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Hint: In the above given question we are asked to write the additive inverse and multiplicative inverse of a rational number. For additive inverse, the sum of rational number and its inverse should be equal to zero. For multiplicative inverse, the multiplication of a rational number with its reciprocal should be equal to 1.
Complete step-by-step answer:
Let us assume a rational number be $a$.
Now, we know that the additive inverse of a number $a$ is the number that, when added to $a$, yields zero
$ \Rightarrow a + ( - a) = 0$
Therefore, the additive inverse of $a$ is $ - a$.
Also, we know that the reciprocal of a number or the multiplicative inverse obtained is such that when it is multiplied with the original number the value equals 1.
$ \Rightarrow a \times \dfrac{1}{a} = 1$
Therefore, the multiplicative inverse of $a$ is $\dfrac{1}{a}$ .
Hence, the additive inverse of $a$ is $ - a$ and the multiplicative inverse of $a$ is $\dfrac{1}{a}$.
Note: When we face such types of problems, we must have adequate knowledge of rational numbers and its properties. Here, in this question, we have assumed a rational number and with the aid of the properties of additive inverse and multiplicative inverse the required solution can be obtained.
Complete step-by-step answer:
Let us assume a rational number be $a$.
Now, we know that the additive inverse of a number $a$ is the number that, when added to $a$, yields zero
$ \Rightarrow a + ( - a) = 0$
Therefore, the additive inverse of $a$ is $ - a$.
Also, we know that the reciprocal of a number or the multiplicative inverse obtained is such that when it is multiplied with the original number the value equals 1.
$ \Rightarrow a \times \dfrac{1}{a} = 1$
Therefore, the multiplicative inverse of $a$ is $\dfrac{1}{a}$ .
Hence, the additive inverse of $a$ is $ - a$ and the multiplicative inverse of $a$ is $\dfrac{1}{a}$.
Note: When we face such types of problems, we must have adequate knowledge of rational numbers and its properties. Here, in this question, we have assumed a rational number and with the aid of the properties of additive inverse and multiplicative inverse the required solution can be obtained.
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