Question Number 213725 by zakirullah last updated on 14/Nov/24
$${please}\:{prove}\:\frac{\mathrm{1}}{{x}}\:=\:{x}^{−\mathrm{1}} \\ $$
Answered by Rasheed.Sindhi last updated on 14/Nov/24
$$\:\:\:\:\:\frac{\mathrm{1}}{{x}}=\frac{{x}^{\mathrm{0}} }{{x}^{\mathrm{1}} }={x}^{\mathrm{0}−\mathrm{1}} ={x}^{−\mathrm{1}} \\ $$
Answered by Rasheed.Sindhi last updated on 14/Nov/24
$${x}^{−\mathrm{1}} ={x}^{\mathrm{0}−\mathrm{1}} =\frac{{x}^{\mathrm{0}} }{{x}^{\mathrm{1}} }=\frac{\mathrm{1}}{{x}} \\ $$
Answered by Ghisom last updated on 14/Nov/24
$$\mathrm{we}\:\mathrm{define}\:\mathrm{the}\:\mathrm{symbol}\:{a}^{{n}} \\ $$$${a}^{{n}} :=\underset{{n}\:\mathrm{times}} {\underbrace{{a}×{a}×{a}×…}} \\ $$$${a}×{b}={c}\:\Leftrightarrow\:{a}={b}\boldsymbol{\div}{c} \\ $$$$\Rightarrow\:{a}^{{m}} ×{a}^{{n}} ={a}^{{m}+{n}} \:\Leftrightarrow\:{a}^{{m}+{n}} \boldsymbol{\div}{a}^{{n}} ={a}^{{m}} \\ $$$$\mathrm{this}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as} \\ $$$${a}^{{m}+{n}} \boldsymbol{\div}{a}^{{n}} ={a}^{\left({m}+{n}\right)−{n}} \\ $$$$\mathrm{generally} \\ $$$${a}^{{m}} \boldsymbol{\div}{a}^{{n}} ={a}^{{m}−{n}} \\ $$$$\mathrm{what}\:\mathrm{happens}\:\mathrm{when}\:{m}={n}? \\ $$$${a}^{{m}} \boldsymbol{\div}{a}^{{m}} ={a}^{{m}−{m}} ={a}^{\mathrm{0}} \\ $$$$\mathrm{but}\:{a}^{{m}} \boldsymbol{\div}{a}^{{m}} =\mathrm{1}\:\Rightarrow\:{a}^{\mathrm{0}} =\mathrm{1} \\ $$$$\mathrm{what}\:\mathrm{happens}\:\mathrm{when}\:{m}<{n}? \\ $$$$\mathrm{let}\:{n}={m}+{k} \\ $$$${a}^{{m}} \boldsymbol{\div}{a}^{{m}+{k}} ={a}^{{m}−\left({m}+{k}\right)} ={a}^{−{k}} \\ $$$$\mathrm{an}\:\mathrm{example}: \\ $$$$\mathrm{2}^{\mathrm{3}} \boldsymbol{\div}\mathrm{2}^{\mathrm{5}} =\mathrm{2}^{−\mathrm{2}} \:\left(?\right) \\ $$$$\mathrm{2}^{\mathrm{3}} \boldsymbol{\div}\mathrm{2}^{\mathrm{5}} =\mathrm{8}\boldsymbol{\div}\mathrm{32}=\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\mathrm{2}^{−\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$${a}^{−{m}} =\frac{\mathrm{1}}{{a}^{{m}} } \\ $$