Problem 34 - College Algebra

Question

In Exercises 25-34, use mathematical induction to prove that each statement is true for every positive integer \(n.\) $$\left(\frac{a}{b}\right)^{n}=\frac{a^{n}}{b^{n}}$$

Step-by-Step Solution

Answer

Using mathematical induction, we have proved that the statement \(\left(\frac{a}{b}\right)^{n}=\frac{a^{n}}{b^{n}}\) is true for every positive integer \( n \).

Step 1: Setup

Define \( P(n) \) as \(\left(\frac{a}{b}\right)^{n}=\frac{a^{n}}{b^{n}}\). We begin by establishing the base case.

Step 2: Base Case

We start with \( n=1 \). According to the definition of \( P(n) \), \( P(1) \) would be \(\left(\frac{a}{b}\right)^{1}=\frac{a^{1}}{b^{1}}\) which simplifies to \(\frac{a}{b} = \frac{a}{b}\). Hence, \( P(1) \) is true.

Step 3: Induction Hypothesis

Now, we assume the statement \( P(k) \) to be true for some arbitrary positive integer \( k \), i.e., we assume that \(\left(\frac{a}{b}\right)^{k}=\frac{a^{k}}{b^{k}}\).

Step 4: Induction Step

We need to show the statement \( P(k+1) \) is true, i.e., \(\left(\frac{a}{b}\right)^{k+1}=\frac{a^{k+1}}{b^{k+1}}\). If we expand the left-hand side using the rules of exponents, we obtain \(\left(\frac{a}{b}\right)^{k}\cdot \left(\frac{a}{b}\right)^{1}.\) According to our induction hypothesis, \(\left(\frac{a}{b}\right)^{k} = \frac{a^{k}}{b^{k}}\), so substitute to get \(\frac{a^{k}}{b^{k}} \cdot \frac{a}{b} = \frac{a^{k}}{b^{k}}\cdot \frac{a^{1}}{b^{1}} = \frac{a^{k+1}}{b^{k+1}}. \) Hence, the statement \( P(k+1) \) is also true if \( P(k) \) is true.

Key Concepts

Exponents

Exponents are a fundamental concept in mathematics, providing a way to represent repeated multiplication concisely. When a number or expression is raised to a power, it indicates the number of times it is multiplied by itself. In the expression \(x^n\), \(x\) is known as the base, and \(n\) is the exponent. Understanding how exponents work is crucial for comprehending mathematical induction, as they often feature in problems involving sequences and powers.

Key properties include:

• Product of powers: \(x^m \cdot x^n = x^{m+n}\)
• Power of a power: \((x^m)^n = x^{mn}\)
• Power of a product: \((xy)^n = x^n \cdot y^n\)

These rules will be handy as you follow through a proof involving exponents and show how such expressions simplify, aiding in understanding the concept of powers within mathematical induction.

Base Case

The base case is the first step in a mathematical induction proof, ensuring that the statement holds for the initial value, typically \(n=1\). This step verifies that the assertion is true at the starting point, forming the foundation for further steps.

In our exercise, the base case involves checking the proposition \(P(1)\):

• The expression is \(\left(\frac{a}{b}\right)^1 = \frac{a^1}{b^1}\).
• Simplifying, we find \(\frac{a}{b} = \frac{a}{b}\), confirming the statement is true.

This confirmation is crucial because it establishes the validity of the statement for \(n=1\), setting the stage for the induction hypothesis and subsequent proof steps.

Induction Hypothesis

The induction hypothesis assumes that the statement we want to prove is true for some arbitrary positive integer \(k\). It bridges the base case to the induction step by providing a stepping stone that connects both.

Here's how the induction hypothesis is applied:

• Assume \(P(k)\) is true: \(\left(\frac{a}{b}\right)^k = \frac{a^k}{b^k}\).
• This assumption is not proven during this step; rather, it is a temporary acceptance to aid in proving the next step.

By assuming the statement is true for \(k\), we prepare to show that this truth extends to \(k+1\). This is a critical aspect of mathematical induction as it forms the logical continuum necessary for the proof.

Induction Step

The induction step is where the magic of mathematical induction happens. Here, you demonstrate that if the hypothesis holds for an arbitrary integer \(k\), then it must also hold for \(k+1\). This step effectively proves that the statement is true for all integers following the base case.

For our specific problem, the induction step involves these transformations:

• We want to show \(P(k+1): \left(\frac{a}{b}\right)^{k+1} = \frac{a^{k+1}}{b^{k+1}}\).
• Using the existing hypothesis \(\left(\frac{a}{b}\right)^k = \frac{a^k}{b^k}\), we can expand \(\left(\frac{a}{b}\right)^{k+1}\) as \(\left(\frac{a}{b}\right)^k \cdot \left(\frac{a}{b}\right)\).
• Replace \(\left(\frac{a}{b}\right)^k\) using the hypothesis: \(\frac{a^k}{b^k} \cdot \frac{a}{b} = \frac{a^{k+1}}{b^{k+1}}\).

This successfully shows that if the statement is true for \(n=k\), then it is also true for \(n=k+1\). Thus, by mathematical induction, it holds for all positive integers \(n\). This step is vital because it forms the ongoing logical structure that completes the induction proof.