity $1.$

Therefore the multiplicative inverse of rational numbers $\displaystyle \dfrac { 8 }{ 13 } ,-\dfrac { 13 }{ 11 } ,\dfrac { 12 }{ 17 } ,-\dfrac { 101 }{ 100 } ,\dfrac { 26 }{ 23 }$

is $\dfrac { 13 }{ 8 } ,-\dfrac { 11 }{ 13 } ,\dfrac { 17 }{ 12 } ,-\dfrac { 100 }{ 101 } ,\dfrac { 23 }{ 26 }$ respectively.

For each set of rational numbers , given below , verify the associative property of addition of rational numbers .
$\dfrac{-2}{5} , \dfrac{4}{15} and \dfrac{-7}{10}$

To prove

$\dfrac{-2}{5} + \left ( \dfrac{4}{15} + \dfrac{-7}{10} \right ) = \left ( \dfrac{-2}{5} + \dfrac{4}{15} \right ) + \dfrac{-7}{10}$
Taking LCM
2 15,10
3 15,5
5 5,5
1,1
LHS

$= \dfrac{-2}{5} + \left ( \dfrac{4}{15} + \dfrac{-7}{10} \right )$
TAking LCM

$LHS = \dfrac{-2}{5} + \left ( \dfrac{4 \times2}{15 \times 2} + \dfrac{-7 \times 3}{10 \times 3} \right )$

$\dfrac{-2}{5} + \left ( \dfrac{8 - 21 }{30} \right )$
RHS

$= \dfrac{-2}{5} = \dfrac{-2 \times 6}{5 \times6} - \dfrac{13}{30} = \dfrac{-2 \times 6}{5 \times6} (∵LCM of 5 and 30=30)$
RHS

$= \left ( \dfrac{-2 \times 3 }{5 \times 3} + \dfrac{4 \times 1}{15 \times 1} \right ) + \dfrac{-7}{10}$
Taking LCM
3 5,15
5 5,5
1,1
Hence, the associative property for the addition of rational numbers is verified.

Write the additive inverse (negative ) of :
$\dfrac{-3}{8}$

The additive inverse of

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

$\dfrac{3}{8}$

For each set of rational numbers , given below , verify the associative property of addition of rational numbers .
$- 1 \dfrac{5}{6} and \dfrac{-2}{3}$

To prove

$\dfrac{-1}{1} + \left ( \dfrac{5}{6} + \dfrac{-2}{3} \right ) = \left ( \dfrac{-1}{1} + \dfrac{5}{6} \right ) + \dfrac{-2}{3}$
LHS

$= \dfrac{-1}{1} + \left ( \dfrac{5}{6} + \dfrac{-2}{3} \right )$
∴LCM of 6 and 3 = 6

$LHS = \dfrac{-1}{1} + \left ( \dfrac{5 \times 1}{6 \times1} + \dfrac{-2 \times 2}{3 \times 2} \right )$

$= \dfrac{-1}{1} + \dfrac{1}{6}$
Taking LCM
2 3,6
3 3,3
1,1
∴LCM of 6 and 3 = 6
Hence, the associative property for the addition of rational numbers is verified.

For each pair of , verify commutative property of addition of rational numbers.
$- 2 and \dfrac{3}{-5}$

$- 2 and \dfrac{3}{-5}$
to prove

$\dfrac{-2}{1} + \dfrac{-3}{5} = \dfrac{-3}{5} + \dfrac{-2}{1}$
LHS

$= \dfrac{-2}{1} + \dfrac{-3}{5}$

$=\dfrac{-2 \times 5 }{1 \times 5} + \dfrac{-3 \times 1}{5 \times 1}(∵LCM of 1 and 5=5)$

$= \dfrac{-10 - 3}{5} = \dfrac{-13}{5}$
RHS

$= \dfrac{-3}{5} + \dfrac{-2}{1}$

$= \dfrac{-3 \times 1}{5 \times 1} + \dfrac{-2 \times 5}{1 \times 5}(∵LCM of 1 and 5=5)$

$∴ RHS = LHS$
Hence, the commutative property for the addition of rational numbers is verified.

For each pair of , verify commutative property of addition of rational numbers.
$\dfrac{2}{-5} + \dfrac{11}{-15} = \dfrac{11}{-15} + \dfrac{2}{-5}$

$\dfrac{2}{-5} + \dfrac{11}{-15} = \dfrac{11}{-15} + \dfrac{2}{-5}$
LHS

$= \dfrac{2}{-5} + \dfrac{11}{-15}$

$\therefore LCM of 5 and 15=15$

$LHS = \dfrac{-2 \times 3}{5 \times 3} - \dfrac{11 \times1}{15 \times 1}$

$= \dfrac{-6 -11}{17 \times 15}$
RHS

$= \dfrac{11}{15} + \dfrac{2 }{-5}$

$\dfrac{-11 \times 1}{15 \times 1} - \dfrac{2 \times 3}{5 \times 3}$

$= \dfrac{-11 -6}{15} = \dfrac{- 17}{15} \therefore RHS = LHS$

$i.e \dfrac{2}{-5} = \dfrac{11}{-15} = \dfrac{11}{-15} = \dfrac{11}{-15} + \dfrac{2}{5}$
Hence, the commutative property for the addition of rational numbers is verified.

For each set of rational numbers , given below , verify the associative property of addition of rational numbers .
$\dfrac{-7}{9} , \dfrac{2}{-3} and \dfrac{-5}{18}$

$\dfrac{-7}{9} , \dfrac{2}{-3} and \dfrac{-5}{18}$

$\frac{-7}{9} + \left ( \frac{3}{-3} + \frac{-5}{18} \right )$

LHS

$= \dfrac{-7}{9} + \left ( \dfrac{2}{-3} + \dfrac{-5}{18} \right )$

Taking LCM

2 3,18

3 3,9

5 3,3

1,1

∴LCM of 3 and

$18 = 2 \times 3 \times 3 = 18$

Taking LCM

3 3,9

3 3,3

1,1

∴LCM of 3 and 9=3

RHS

$= \left ( \dfrac{-7}{9} + \dfrac{2}{-3} \right ) + \dfrac{-5}{18}$

$\therefore LCM of 3 and 9 = 3$

$= \dfrac{-7 - 6}{9} + \dfrac{-5}{18}$

$= \dfrac{-26 -5}{18} = \dfrac{-31}{18} \therefore RHS = LHS$

Hence, the associative property for the addition of rational numbers is verified.

For the given pair of rational numbers, verify commutative property of addition of rational numbers.
$3$ and $\dfrac{-2}{7}$

Given rational numbers are

$3$ and

$\dfrac{-2}{7}$
To prove :

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

$[\because$ as per commutative property of two rational numbers

$a$ and

$b$ ,

$a+b=b+a]$
LHS

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

$= \dfrac{3 \times 7}{1 \times 7} - \dfrac{2 \times 1}{7 \times 1} (\therefore \text{ LCM of 1 and 7} = 7 )$

$= \dfrac{21-2}{7} = \dfrac{19}{7}$
RHS

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

$= \dfrac{-2 \times 1}{7 \times 1} +\dfrac{3 \times 7}{1\times 7} (∵\text{LCM of 1 and 7}=7)$

$= \dfrac{-2}{7} +\dfrac{21}{7}= \dfrac{19}{7}$

$\implies$ LHS

$=$ RHS
Hence, the commutative property for the addition of rational numbers is verified.

For each set of rational numbers , given below , verify the associative property of addition of rational numbers .
$\dfrac{1}{2} , \dfrac{2}{3} and -\dfrac{1}{6}$

To prove

$\dfrac{1}{2} + \left ( \dfrac{2}{3} + \dfrac{-1}{6} \right ) = \left ( \dfrac{1}{2} + \dfrac{2}{3} \right ) + \dfrac{-1}{6}$
LHS

$=\dfrac{1}{2} +\left ( \dfrac{2}{3} + \dfrac{-1}{6}\right )$ Taking LCM
2 3,6
3 3,3
1,1
∴LCM of 3 and 6=6
LHS

$= \dfrac{1}{2} + \left ( \dfrac{2 \times 2}{3 \times 2} + \dfrac{-1 \times 1}{6 \times 1} \right )$
Hence, the associative property for the addition of rational numbers is verified.

Write the additive inverse (negative ) of :
$\dfrac{4}{-9}$

The additive inverse of

$\dfrac{4}{-9}$ =

$\dfrac{4}{9}$

Write the additive inverse (negative ) of :
-2

The additive inverse of -2=2

Rational Numbers - Class 8 Maths - Extra Questions

Write the additive inverse of each of the following rational numbers:3−11\dfrac{3}{-11}. If it is 311\dfrac{3}{11} enter 11 else enter 00

Represent 74\dfrac{7}{4} on the number line.

Write the additive inverse of each of the following rational numbers:−217\dfrac{-2}{17}

By what number should (5)−5\left( 5\right) ^{ -5 } be multiplied so that the product is 1?

Write the negative (additive inverse) of each of the following:−25\dfrac{-2}{5}

Write the additive inverse of −11−25\dfrac{-11}{-25}

Represent the following rational numbers on the number line.4354\dfrac {3}{5}.

Represent the following rational numbers on the number line.−9/7-9/7.

Write the additive inverse (negative ) of :−13 - \dfrac{1}{3}

Fill in the blanks :58 \dfrac{5}{8} and −310 \dfrac{-3}{10} are on the ............ sides of zero.

Write the negative (additive inverse) of 7−9\dfrac{7}{-9}.If the answer is ab\dfrac ab, then the value of (a+b)(a+b) is:-

Fill in the blank:(−5)×(......)=0(-5) \times (......) = 0

Find the multiplicative inverse (i.e. reciprocal of :−1-1.

Represents each of the following rational numbers on the number line:(−1/3)(-1/3)

Find the additive inverse of:314\cfrac3{14}

Represents each of the following rational numbers on the number line:−439\cfrac{-43}9

Represents each of the following rational numbers on the number line:(37/8)(37/8)

Find the additive inverse of:−1115\cfrac{-11}{15}

Find the additive inverse of:15−4\cfrac{15}{-4}

Find the additive inverse of:−18−13\cfrac{-18}{-13}

Fill the additive inverse of:−9-9

Represents each of the following rational numbers on the number line:36−5\cfrac{36}{-5}

Find the additive inverse of:55

Find the additive inverse of:1−6\cfrac{1}{-6}

Fill in the blanks to make the statements true.Additive inverse of 23\dfrac{2}{3} is ________.

Fill in the blanks to make the statements true.On a number line, 43\dfrac{4}{3} is to the ________ of zero (0)(0).

Fill in the blanks to make the statements true.On a number line, −12\dfrac{-1}{2} is to the ________ of zero (0).(0).

Find a rational number exactly halfway between:−13\dfrac{-1}{3} and 13\dfrac{1}{3}

Find the multiplicative inverse or reciprocal of each of the following:00

How many rational number p/q are there between 0 and 1 for which q<P ?

Which integers have a multiplication inverse?

Fill n the blanks :−−5−12+ - \dfrac{-5}{-12} + its Additive inverse = _____

Write the additive inverse (negative ) of :−31 \dfrac{-3}{1}

Write the additive inverse (negative ) of :413 \dfrac{4}{13}

Write the additive inverse (negative ) of :−75 \dfrac{-7}{5}

Fill n the blanks :If ab \dfrac{a}{b} is additive inverse of −cd′ \dfrac{-c}{d'} then −cd\dfrac{-c}{d} is additive inverese of ________ ?

Find the additive inverse of each of the following:−18-18.

Find the additive inverse of the following:13\dfrac {1}{3}

Fill in the blank:{(−5)×3}(−6)=(......)×{3×(−6)}\{(-5) \times 3\} (-6)= (......) \times \{3 \times (-6)\}

Find the multiplicative inverse of 02.\dfrac02.

Write the additive inverse (negative) of:0

Enter 1 if true else enter 0.The numbers 11 and −1 -1 are their own reciprocals.

Can you find a rational number whose multiplicative inverse is −1-1 ?

Find the multiplicative inverse (i.e. reciprocal of :−1/8-1/8.

Fill n the blanks :Additive inverse of −5−12 \dfrac{-5}{-12} = ________

23×621=621×\displaystyle\frac{2}{3}\times\displaystyle\frac{6}{21}=\displaystyle\frac{6}{21}\times 23\displaystyle\frac{2}{3}If true then enter 11 and if false then enter 00

The addition of rational numbers is commutative.If true then enter 11 and if false then enter 00.

If −59\cfrac{-5}{9} is additive inverse of 5p\cfrac{5}{p}, then pp is equal to

The additive inverse of 28\dfrac{2}{8} is −2q\dfrac{-2}{q} , qq is

The rational number that is equal to its negative.

Represent the number on the number line: 74\dfrac{7}{4}

Verify that: −(−x)=x-(-x)=x for x=1115x=\displaystyle\frac{11}{15}.

Substraction of rational numbers is associative.If true then enter 11 and if false then enter 00

1829×4435=−4435×1829\displaystyle\frac{18}{29}\times\displaystyle\frac{44}{35}=\displaystyle\frac{-44}{35}\times\displaystyle\frac{18}{29}If true then enter 11 and if false then enter 00

9−13+4−16=4−16+\displaystyle\frac{9}{-13}+\displaystyle\frac{4}{-16}=\displaystyle\frac{4}{-16}+ 9−13\displaystyle\frac{9}{-13}.If true then enter 11 and if false then enter 00

Multiplication distributes over substraction of rational numbers.If true then enter 11 and if false then enter 00

If (−1114×613)+(−1114×713)=m14\begin{pmatrix}\displaystyle\frac{-11}{14}\times\displaystyle\frac{6}{13}\end{pmatrix}+\begin{pmatrix}\displaystyle\frac{-11}{14}\times\displaystyle\frac{7}{13}\end{pmatrix}=\dfrac{m}{14}.Then find −m-m

−1718×18−17=...........\displaystyle\frac{-17}{18}\times\displaystyle\frac{18}{-17}=............

The reciprocal of a positive rational number is negative.If true then enter 11 and if false then enter 00

If additive inverse of 418\dfrac{4}{18} is added to it then the result will be ___.

Find the multiplicative inverse of the following.(i) −13-13(ii) 15\dfrac{1}{5}(iii) −58×−37\dfrac{-5}{8}\times\dfrac{-3}{7}(iv) −1×−25-1\times\dfrac{-2}{5}(v) −1-1

State what property allows you to compute 13×(6×43)\dfrac{1}{3}\times\begin{pmatrix}6\times\dfrac{4}{3}\end{pmatrix} as (13×6)×43\begin{pmatrix}\dfrac{1}{3}\times6\end{pmatrix}\times\dfrac{4}{3}.

Is 0.30.3 the multiplicative inverse of 3133\dfrac{1}{3}? Why or why not?

Name the property indicated in the following:817+−817=0\dfrac {8}{17} + \dfrac {-8}{17} = 0

Identify the property in the following statements:2⋅8=8⋅22 \cdot 8 = 8\cdot 2

Identify the property in the following statements:2+(3+4)=(2+3)+42 + (3 + 4) = (2 + 3) + 4

Represent −211\dfrac{-2}{11}, −511\dfrac{-5}{11}, −911\dfrac{-9}{11} on the number line.

Check the commutative property of multiplication for −713,2527\dfrac {-7}{13}, \dfrac {25}{27}

Prove that (−1)×(−1)=1(-1)\times (-1)=1

Write the rational number that does not have any reciprocal.

Check the commutative property of multiplication for the following pairs:−89,−1719\dfrac {-8}{9}, \dfrac {-17}{19}

Check the commutative property of addition for −79,−1819\dfrac {-7}{9}, \dfrac {-18}{19}

Find the additive inverse of each of the following numbers:85,610,−38,−163,−41\dfrac {8}{5}, \dfrac {6}{10}, \dfrac {-3}{8}, \dfrac {-16}{3}, \dfrac {-4}{1}

Check the commutative property of multiplication for 224,34\dfrac {22}{4}, \dfrac {3}{4}

Write the additive inverse of the given rational number:−2−3\dfrac{-2}{-3}

Write the additive inverse of 7−13\dfrac{7}{-13}.

Mark the following rational numbers on the number line.103\cfrac { 10 }{ 3 }

Mark the following rational numbers on the number line.32\cfrac { 3 }{ 2 }

Name the property used in each of the following:−12+0=−12=0+−12\dfrac {-1}{2}+0=\dfrac {-1}{2}=0+\dfrac {-1}{2}

Write the additive inverse of each of the following rational numbers:
$\dfrac{3}{-11}$ .

If it is $\dfrac{3}{11}$ enter $1$ else enter $0$

Represent $\dfrac{7}{4}$ on the number line.

Write the additive inverse of each of the following rational numbers:
$\dfrac{-2}{17}$

By what number should $\left( 5\right) ^{ -5 }$ be multiplied so that the product is 1?

Write the negative (additive inverse) of each of the following:
$\dfrac{-2}{5}$

Write the additive inverse of $\dfrac{-11}{-25}$

Represent the following rational numbers on the number line.
$4\dfrac {3}{5}$ .

Represent the following rational numbers on the number line.
$-9/7$ .

Write the additive inverse (negative ) of :
$- \dfrac{1}{3}$

Fill in the blanks :
$\dfrac{5}{8}$ and $\dfrac{-3}{10}$ are on the ............ sides of zero.

Write the negative (additive inverse) of $\dfrac{7}{-9}$ .

If the answer is $\dfrac ab$ , then the value of $(a+b)$ is:-

Fill in the blank:
$(-5) \times (......) = 0$

Find the multiplicative inverse (i.e. reciprocal of :
$-1$ .

Represents each of the following rational numbers on the number line:
$(-1/3)$

Find the additive inverse of:
$\cfrac3{14}$

Represents each of the following rational numbers on the number line:
$\cfrac{-43}9$

Represents each of the following rational numbers on the number line:
$(37/8)$

Find the additive inverse of:
$\cfrac{-11}{15}$

Find the additive inverse of:
$\cfrac{15}{-4}$

Find the additive inverse of:
$\cfrac{-18}{-13}$

Fill the additive inverse of:
$-9$

Represents each of the following rational numbers on the number line:
$\cfrac{36}{-5}$

Find the additive inverse of:
$5$

Find the additive inverse of:
$\cfrac{1}{-6}$

Fill in the blanks to make the statements true.
Additive inverse of $\dfrac{2}{3}$ is ________.

Fill in the blanks to make the statements true.
On a number line, $\dfrac{4}{3}$ is to the ________ of zero $(0)$ .

Fill in the blanks to make the statements true.
On a number line, $\dfrac{-1}{2}$ is to the ________ of zero $(0).$

Find a rational number exactly halfway between:
$\dfrac{-1}{3}$ and $\dfrac{1}{3}$

Find the multiplicative inverse or reciprocal of each of the following:
$0$

Fill n the blanks :
$- \dfrac{-5}{-12} +$ its Additive inverse = _____

Write the additive inverse (negative ) of :
$\dfrac{-3}{1}$

Write the additive inverse (negative ) of :
$\dfrac{4}{13}$

Write the additive inverse (negative ) of :
$\dfrac{-7}{5}$

Fill n the blanks :
If $\dfrac{a}{b}$ is additive inverse of $\dfrac{-c}{d'}$ then $\dfrac{-c}{d}$ is additive inverese of ________ ?

Find the additive inverse of each of the following:
$-18$ .

Find the additive inverse of the following:
$\dfrac {1}{3}$

Fill in the blank:
$\{(-5) \times 3\} (-6)= (......) \times \{3 \times (-6)\}$

Find the multiplicative inverse of $\dfrac02.$

Write the additive inverse (negative) of:
0

Enter 1 if true else enter 0.
The numbers $1$ and $-1$ are their own reciprocals.

Can you find a rational number whose multiplicative inverse is $-1$ ?

Find the multiplicative inverse (i.e. reciprocal of :
$-1/8$ .

Fill n the blanks :
Additive inverse of $\dfrac{-5}{-12}$ = ________

$\displaystyle\frac{2}{3}\times\displaystyle\frac{6}{21}=\displaystyle\frac{6}{21}\times$ $\displaystyle\frac{2}{3}$

If true then enter $1$ and if false then enter $0$

The addition of rational numbers is commutative.
If true then enter $1$ and if false then enter $0$ .

If $\cfrac{-5}{9}$ is additive inverse of $\cfrac{5}{p}$ , then $p$ is equal to

The additive inverse of $\dfrac{2}{8}$ is $\dfrac{-2}{q}$ , $q$ is

Represent the number on the number line: $\dfrac{7}{4}$

Verify that: $-(-x)=x$ for $x=\displaystyle\frac{11}{15}$ .

Substraction of rational numbers is associative.
If true then enter $1$ and if false then enter $0$

$\displaystyle\frac{18}{29}\times\displaystyle\frac{44}{35}=\displaystyle\frac{-44}{35}\times\displaystyle\frac{18}{29}$
If true then enter $1$ and if false then enter $0$

$\displaystyle\frac{9}{-13}+\displaystyle\frac{4}{-16}=\displaystyle\frac{4}{-16}+$ $\displaystyle\frac{9}{-13}$ .
If true then enter $1$ and if false then enter $0$

Multiplication distributes over substraction of rational numbers.
If true then enter $1$ and if false then enter $0$

If $\begin{pmatrix}\displaystyle\frac{-11}{14}\times\displaystyle\frac{6}{13}\end{pmatrix}+\begin{pmatrix}\displaystyle\frac{-11}{14}\times\displaystyle\frac{7}{13}\end{pmatrix}=\dfrac{m}{14}$ .Then find $-m$

$\displaystyle\frac{-17}{18}\times\displaystyle\frac{18}{-17}=...........$ .

The reciprocal of a positive rational number is negative.
If true then enter $1$ and if false then enter $0$

If additive inverse of $\dfrac{4}{18}$ is added to it then the result will be ___.

Find the multiplicative inverse of the following.
(i) $-13$

(ii) $\dfrac{1}{5}$

(iii) $\dfrac{-5}{8}\times\dfrac{-3}{7}$

(v) $-1$

State what property allows you to compute $\dfrac{1}{3}\times\begin{pmatrix}6\times\dfrac{4}{3}\end{pmatrix}$ as $\begin{pmatrix}\dfrac{1}{3}\times6\end{pmatrix}\times\dfrac{4}{3}$ .

Is $0.3$ the multiplicative inverse of $3\dfrac{1}{3}$ ? Why or why not?

Name the property indicated in the following:
$\dfrac {8}{17} + \dfrac {-8}{17} = 0$

Identify the property in the following statements:
$2 \cdot 8 = 8\cdot 2$

Identify the property in the following statements:
$2 + (3 + 4) = (2 + 3) + 4$

Represent $\dfrac{-2}{11}$ , $\dfrac{-5}{11}$ , $\dfrac{-9}{11}$ on the number line.

Check the commutative property of multiplication for $\dfrac {-7}{13}, \dfrac {25}{27}$

Prove that $(-1)\times (-1)=1$

Check the commutative property of multiplication for the following pairs:
$\dfrac {-8}{9}, \dfrac {-17}{19}$

Check the commutative property of addition for $\dfrac {-7}{9}, \dfrac {-18}{19}$

Find the additive inverse of each of the following numbers:
$\dfrac {8}{5}, \dfrac {6}{10}, \dfrac {-3}{8}, \dfrac {-16}{3}, \dfrac {-4}{1}$

Check the commutative property of multiplication for $\dfrac {22}{4}, \dfrac {3}{4}$

Write the additive inverse of the given rational number:
$\dfrac{-2}{-3}$

$\dfrac{7}{-13}$ .

Mark the following rational numbers on the number line.
$\cfrac { 10 }{ 3 }$

Mark the following rational numbers on the number line.
$\cfrac { 3 }{ 2 }$

Name the property used in each of the following:
$\dfrac {-1}{2}+0=\dfrac {-1}{2}=0+\dfrac {-1}{2}$

Mark the following rational numbers on the number line.
$\cfrac { 3 }{ 4 }$

Write the multiplicate inverse of
$\dfrac { 3 }{ 2 }$

Write the additive inverse of $\left( \dfrac { -3 }{ 10 } +\dfrac { 6 }{ 10 } \right)$

Write the multiplicate inverse of:
$-5\dfrac { 6 }{ 7 }$

Tell what property allows you to compute $\frac{1}{3} \times \left( {6 \times \frac{4}{3}} \right)$ as $\left( {\frac{1}{3} \times 6} \right) \times \frac{1}{4}$