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} $.
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 $\frac{19}{6}$.
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