# 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) $\frac{19}{-6}$.

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To do:

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

Solution:

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 $\frac{19}{6}$.

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