Fiind the value of each of the following:

Solution:

(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:

$\left(\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right)^{-1}=(3-4)^{-1} \quad \ldots\left(a^{-1}=1 / a\right)$

$=(-1)^{-1}$

$=-1 \quad \cdots\left(a^{-1}=1 / a\right)$