Fiind the value of each of the following:
(i) 3−1 + 4−1
(ii) (30 + 4−1) × 22
(iii) (3−1 + 4−1 + 5−1)0
(iv) $\left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}$
(i) We know from the property of powers that for every natural number a, a−1 = 1/a. Then:
$3^{-1}+4^{-1}=\frac{1}{3}+\frac{1}{4} \quad \ldots\left(a^{-1}=1 / a\right)$
$=\frac{4+3}{12}$
$=\frac{7}{12}$
(ii) We know from the property of powers that for every natural number a, a−1 = 1/a.
Moreover, a0 is 1 for every natural number a not equal to 0. Then:
$\left(3^{0}+4^{-1}\right) \times 2^{2}$
$=\left(1+\frac{1}{4}\right) \times 4 \quad\left[\right.$ as, $\left.\mathrm{a}^{-1}=\frac{1}{\mathrm{a}} ; \mathrm{a}^{0}=1\right]$
$=\frac{5}{4} \times 4$
$=5$
(iii) We know from the property of powers that for every natural number a, a−1 = 1/a.
Moreover, a0 is 1 for every natural number a not equal to 0. Then:
$\left(3^{-1}+4^{-1}+5^{-1}\right)=1 \quad \cdots$ (Ignore the expression inside the bracket and use $a^{0}=1$ immediately. $)$
(iv) We know from the property of powers that for every natural number a, a−1 = 1/a. Then:
$\left(\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right)^{-1}=(3-4)^{-1} \quad \ldots\left(a^{-1}=1 / a\right)$
$=(-1)^{-1}$
$=-1 \quad \cdots\left(a^{-1}=1 / a\right)$