Problem 33 - College Algebra

Question

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

Step-by-Step Solution

Answer

By using the mathematical induction principle, the equation \((ab)^n = a^n b^n\) is proven to be true for all positive integers \(n\). The equation is initially proven to hold for \(n=1\). This is known as the base case. We then assume the equation is true for a general positive integer \(k\), and then prove it's also true for \(k+1\). This is known as the inductive step. Therefore the equation holds for all positive integers \(n\).

Step 1: Base Case

The first step in mathematical induction is to validate the base case. In this instance, our base case corresponds to \(n=1\). As a result, we need to prove that the formula holds true for \(n=1\). The left-hand side, \((ab)^1\), is equal to \(ab\). Then, the right-hand side, \(a^1 * b^1\), is also equal to \(ab\). Hence, for \(n=1\), \((ab)^n = a^n * b^n\), as both sides equate to \(ab\).

Step 2: Inductive Step - Induction Hypothesis

Now that we have our base case, the next step in mathematical induction is to make an induction hypothesis. We will assume that the statement is correct for a random positive integer \(k\), i.e., \((ab)^k = a^k * b^k\). This is not a proven fact, but an assumption that we will then leverage to prove that the statement is correct for \(k + 1\).

Step 3: Inductive Step - Inductive Proof

We need now to prove that if the statement is true for a number \(k\), it will also be true for \(k+1\). Substituting \(k+1\) instead of \(n\) in our equation, we have \((ab)^{k+1} = a^{k+1} b^{k+1}\). This simplifies to \((ab)^k * (ab) = a^k * b^k * a * b\). According to our inductive hypothesis, \((ab)^k = a^k * b^k\), thus we can substitute \(a^k * b^k\) instead of \((ab)^k\) in the equation. This yields \(a^k * b^k * a * b = a^k * a * b^k * b\), because multiplication is commutative, which simplifies to \(a^{k+1} b^{k+1} = a^{k+1} b^{k+1}\), proving the statement to be true. Therefore, the statement is also true for the integer following \(k\). The exercise is successfully solved using mathematical induction, and the statement is proven to be true for every positive integer \(n\).

Key Concepts

Base Case

The base case is the starting point in mathematical induction. In this process, we need to show that the given statement is true for the initial value of the variable, usually when the variable equals one. In our exercise, the statement involves exponentiation and the product of two variables:

• The expression on the left side is \((ab)^n\).
• The expression on the right is \(a^n \times b^n\).

To validate the base case for \(n=1\), substitute 1 into the equation. Both sides become \(ab\), as \((ab)^1 = ab\) and \(a^1 \times b^1 = ab\). Since these two expressions are equal, this confirms that our base case holds true.
Successfully validating the base case is a crucial first step in mathematical induction.

Inductive Step

After proving the base case, we move on to the inductive step. This is where the power of induction comes into play, allowing us to extend the validity of the statement from one number to all integers greater than one. The inductive step involves two main components:

• First, the induction hypothesis: here, we assume the statement is true for some integer \(k\). In mathematical terms, we assume \((ab)^k = a^k \times b^k\) is correct.
• Next, we prove that if the statement is true for \(k\), it must also be true for \(k+1\). We make this leap by plugging \(k+1\) into our equation, resulting in \((ab)^{k+1} = a^{k+1} \times b^{k+1}\).

The actual magic happens when substituting the induction hypothesis into the equation for \(k+1\), transforming \((ab)^k \times ab\) into \(a^k \times b^k \times a \times b\). This simplifies through the commutative property of multiplication, ensuring both sides of the equation match.

Exponentiation

Exponentiation plays a key role in our exercise. It is the process of raising a number or expression to a power, represented with the exponent symbol \(^n\). When dealing with exponentiation in this exercise, the primary focus is:

• Recognizing patterns: \((ab)^n\) implies both \(a\) and \(b\) are multiplied together, then raised to the power \(n\).
• Breaking it down: \(a^n \times b^n\) is the result of separately raising both \(a\) and \(b\) to the power \(n\), then multiplying the results.

Understanding the act of exponentiation helps us comprehend why the statement is true. The exercise demonstrates that the power properties apply equally within the constraints of mathematical induction, allowing us to prove that \((ab)^n = a^n \times b^n\) is valid for all positive integers. These properties ensure we can deal with exponents in an algebraic context effectively.