# Find the additive inverse of each of the following:

Question:

Find the additive inverse of each of the following:

(i) $\frac{1}{3}$

(ii) $\frac{23}{9}$

(iii) $-18$

(iv) $\frac{-17}{8}$

(v) $\frac{15}{-4}$

(vi) $\frac{-16}{-5}$

(vii) $\frac{-3}{11}$

(viii) 0

(ix) $\frac{19}{-6}$

(x) $\frac{-8}{-7}$

Solution:

The additive inverse of $\frac{a}{b}$ is $\frac{-a}{b}$. Therefore, $\frac{a}{b}+\left(\frac{-a}{b}\right)=0$

(i) Additive inverse of $\frac{1}{3}$ is $\frac{-1}{3}$.

(ii) Additive inverse of $\frac{23}{9}$ is $\frac{-23}{9}$.

(iii) Additive inverse of $-18$ is $18 .$

(iv) Additive inverse of $\frac{-17}{8}$ is $\frac{17}{8}$.

(v) In the standard form, we write $\frac{15}{-4}$ as $\frac{-15}{4}$.

Hence, its additive inverse is $\frac{15}{4}$.

(vi) We can write:

$\frac{-16}{-5}=\frac{-16 \times(-1)}{-5 \times(-1)}=\frac{16}{5}$

Hence, its additive inverse is $\frac{-16}{5}$.

(vii) Additive inverse of $\frac{-3}{11}$ is $\frac{3}{11}$.

(viii) Additive inverse of 0 is $0 .$

(ix) In the standard form, we write $\frac{19}{-6} \operatorname{as} \frac{-19}{6}$.

Hence, its additive inverse is $\frac{19}{6}$.

(x) We can write:

$\frac{-8}{-7}=\frac{-8 \times(-1)}{-7 \times(-1)}=\frac{8}{7}$

Hence, its additive inverse is $\frac{-8}{7}$.