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

AcademicMathematicsNCERTClass 8

To do:

We have to write the additive inverse of the given rational numbers.

Solution:

Additive Inverse:

The number in the set of real numbers that when added to a given number will give zero. 

(i) Let the additive inverse of the given rational number be $x$.

Therefore,

$x+\frac{2}{8}=0$

$x=0-(\frac{2}{8})$

$=0-\frac{2}{8}$   

$=-\frac{2}{8}$

The additive inverse of the given rational number is $-\frac{2}{8}$.    

(ii) Let the additive inverse of the given rational number be $x$.

Therefore,

$x+\frac{-5}{9}=0$

$x=0-(\frac{-5}{9})$

$=0+\frac{5}{9}$   

$=\frac{5}{9}$

The additive inverse of the given rational number is $\frac{5}{9}$.     

(iii) Let the additive inverse of the given rational number be $x$.

Therefore,

$x+\frac{-6}{-5}=0$

$x=0-(\frac{-6}{-5})$

$=0-\frac{6}{5}$   

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

The additive inverse of the given rational number is $-\frac{6}{5}$.      

(iv) Let the additive inverse of the given rational number be $x$.

Therefore,

$x+\frac{2}{-9}=0$

$x=0-(\frac{2}{-9})$

$=0+\frac{2}{9}$   

$=\frac{2}{9}$

The additive inverse of the given rational number is $\frac{2}{9}$.      

(v) Let the additive inverse of the given rational number be $x$.

Therefore,

$x+\frac{19}{-6}=0$

$x=0-(\frac{19}{-6})$

$=0+\frac{19}{6}$   

$=\frac{19}{6}$

The additive inverse of the given rational number is $\frac{19}{6}$.      

raja
Updated on 10-Oct-2022 13:47:38

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