Write the negative (additive inverse) of each of the following:
(i) $ \frac{-2}{5} $
(ii) $ \frac{7}{-9} $
(iii) $ \frac{-16}{13} $
(iv) $ \frac{-5}{1} $
(v) 0
(vi) 1
(vii) $ -1 $

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,

$x+\frac{-2}{5}=0$

$x=0-(\frac{-2}{5})$

$=0+\frac{2}{5}$   

$=\frac{2}{5}$

The additive inverse of the given rational number is $\frac{2}{5}$.    

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

Therefore,

$x+\frac{7}{-9}=0$

$x=0-(\frac{-7}{9})$

$=0+\frac{7}{9}$   

$=\frac{7}{9}$

The additive inverse of the given rational number is $\frac{7}{9}$.     

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

Therefore,

$x+\frac{-16}{13}=0$

$x=0-(\frac{-16}{13})$

$=0+\frac{16}{13}$   

$=\frac{16}{13}$

The additive inverse of the given rational number is $\frac{16}{13}$.      

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

Therefore,

$x+\frac{-5}{1}=0$

$x=0-(\frac{-5}{1})$

$=0+\frac{5}{1}$   

$=\frac{5}{1}$

The additive inverse of the given rational number is $\frac{5}{1}$.      

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

Therefore,

$x+\frac{-5}{1}=0$

$x=0-(\frac{-5}{1})$

$=0+\frac{5}{1}$   

$=\frac{5}{1}$

The additive inverse of the given rational number is $\frac{5}{1}$.      

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

Therefore,

$x+1=0$

$x=0-1$

$=-1$

The additive inverse of the given rational number is $-1$.  

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

Therefore,

$x+(-1)=0$

$x=0-(-1)$

$x=0+1$

$=1$

The additive inverse of the given rational number is $1$.   

Tutorialspoint

Updated on: 10-Oct-2022

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