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

Last updated date: 29th Mar 2023
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Hint-This question can be solved by knowing the concept of additive inverse. Additive inverse is a number which is added to a particular number to obtain zero.

We know that,
if $a$ is any rational number then,
$a + \left( { - a} \right) = 0$
$- a$ is called additive inverse of $a$ an vice versa.
Here,
$\left( i \right)$ Additive inverse of $\dfrac{2}{8}$ is $\left( { - \dfrac{2}{8}} \right)$
$\dfrac{2}{8} + \left( {\dfrac{{ - 2}}{8}} \right) = 0$
Therefore, the additive inverse of $\dfrac{2}{8}$ is $\left( { - \dfrac{2}{8}} \right)$.
$\left( {ii} \right)$ Additive inverse of $\dfrac{{ - 5}}{9}$ is $\left( {\dfrac{5}{9}} \right)$
$\dfrac{{ - 5}}{9} + \left( {\dfrac{5}{9}} \right) = 0$
Therefore, the additive inverse of $\dfrac{{ - 5}}{9}$ is $\left( {\dfrac{5}{9}} \right)$.
$\left( {iii} \right)$ Additive inverse of $\dfrac{{ - 6}}{{ - 5}}$ is $\dfrac{{ - 6}}{5}$
$\dfrac{{ - 6}}{{ - 5}} + \left( {\dfrac{{ - 6}}{5}} \right) = \dfrac{6}{5} + \left( {\dfrac{{ - 6}}{5}} \right) = 0$
Therefore, Additive inverse of $\dfrac{{ - 6}}{{ - 5}}$ is $\dfrac{{ - 6}}{5}$
$\left( {iv} \right)$ Additive inverse of $\dfrac{2}{{ - 9}}$ is $\left( {\dfrac{2}{9}} \right)$
$\dfrac{2}{{ - 9}} + \left( {\dfrac{2}{9}} \right) = 0$
Therefore, Additive inverse of $\dfrac{2}{{ - 9}}$ is $\left( {\dfrac{2}{9}} \right)$.
$\left( v \right)$ Additive inverse of $\dfrac{{19}}{{ - 6}}$ is $\left( {\dfrac{{19}}{6}} \right)$
$\dfrac{{19}}{{ - 6}} + \left( {\dfrac{{19}}{6}} \right) = 0$
Therefore, Additive inverse of $\dfrac{{19}}{{ - 6}}$ is $\left( {\dfrac{{19}}{6}} \right)$.

Note- Whenever we face such types of questions the key concept is that if $a$ is any rational number then, $a + \left( { - a} \right) = 0$ and $\left( { - a} \right)$ is called additive inverse of $a$ an vice versa. In this question we did the same.