Problem 31 - Precalculus with Limits

Question

In Exercises 31 - 42, use mathematical induction to prove the property for all positive integers \( n \). \( \left(ab\right)^n = a^n b^n \)

Step-by-Step Solution

Answer

Using mathematical induction, we successfully proved the property \( \left(ab\right)^n = a^n b^n \) holds for all positive integers \( n \).

Step 1: Base Case

Let's first verify the base case \( n = 1 \). On the left side, we have \( \left(ab\right)^1 = ab \), and on the right side we have \( a^1 \cdot b^1 = ab \), so the formula holds for the base case as both sides are equal.

Step 2: Inductive Hypothesis

Now let's assume the property to be true for some positive integer \( k \). That is, \( \left(ab\right)^k = a^k b^k \). This is known as the inductive hypothesis.

Step 3: Inductive Step

The final step in mathematical induction is to show that the property holds for \( n = k + 1 \), given that it holds for \( n = k \). That is we need to show \( \left(ab\right)^{k+1} = a^{k+1} b^{k+1} \). Simplifying the left side, we have \( \left(ab\right)^{k+1} = \left(ab\right)^k \cdot \left(ab\right) = a^kb^k \cdot ab \). By using the hypothesis assumption, we simplify it to \( a^{k+1}b^{k+1} \), which matches with the right side.

Key Concepts

Positive Integers

In mathematical induction, we deal with positive integers. These are numbers that are greater than zero and do not include fractions or decimals. Essentially, they are whole numbers like 1, 2, 3, and so on. When a property is to be proven for all positive integers, it means we want to show that every number in this list satisfies the given statement. Understanding positive integers is crucial when using mathematical induction. This is because the method allows us to logically validate that if a property holds for one positive integer, then it must hold for the next one, and by extension, all positive integers. This step-by-step progression forms the heart of why induction is a powerful mathematical tool.

Base Case

The base case is the starting point of mathematical induction. It involves verifying that the given property holds for the smallest positive integer, usually 1. In our original exercise, this means checking whether \( (ab)^1 = a^1 b^1 \). For the base case to succeed, both sides of the equation must be equal. This step is crucial since missing or incorrect base cases might lead to faulty conclusions subsequently. Think of the base case as the foundation of a tall building. If it is stable, we can confidently add more floors. Thus, proving the base case ensures that we can begin using induction with a strong starting point.

Inductive Hypothesis

Once we establish the base case, we move on to the next phase known as the inductive hypothesis. This requires us to assume that the property is true for some arbitrary positive integer, say \( n = k \). In our case, the assumption is \( (ab)^k = a^k b^k \). It's important to note here that this is purely an assumption and not a proven fact yet.This step acts like a bridge, joining the base case and the next step of proving the next number. Think of the inductive hypothesis as a hypothesis in a science experiment. It's a carefully crafted assumption that helps us explore further and justifies the progression from one integer to the next.

Inductive Step

In the inductive step, we prove that if the property holds for \( n = k \), then it must also hold for \( n = k + 1 \). This step relies heavily on our inductive hypothesis. For our particular problem, we aim to prove \( (ab)^{k+1} = a^{k+1} b^{k+1} \).To demonstrate this, we start from \( (ab)^{k+1} \). Using the previous assumption \( (ab)^k = a^k b^k \), we can simplify \( (ab)^{k+1} = (ab)^k \cdot (ab) \) into \( a^k b^k \cdot a b \), which further simplifies straightforwardly to \( a^{k+1} b^{k+1} \). The inductive step solidifies the entire process of mathematical induction. It's akin to taking a leap forward, equipped with the stepping stone provided by the inductive hypothesis. Completing this step means assuring that our property holds not just for one instance but progresses universally, justifying its proof for all positive integers.