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) $\frac{-8}{-7}$
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 $\frac{-8}{7}$.