Question

1. What is the additive inverse of 
$\sf\frac{5}{13}$
​

2. Write the multiplicative inverse of 
$\sf\frac{-2}{3}$


3. Which number has no reciprocal?

4. Find the value :-

$\sf \:(i) \: \frac{2}{5} \:+ \: \frac{3}{4}$

$\sf (ii) \: \frac{-2}{7} \: + \: \frac{4}{3}$


$\sf (iii) \: \frac{4}{3} \:-\: \frac{2}{3} \: + \: \frac{3}{11}$
​

Answer 1

Answer:

What is the additive inverse of  $\sf\frac{5}{13}$

= $\sf\frac{5}{13}$ + ( $\sf\frac{-5}{13}$ ) = 0

so $\sf\frac{-5}{13}$ is the additive inverse of $\sf\frac{5}{13}$

2. Write the multiplicative inverse of  $\sf\frac{-2}{3}$

$\sf\frac{-2}{3}$ × $\sf\frac{-3}{2}$

= $\sf\frac{-1~ × ~-1 }{1~×~1}$

= 1

so, $\sf\frac{-3}{2}$ is Write the multiplicative inverse of  $\sf\frac{-2}{3}$

3. Which number has no reciprocal?

  $\bold{Zero~dont~have~reciprocal}$

4. Find the value :-

$\sf \:(i) \: \frac{2}{5} \:+ \: \frac{3}{4}$

$\sf \: \frac{2}{5} + \frac{3}{4} \\\sf \color{teal}{take \: lcm \: of \: denominator \: 5 \: and \: 4} \\ = \sf\frac {(2 \times 4) + (3 \times 5)}{20} \\ = \sf \frac{23}{20}$

$\sf (ii) \: \frac{-2}{7} \: + \: \frac{4}{3}$

$\sf \frac{ - 2}{7} + \frac{4}{3} \\ \sf \color{teal}{take \: lcm \: of \: denominators \: 3 \: and \: 7}\\ \sf = \frac{ (- 2 \times 3) + (4 \times 7)}{21} \\ = \frac{22}{21}$

$\sf (iii) \: \frac{4}{3} \:-\: \frac{2}{3} \: + \: \frac{3}{11}$

$\sf \frac{4}{3} - \frac{2}{3} + \frac{3}{11} \\ \sf \color{teal}{take \: lcm \: of \: denominators \: 3 \: and \: 11} \\ = \sf\frac{(4 \times 11 )+ (2 \times 11) +( 3 \times 3)}{33} \\ = \sf \: \frac{75}{11}$

Answer 2

$\large\underline{\bold{Question - 1}}$

$\sf \:What \: is \: the \: additive \: inverse \: of  \: \dfrac{5}{13}$

$\large\underline{\bold{Solution-}}$

Basic Concept :- An additive inverse of a number is defined as the value, which on adding with the given number results in zero value. It is the value we add to a number to give resultant zero.

Let, a is any real number, then its additive inverse will be - a so that, a + (-a) = a – a = 0.

$\sf \: Hence \: additive \: inverse \: of  \: \dfrac{5}{13} = \bf \: - \: \dfrac{5}{13}$

$\large\underline{\bold{Question - 2}}$

$\sf \: Write \: the \: multiplicative \: inverse \: of \: - \: \dfrac{2}{3}$

$\large\underline{\bold{Solution-}}$

Basic concept :- The multiplicative inverse of a number x (should be non - zero) is given by 1/x, such that when it is multiplied by given number, it results in value equal to 1. It is also called Reciprocal of a given real number.

If p/q is a fraction, then the multiplicative inverse of p/q should be such that, when it is multiplied to the fraction, then the result should be 1. Hence, q/p is the multiplicative inverse of fraction p/q.

$\sf \: Hence \:multiplicative \: inverse \: of \: - \: \dfrac{2}{3} = \bf \: - \: \dfrac{3}{2}$

$\large\underline{\bold{Question - 3}}$

$\sf \: Which \: number \: has \: no \: reciprocal?$

$\large\underline{\bold{Solution-}}$

$\bf \: 0 \: \sf \: has \: no \: reciprocal \: as \: \dfrac{1}{0} \: is \: not \: defined.$

$\large\underline{\bold{Question - 4}}$

$\bf \: Find \: the \: value \: of \:$

$\bf \: (i) \: \sf \: \dfrac{2}{5} + \dfrac{3}{4}$

Firstly, We have to find the LCM of 5 and 4

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:4\:, \: 5 \:\:}}}\\ {\underline{\sf{2}}}& \underline{\sf{\:\:2\:, \: 5 \:\:}} \\\underline{\sf{5}}&\underline{\sf{\:\:1\:, \: 5 \:\:}}\\\underline{\sf{}}&{\sf{\:\:1\:, \: 1 \:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}$

So, LCM of 5 and 4 is 20

As we know that,

To find the operation of additions or subtraction in rational numbers, we will divide the LCM by denominator and then we will multiply the outcome with numerator.

So,

$= \: \sf \: \dfrac{2 \times 4 + 3 \times 5}{20}$

$= \: \sf \: \dfrac{8 + 15}{20}$

$= \: \sf \: \dfrac{23}{20}$

$\bf \: (ii) \: \sf \: - \: \dfrac{2}{7} + \dfrac{4}{3}$

Firstly, We have to find the LCM of 7 and 3

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{3}}}&{\underline{\sf{\:\:7\:, \: 3 \:\:}}}\\ {\underline{\sf{7}}}& \underline{\sf{\:\:7\:, \: 1 \:\:}} \\{\sf{}}&\underline{\sf{\:\:1\:, \: 1 \:\:}}\end{array}\end{gathered}\end{gathered}\end{gathered}$

So, LCM of 3 and 7 is 21

As we know that,

To find the operations of addition or subtraction of rational numbers, we will divide the LCM by enominator and then we will multiply the outcome with numerator.

So,

$= \: \sf \: \dfrac{ - 2 \times 3 + 4 \times 7}{21}$

$= \: \sf \: \dfrac{ - 6 + 28}{21}$

$= \: \sf \: \dfrac{22}{21}$

$\bf \: (iii) \: \sf \: \dfrac{4}{3} \:-\: \dfrac{2}{3} \: + \: \dfrac{3}{11}$

Firstly, We have to find the LCM of 3 and 11.

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{3}}}&{\underline{\sf{\:\:11\:, \: 3 \:\:}}}\\ {\underline{\sf{11}}}& \underline{\sf{\:\:11\:, \: 1 \:\:}} \\{\sf{}}&\underline{\sf{\:\:1\:, \: 1 \:\:}}\end{array}\end{gathered}\end{gathered}\end{gathered}$

So, LCM of 11 and 3 is 33.

As we know that,

To find the operations of addition or subtraction of rational numbers, we will divide the LCM by enominator and then we will multiply the outcome with numerator.

$= \: \sf \: \dfrac{4 \times 11 - 2 \times 11 + 3 \times 3}{33}$

$= \: \sf \: \dfrac{44 - 22 + 9}{33}$

$= \: \sf \: \dfrac{22 + 9}{33}$

$= \: \sf \: \dfrac{31}{33}$

Note :-

Additive Inverse and Additive Identity both are different.

• Additive inverse is the number which ia added to the given number to get the result as zero.

• Additive identity is the value that is added to number to get the original number. Additive identity is always 0.