Chapter 5.2, Problem 58WE

To determine

To Prove: $\frac{a^{-m}}{a^{-n}}=a^{\left(-m\right)-\left(-n\right)}$

Explanation of Solution

Given information:

Use exponent laws for positive exponents.

  $m,n$ are positive integers with $m>n$ .

Concept:

  $\frac{a}{b}\cdot \frac{c}{d}=\frac{a\cdot c}{b\cdot d}$

Using law of exponents

  $\begin{array}{l}{a}^{-m}=\frac{1}{a^m}\\ a^m\cdot a^n=a^{m+n}\\ \frac{a^m}{a^n}=a^{m-n}\\ a^0=1\end{array}$

  ${\left(a^n\right)}^m=a^{mn}$

Where $a$ be any value.

  $\frac{a^{-m}}{a^{-n}}=a^{\left(-m\right)-\left(-n\right)}$

Taking left hand side.

Left hand side $=\frac{a^{-m}}{a^{-n}}$

  $\Rightarrow$ Left hand side $=a^{-m}\cdot \frac{1}{a^{-n}}$

  $\Rightarrow$ Left hand side $=\frac{1}{a^m}\cdot \frac{1}{\left(\frac{1}{a^n}\right)}$

  $\Rightarrow$ Left hand side $=\frac{1}{a^m}\cdot \frac{a^n}{1}$

  $\Rightarrow$ Left hand side $=\frac{a^n}{a^m}$

Using exponents laws

We get

  $\Rightarrow$ Left hand side $=a^{n-m}$

  $\Rightarrow$ Left hand side $=a^{-m+n}$

  $\Rightarrow$ Left hand side $=a^{\left(-m\right)-\left(-n\right)}$

  $\Rightarrow$ Left hand side $=$ Right hand side.

Therefore,

  $\frac{a^{-m}}{a^{-n}}=a^{\left(-m\right)-\left(-n\right)}$

Is proved.