Question:
Find the value of:
(i) $\left(\frac{5}{8}\right)^{-1}$
(ii) $\left(\frac{-4}{9}\right)^{-1}$
(iii) $(-7)^{-1}$
(iv) $\left(\frac{1}{-3}\right)^{-1}$
Solution:
We know that $a^{-1}=\frac{1}{a}$ or $a^{-1} \times a=1$
(i) $\left(\frac{5}{8}\right)^{-1}=\frac{8}{5}$
$\because \frac{5}{8} \times\left(\frac{5}{8}\right)^{-1}=1$
(ii) $\left(\frac{-4}{9}\right)^{-1}=\frac{9}{-4}=\frac{-9}{4}$
$\because \frac{-4}{9} \times\left(\frac{-4}{9}\right)^{-1}=1$
(iii) $(-7)^{-1}=\frac{1}{-7}=\frac{-1}{7}$
$\because-7 \times(-7)^{-1}=1$
(iv) $(-3)^{-1}$
$(-3)^{-1}=\frac{1}{-3}=\frac{-1}{3}$
$\because(-3)^{-1} \times(-3)=1$