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
$${x}^{−\mathrm{1}} ={x}^{\mathrm{0}−\mathrm{1}} =\frac{{x}^{\mathrm{0}} }{{x}^{\mathrm{1}} }=\frac{\mathrm{1}}{{x}} \\ $$
Commented by zakirullah last updated on 15/Nov/24
$${very}\:{true}\:{but}\:{some}\:{students}\:{asked}\:{about}\: \\ $$$${the}\:{power}\:{or}\:{exponent}\:{that}\:{why}\:{they}\:{change}\:{its} \\ $$$${sign}\:{when}\:{it}\:{go}\:{from}\:{nomerator}\:{to}\:{denominator}\:{and} \\ $$$${from}\:{denominator}\:{to}\:{nomerator}? \\ $$$${please}\:{give}\:{me}\:{any}\:{suggistion}. \\ $$
Commented by Rasheed.Sindhi last updated on 15/Nov/24
$$\left(\mathrm{1}\right) \\ $$$${a}^{\mathrm{0}} =\mathrm{1}\:\left({definition}\right) \\ $$$$\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}.{a}^{−\mathrm{1}} }{{a}.{a}^{−\mathrm{1}} } \\ $$$$\:\left({numerator}\:{and}\:{denominator}\:{may}\right. \\ $$$$\left.\:{be}\:{multiplied}\:{by}\:{same}\:{number}\right) \\ $$$$\:\:\:\:=\frac{{a}^{−\mathrm{1}} }{{a}^{\mathrm{1}+\left(−\mathrm{1}\right)} }=\frac{{a}^{−\mathrm{1}} }{{a}^{\mathrm{0}} }=\frac{{a}^{−\mathrm{1}} }{\mathrm{1}}={a}^{−\mathrm{1}} \\ $$$$\left(\mathrm{2}\right) \\ $$$$\begin{cases}{{a}^{{x}} ×{a}^{{y}} ={a}^{{x}+{y}} }\\{\frac{{a}^{{x}} }{{a}^{{y}} }={a}^{{x}−{y}} \:\left(\ast\right)}\end{cases}\left({As}\:{Rules}\right) \\ $$$$\mathrm{1}=\frac{{a}}{{a}}={a}^{\mathrm{1}−\mathrm{1}} ={a}^{\mathrm{0}} \Rightarrow{a}^{\mathrm{0}} =\mathrm{1} \\ $$$${Now}, \\ $$$$\frac{\mathrm{1}}{{a}^{{n}} }=\frac{{a}^{\mathrm{0}} }{{a}^{{n}} }={a}^{\mathrm{0}−{n}} ={a}^{−{n}} \\ $$$$\left(\ast\right)\:{a}^{{x}} ×{a}^{{y}} ={a}^{{x}+{y}} \Rightarrow{a}^{{x}} =\frac{{a}^{{x}+{y}} }{{a}^{{y}} }={a}^{\left({x}+{y}\right)−{y}} \\ $$$$\:\:\:\:\:\Rightarrow\frac{{a}^{{m}} }{{a}^{{n}} }={a}^{{m}−{n}} \\ $$
Answered by Ghisom last updated on 14/Nov/24
$$\mathrm{we}\:\mathrm{define}\:\mathrm{the}\:\mathrm{symbol}\:{a}^{{n}} \\ $$$${a}^{{n}} :=\underset{{n}\:\mathrm{times}} {\underbrace{{a}×{a}×{a}×…}} \\ $$$${a}×{b}={c}\:\Leftrightarrow\:{a}={b}\boldsymbol{\div}{c} \\ $$$$\Rightarrow\:{a}^{{m}} ×{a}^{{n}} ={a}^{{m}+{n}} \:\Leftrightarrow\:{a}^{{m}+{n}} \boldsymbol{\div}{a}^{{n}} ={a}^{{m}} \\ $$$$\mathrm{this}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as} \\ $$$${a}^{{m}+{n}} \boldsymbol{\div}{a}^{{n}} ={a}^{\left({m}+{n}\right)−{n}} \\ $$$$\mathrm{generally} \\ $$$${a}^{{m}} \boldsymbol{\div}{a}^{{n}} ={a}^{{m}−{n}} \\ $$$$\mathrm{what}\:\mathrm{happens}\:\mathrm{when}\:{m}={n}? \\ $$$${a}^{{m}} \boldsymbol{\div}{a}^{{m}} ={a}^{{m}−{m}} ={a}^{\mathrm{0}} \\ $$$$\mathrm{but}\:{a}^{{m}} \boldsymbol{\div}{a}^{{m}} =\mathrm{1}\:\Rightarrow\:{a}^{\mathrm{0}} =\mathrm{1} \\ $$$$\mathrm{what}\:\mathrm{happens}\:\mathrm{when}\:{m}<{n}? \\ $$$$\mathrm{let}\:{n}={m}+{k} \\ $$$${a}^{{m}} \boldsymbol{\div}{a}^{{m}+{k}} ={a}^{{m}−\left({m}+{k}\right)} ={a}^{−{k}} \\ $$$$\mathrm{an}\:\mathrm{example}: \\ $$$$\mathrm{2}^{\mathrm{3}} \boldsymbol{\div}\mathrm{2}^{\mathrm{5}} =\mathrm{2}^{−\mathrm{2}} \:\left(?\right) \\ $$$$\mathrm{2}^{\mathrm{3}} \boldsymbol{\div}\mathrm{2}^{\mathrm{5}} =\mathrm{8}\boldsymbol{\div}\mathrm{32}=\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\mathrm{2}^{−\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$${a}^{−{m}} =\frac{\mathrm{1}}{{a}^{{m}} } \\ $$
Commented by zakirullah last updated on 15/Nov/24
$${well}−{done}\:{sir}\:{G} \\ $$