Problem 25 - A Graphical Approach to Precalculus with Limits

Question

Prove each statement by mathematical induction. \(\left(a^{m}\right)^{n}=a^{m n}(\text { Assume that } a \text { and } m\) are constant.)

Step-by-Step Solution

Answer

Proven by induction for all positive integers \(n\).

Step 1: Understanding the Problem

We need to prove the expression \(\left(a^{m}\right)^{n}=a^{m n}\) using mathematical induction. This involves showing the statement is true for an initial value, usually \(n=1\), and then proving that if it's true for an arbitrary \(n=k\), it must be true for \(n=k+1\).

Step 2: Base Case

For \(n=1\), the expression becomes \(\left(a^{m}\right)^{1} = a^{m}\). The right side becomes \(a^{m \cdot 1} = a^m\). Both sides are equal, so the statement holds for \(n=1\).

Step 3: Inductive Hypothesis

Assume that the statement is true for some arbitrary positive integer \(k\), i.e., \(\left(a^{m}\right)^{k} = a^{m \cdot k}\). This assumption will be used to prove the statement for \(n=k+1\).

Step 4: Inductive Step

We need to prove that \(\left(a^{m}\right)^{k+1} = a^{m (k+1)}\). Start with the left side using the inductive hypothesis: \(\left(a^{m}\right)^{k+1} = \left(a^{m}\right)^{k} \cdot a^{m}\). By the inductive hypothesis, \(\left(a^{m}\right)^{k} = a^{m \cdot k}\), so this becomes \(a^{m \cdot k} \cdot a^{m}\). Using the property of exponents \(a^{x} \cdot a^{y} = a^{x+y}\), we find \(a^{m \cdot k} \cdot a^{m} = a^{m \cdot k + m} = a^{m (k+1)}\). Thus, the statement is proven for \(n=k+1\).

Step 5: Conclusion

Since the base case holds and the inductive step is valid, by mathematical induction, \(\left(a^{m}\right)^{n}=a^{m n}\) is true for all positive integers \(n\).

Key Concepts

Exponents

Understanding exponents is crucial to grasping the entire concept of mathematical induction in this exercise. An exponent is a way of expressing repeated multiplication of a number by itself. For instance, when we write \(a^n\), it means we multiply the base \(a\) by itself \(n\) times. Here are some key points about using exponents effectively:

• Base: The number \(a\) which gets multiplied.
• Exponent: The number \(n\) which shows how many times the base is used as a factor.
• Power: The result of this repeated multiplication.

A property that's often useful, such as in this exercise, is the power of a power property: \( \left(a^m\right)^n = a^{m \times n} \). This states that raising a power to another power multiplies the exponents together. This very property is what we aim to prove using mathematical induction in our problem.

Base Case

The base case is the initial step in mathematical induction. It's the foundation on which the rest of the proof builds. In this exercise, we test the equation for \(n=1\). This is often the starting point for induction because if it fails here, the statement cannot hold for all \(n\). Consider our expression with \(n=1\): we get \((a^m)^1 = a^m\). Both sides of the equation are identical, showing that our statement holds true at this base level. This step confirms that our hypothesis is valid for the smallest value of \(n\), preparing the ground for the next phase of induction.

Inductive Step

The inductive step is where we strive to show that if a statement is true for an arbitrary positive integer \(k\), it is also true for \(k+1\). Using the inductive hypothesis, assume \(\left(a^m\right)^k = a^{m \cdot k}\). We must then show that \(\left(a^m\right)^{k+1} = a^{m (k+1)}\).Here's how we do it:

• Start by considering \(\left(a^m\right)^{k+1} = \left(a^m\right)^k \cdot a^m\).
• By our inductive hypothesis, substitute \(\left(a^m\right)^k\) with \(a^{m \cdot k}\).
• This turns into \(a^{m \cdot k} \cdot a^m\).
• Use the property of exponents to combine them into \(a^{m \cdot k + m}\), or equivalently \(a^{m (k+1)}\).

This completes the inductive step, demonstrating that if the statement is true for \(n=k\), it is indeed true for \(n=k+1\). Thus, the proof by induction is successful, confirming the validity of the equation for all natural numbers \(n\).