Write the additive inverse of each of the following rational numbers:
(i) $ \frac{-2}{17} $
(ii) $ \frac{3}{-11} $
(iii) $ \frac{-17}{5} $
(iv) $ \frac{-11}{-25} $
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,
$x+\frac{-2}{17}=0$
$x=0-(\frac{-2}{17})$
$=0+\frac{2}{17}$
$=\frac{2}{17}$
The additive inverse of the given rational number is $\frac{2}{17}$.
(ii) Let the additive inverse of the given rational number be $x$.
Therefore,
$x+\frac{3}{-11}=0$
$x=0-(\frac{3}{-11})$
$=0+\frac{3}{11}$
$=\frac{3}{11}$
The additive inverse of the given rational number is $\frac{3}{11}$.
(iii) Let the additive inverse of the given rational number be $x$.
Therefore,
$x+\frac{-17}{5}=0$
$x=0-(\frac{-17}{5})$
$=0+\frac{17}{5}$
$=\frac{17}{5}$
The additive inverse of the given rational number is $\frac{17}{5}$.
(iv) Let the additive inverse of the given rational number be $x$.
Therefore,
$x+\frac{-11}{-25}=0$
$x=0-(\frac{-11}{-25})$
$=0-\frac{11}{25}$
$=-\frac{11}{25}$
The additive inverse of the given rational number is $-\frac{11}{25}$.
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