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

Answer 1

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 $\frac{1}{3}\mathrm{is}\frac{-1}{3}.$

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

(iii) Additive inverse of -18 is 18.

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

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

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 $\frac{-3}{11}\mathrm{is}\frac{3}{11}.$

(viii) Additive inverse of 0 is 0.

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

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