Chapter 9.4, Problem 36E

To determine

To proof: The expression ${\left(\frac{a}{b}\right)}^n=\frac{a^n}{b^n}$ with help of mathematical induction for integers $n\ge 1$ .

Explanation of Solution

Given information:

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

Formula used:

The steps to prove a statement by mathematical induction are as follows:

Step 1. Show that the statement is true for $n=1$ .

Step 2. Assume that the statement is true for $n=k$ .

Step 3. Prove that the statement is true for $n=k+1$ with help of step 2.

Proof:

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

Denote the expression by $S_n$ .

Recall that the steps to prove a statement by mathematical induction are as follows:

Step 1. Show that the statement is true for $n=1$ .

Step 2. Assume that the statement is true for $n=k$ .

Step 3. Prove that the statement is true for $n=k+1$ with help of step 2.

Substitute $n=1$ in the expression,

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

The above equation holds true.

Therefore, the statement is true for $n=1$ so $S_1$ is true.

Now, assume that the statement if true for $n=k$ .

  ${\left(\frac{a}{b}\right)}^k=\frac{a^k}{b^k}$  ...(1)

Therefore, the statement holds true for $n=k$ so $S_k$ is true.

Now, prove it for $n=k+1$ . Write the expression for $n=k+1$ . Group the terms that involve $S_k$ .

  ${\left(\frac{a}{b}\right)}^{k+1}={\left(\frac{a}{b}\right)}^k\left(\frac{a}{b}\right)$

Substitute the value of $S_k$ from equation (1),

  $\begin{array}{c}{\left(\frac{a}{b}\right)}^{k+1}={\left(\frac{a}{b}\right)}^k\left(\frac{a}{b}\right)\\ =S_k\left(\frac{a}{b}\right)\\ =\frac{a^k}{b^k}\left(\frac{a}{b}\right)\end{array}$

Group the similar terms,

  $\begin{array}{c}{\left(\frac{a}{b}\right)}^{k+1}={\left(\frac{a}{b}\right)}^k\left(\frac{a}{b}\right)\\ =S_k\left(\frac{a}{b}\right)\\ =\frac{a^k}{b^k}\left(\frac{a}{b}\right)\\ =\frac{a^{k+1}}{b^{k+1}}\end{array}$

Result obtained is that $S_k$ implies $S_{k+1}$ that is,

  ${\left(\frac{a}{b}\right)}^{k+1}=\frac{a^{k+1}}{b^{k+1}}$

Hence, by the principle of mathematical induction the statement ${\left(\frac{a}{b}\right)}^n=\frac{a^n}{b^n}$ is true for all integers $n\ge 1$ .