Problem 17 - Precalculus: Mathematical for Calculus

Question

Show that \(8^{n}-3^{n}\) is divisible by 5 for all natural numbers \(n .\)

Step-by-Step Solution

Answer

The expression \( 8^n - 3^n \) is divisible by 5 for all natural numbers \( n \).

Step 1: Base Case

First, we will demonstrate that the statement holds for the smallest natural number, which is 1. Calculate \( 8^1 - 3^1 \). This simplifies to \( 8 - 3 = 5 \), which is divisible by 5. Therefore, the base case is true.

Step 2: Inductive Step Setup

Assume that \( 8^k - 3^k \) is divisible by 5 for some arbitrary natural number \( k \). This is our inductive hypothesis.

Step 3: Inductive Step Proof

We need to show that \( 8^{k+1} - 3^{k+1} \) is also divisible by 5. Begin by expressing this as \( 8^{k+1} - 3^{k+1} = 8 \times 8^k - 3 \times 3^k = 8 \cdot 8^k - 3 \cdot 3^k \).

Step 4: Simplify Expressions Modulo 5

Reduce \( 8 \equiv 3 \mod 5 \) and \( 8^k \equiv 3^k \mod 5 \) due to our hypothesis. Then \( 8\cdot 8^k \equiv 3\cdot 3^k \mod 5 \).

Step 5: Finish Inductive Proof

We see that \( 8^{k+1} - 3^{k+1} \equiv 3\cdot 3^k - 3\cdot 3^k \equiv 0 \mod 5 \). This is divisible by 5, completing the induction step.

Step 6: Conclusion

By mathematical induction, \( 8^n - 3^n \) is divisible by 5 for all natural numbers \( n \).

Key Concepts

Base Case

The base case is the foundation of mathematical induction. It's the starting point that sets the correctness of the entire proof in motion. In our exercise, the base case involves calculating the expression for the smallest possible natural number, which is \( n = 1 \). To establish the base case, we calculated \( 8^1 - 3^1 \), which equals 5. Since 5 is divisible by 5, the base case is satisfied.
You need to prove that the statement's validity begins at this initial point to assert its truth across all natural numbers. It's akin to proving the first domino falls so all others follow. Without a valid base case, the chain reaction in mathematical induction can't start.
To use base cases effectively, ensure you're clear on which instance of \( n \) you're starting with and show the math works for that smallest instance.

Inductive Hypothesis

The inductive hypothesis is a critical assumption made during the process of mathematical induction. In essence, it involves assuming the statement you’re trying to prove is true for some arbitrary natural number, say \( k \). This logical building block allows us to demonstrate that if the statement holds for \( k \), it must also hold for \( k+1 \).
In our exercise, we assume that \( 8^k - 3^k \) is divisible by 5. This assumption is a bridge between the known (base case) and the unknown (goal), which in this case is proving \( 8^{k+1} - 3^{k+1} \) is divisible by 5.
Here's a simple way to think about it:

• You trust that the statement is true for \( k \).
• Build on this trust to argue it must be true for \( k + 1 \).

By following this method, you ensure the induction process moves smoothly from one natural number to the next.

Modulo Arithmetic

Modulo arithmetic simplifies complex expressions during mathematical induction proofs by focusing on the remainders when numbers are divided. It is exceptionally useful in divisibility problems. The term "modulo" means "with respect to the remainder." So, when we say \( a \equiv b \mod m \), it means \( a \) and \( b \) have the same remainder when divided by \( m \).
In our exercise, modulo arithmetic is applied to simplify where \( 8 \equiv 3 \mod 5 \) and consequently, \( 8^k \equiv 3^k \mod 5 \) based on our inductive hypothesis. This step is crucial as it helps break down the complex power expressions to a manageable form.
By using modulo arithmetic, calculations become concise:

• Convert large powers into simpler equivalents.
• Focus on remainders and bypass exhaustive division.
• Check repetition patterns to predict behavior over repetitions.

The simplification not only alleviates tedious calculations but also provides a robust check for divisibility in modulo contexts.