Problem 29 - A Graphical Approach to Precalculus with Limits

Question

Prove each statement by mathematical induction. If \(a>1,\) then \(a^{n}>1\)

Step-by-Step Solution

Answer

By induction, if \(a>1\), then \(a^n>1\) for all integers \(n\geq 1\).

Step 1: Base Case

First, verify that the statement holds for the base case, when \( n = 1 \). Substitute \( n = 1 \) into the expression: \( a^n = a^1 = a \). Since \( a > 1 \), we have \( a > 1 \), so the base case holds.

Step 2: Inductive Hypothesis

Assume that the statement is true for an arbitrary positive integer \( k \), that is, assume \( a^k > 1 \). This assumption is called the inductive hypothesis.

Step 3: Inductive Step

Now, use the inductive hypothesis to prove the statement for \( k + 1 \). We need to show that \( a^{k+1} > 1 \). Start from \( a^{k+1} = a^k \cdot a \). Since we assumed \( a^k > 1 \) and we know \( a > 1 \), it follows that \( a^k \cdot a > 1 \cdot a > 1 \). Thus, \( a^{k+1} > 1 \).

Step 4: Conclusion

Since both the base case holds and the inductive step has been proven, by the principle of mathematical induction, we conclude that \( a^n > 1 \) for all integers \( n \geq 1 \) if \( a > 1 \).

Key Concepts

Base Case

Mathematical induction always kicks off with the base case. This is the starting point for the domino effect that induction creates. In this exercise, we start with a base case of \( n = 1 \). This means we need to verify whether the statement is true when \( n \) equals 1. We substitute \( n = 1 \) into our initial statement, and we get:

• \( a^n = a^1 = a \)

Given that \( a > 1 \), it immediately follows that \( a > 1 \) is true.
This confirms our base case.By successfully establishing that the statement holds for \( n=1 \), we set the stage for further steps of induction.
This step might look straightforward, but it is crucial because without a true base case, we can't proceed to other values of \( n \). A valid base case guarantees that our chain of logic can get off the ground.

Inductive Hypothesis

After establishing the base case, we need to assume the statement is true for a generic case, \( n = k \). This is the inductive hypothesis, essentially the backbone of mathematical induction. Why do we need this assumption? - It serves as a stepping stone for showing that if one case holds true, then the next must also be true. We assume \( a^k > 1 \) for some arbitrary positive integer \( k \).
What does this assumption mean? It means that at some point in our series, the statement correctly holds, and now we need to show that it will hold for the next case, that is, when \( n = k+1 \).
It's a form of mathematical extension where, by assuming this hypothesis, the next part of the induction steps can work to bridge from this assumption to the following case.

Inductive Step

Now comes the inductive step, where we must verify that if the statement is true for \( n = k \), it is also true for \( n = k + 1 \). This is like ensuring that one domino topples the next one in a chain reaction.We start with:

• \( a^{k+1} = a^k \cdot a \)

Given our inductive hypothesis \( a^k > 1 \) and \( a > 1 \), we can multiply these inequalities to get:

• \( a^k \cdot a > 1 \cdot a > 1 \)

Thus, \( a^{k+1} > 1 \).
Here, the domino effect is completed. This step shows that if \( a^k > 1 \) holds, then \( a^{k+1} > 1 \) naturally follows. Without a successful inductive step, the proof cannot claim the statement is true for all \( n \). Together with a true base case and the solid foundation of the inductive hypothesis, we've now sewn up the entire argument: \( a^n > 1 \) for all \( n \geq 1 \).