Chapter 1.2, Problem 94E

(a)

To determine

To prove: the given Laws of exponents for the case in which $m$ and $n$ are positive integers and $m>n$ .

Explanation of Solution

Given information:

Law 2

Calculation:

Consider two positive numbers $m$ and $n$ such that $m$ is greater than $n$ .

Then, prove:

  $\frac{a^m}{a^n}=a^{m-n}$

First, rewrite $\frac{a^m}{a^n}$ as a product and apply the law of negative exponents to obtain:

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

Since, $a^m$ and $a^{-n}$ have a common base $a$ , apply the additive law of exponents and use the fact that $n$ is a positive number to simplify and obtain the desired result:

  $\begin{array}{c}{a}^m\cdot a^{-n}=a^{m+\left(-n\right)}\\ =a^{m-n}\end{array}$

(b)

To determine

To prove: the given Laws of exponents for the case in which $m$ and $n$ are positive integers and $m>n$ .

Explanation of Solution

Given information:

Law 5

Calculation:

Consider, $n$ is a positive number,

Then prove:

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

First rewrite $\frac{a}{b}$ as a product within the expression ${\left(\frac{a}{b}\right)}^n$

  ${\left(\frac{a}{b}\right)}^n={\left(a\cdot b^{-1}\right)}^n$

The law for powers of a product states that for ant two numbers $x$ and $y$ .

  ${\left(x\cdot y\right)}^n=x^n\cdot y^n$

Now, evaluate the expression ${\left(a\cdot b^{-1}\right)}^n$ using the law for powers of a product and apply the multiplicative law for exponents to ${\left(b^{-1}\right)}^n$ .

  $\begin{array}{c}\left(a\cdot b^{-1}\right)=a^n\cdot {\left(b^{-1}\right)}^n\\ =a^n\cdot b^{-n}\end{array}$

Finally, rewrite $b^{-n}$ using the law for negative exponents to obtain the desired result.

  $\begin{array}{c}{a}^n\cdot b^{-n}=a^n\cdot \frac{1}{b^n}\\ =\frac{a^n}{b^n}\end{array}$

(c)

To determine

To prove: the given Laws of exponents for the case in which $m$ and $n$ are positive integers and $m>n$ .

Answer to Problem 94E

Explanation of Solution

Given information:

Law 6

Calculation:

Consider, $n$ is a positive number,

Then prove:

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

First rewrite the expression ${\left(\frac{a}{b}\right)}^{-n}$ as a product of two exponents.

  ${\left(\frac{a}{b}\right)}^{-n}={\left({\left(\frac{a}{b}\right)}^{-1}\right)}^n$

Therefore, for any two numbers $x,y,$ and $c$ ,

  ${\left(\frac{x}{y}\right)}^c=\frac{x^c}{y^c}$

Apply the law to the expression ${\left(\frac{a}{b}\right)}^{-1}$ to obtain.

  ${\left({\left(\frac{a}{b}\right)}^{-1}\right)}^n={\left(\frac{a^{-1}}{b^{-1}}\right)}^n$

Rewrite as a product and apply the law of negative exponents to $a^{-1}$ and the multiplicative law of exponents to ${\left(b^{-1}\right)}^{-1}$ , giving us the desired result.

  $\begin{array}{c}{\left(\frac{a^{-1}}{b^{-1}}\right)}^n={\left(a^{-1}\cdot {\left(b^{-1}\right)}^{-1}\right)}^n\\ ={\left(a^{-1}\cdot b\right)}^n\\ ={\left(\frac{b}{a}\right)}^n\end{array}$