Question

Express each of the following as a rational number:

(i) 5⁻³
(ii) (−2)⁻⁵
(iii) ${\left(\frac{1}{4}\right)}^{-4}$
(iv) ${\left(\frac{-3}{4}\right)}^{-3}$
(v) ${\left(-3\right)}^{-1}\times {\left(\frac{1}{3}\right)}^{-1}$
(vi) ${\left(\frac{5}{7}\right)}^{-1}\times {\left(\frac{7}{4}\right)}^{-1}$
(vii) (5⁻¹−7⁻¹)⁻¹
(viii) ${\left\{{\left(\frac{4}{3}\right)}^{-1}-{\left(\frac{1}{4}\right)}^{-1}\right\}}^{-1}$
(ix) $\left\{{\left(\frac{3}{2}\right)}^{-1}÷{\left(\frac{-2}{5}\right)}^{-1}\right\}$
(x) ${\left(\frac{23}{25}\right)}^0$

Answer 1

Note: $\left[{\left(\frac{a}{b}\right)}^{-1}={\left(\frac{b}{a}\right)}^1\right]$


(i) 5⁻³ = ${\left(\frac{5}{1}\right)}^{-3}={\left(\frac{1}{5}\right)}^3=\frac{{\left(1\right)}^3}{{\left(5\right)}^3}=\frac{1}{125}$


(ii) (−2)⁻⁵ = ${\left(\frac{-2}{1}\right)}^{-5}={\left(\frac{1}{-2}\right)}^5=\frac{{\left(1\right)}^5}{{\left(-2\right)}^5}=\frac{1\times -1}{-32\times -1}=\frac{-1}{32}$


(iii) ${\left(\frac{1}{4}\right)}^{-4}={\left(\frac{4}{1}\right)}^4=\frac{{\left(4\right)}^4}{{\left(1\right)}^4}=\frac{256}{1}=256$


(iv) ${\left(\frac{-3}{4}\right)}^{-3}={\left(\frac{4}{-3}\right)}^3$ = $\frac{{\left(4\right)}^3}{{\left(-3\right)}^3}=\frac{64}{-27}=\frac{64\times -1}{-27\times -1}=\frac{-64}{27}$


(v) ${\left(-3\right)}^{-1}\times {\left(\frac{1}{3}\right)}^{-1}={\left(\frac{1}{-3}\right)}^1\times {\left(\frac{3}{1}\right)}^1={\left(\frac{1\times 3}{-3\times 1}\right)}^1={\left(\frac{3}{-3}\right)}^1=\frac{1}{-1}=\frac{1\times -1}{-1\times -1}=\frac{-1}{1}=-1$


(vi) ${\left(\frac{5}{7}\right)}^{-1}\times {\left(\frac{7}{4}\right)}^{-1}={\left(\frac{7}{5}\right)}^1\times {\left(\frac{4}{7}\right)}^1={\left(\frac{7\times 4}{5\times 7}\right)}^1=\frac{4}{5}$


(vii) ${\left(5^{-1}-7^{-1}\right)}^{-1}={\left(\frac{1}{5}-\frac{1}{7}\right)}^{-1}={\left(\frac{7-5}{35}\right)}^{-1}$
= ${\left(\frac{2}{35}\right)}^{-1}={\left(\frac{35}{2}\right)}^1=\frac{35}{2}$

(viii) ${\left\{{\left(\frac{4}{3}\right)}^{-1}-{\left(\frac{1}{4}\right)}^{-1}\right\}}^{-1}$ = ${\left\{{\left(\frac{3}{4}\right)}^1-{\left(\frac{4}{1}\right)}^1\right\}}^{-1}={\left(\frac{3}{4}-\frac{4}{1}\right)}^{-1}$
$$\begin{gathered} ={\left(\frac{3-16}{4}\right)}^{-1}={\left(\frac{-13}{4}\right)}^{-1} \\ ={\left(\frac{4}{-13}\right)}^1=\left(\frac{4\times -1}{-13\times -1}\right) \\ =\frac{-4}{13} \end{gathered}$$

(ix) $\left\{{\left(\frac{3}{2}\right)}^{-1}÷{\left(\frac{-2}{5}\right)}^{-1}\right\}=\left\{{\left(\frac{2}{3}\right)}^1÷{\left(\frac{5}{-2}\right)}^1\right\}$
$$\begin{gathered} =\left(\frac{2}{3}\times \frac{-2}{5}\right) \\ =\frac{-4}{15} \end{gathered}$$

(x) ${\left(\frac{23}{25}\right)}^0=1$ [since a⁰ = 1 for every integer a]