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