Problem 16 - Algebra and Trigonometry

Question

Show that \(5^{n}-1\) is divisible by 4 for all natural numbers \(n\)

Step-by-Step Solution

Answer

\(5^n - 1\) is divisible by 4 for all natural numbers \(n\).

Step 1: Base Case Verification

To show divisibility by 4 for all natural numbers using mathematical induction, first verify the base case. Let's substitute \( n = 1 \):\[ 5^1 - 1 = 4 \]\( 4 \) is divisible by \( 4 \), so the base case \( n = 1 \) is true.

Step 2: Inductive Hypothesis

Assume that the statement is true for some arbitrary natural number \( k \). That is, assume \( 5^k - 1 \) is divisible by \( 4 \). This means \( 5^k - 1 = 4m \) for some integer \( m \). We need to prove it holds for \( k+1 \).

Step 3: Inductive Step

Now consider \( 5^{k+1} - 1 \).Rewrite this as:\[ 5^{k+1} - 1 = 5^k \cdot 5 - 1 = 5 \cdot 5^k - 1 \]Substitute the inductive hypothesis \( 5^k = 4m + 1 \):\[ 5 \cdot (4m + 1) - 1 = 20m + 5 - 1 = 20m + 4 \]This expression can be factored as:\[ 4 (5m + 1) \]Thus, \( 5^{k+1} - 1 \) is divisible by \( 4 \). This completes the inductive step.

Step 4: Conclusion

Since we've verified the base case and shown the inductive step holds, by mathematical induction, \( 5^n - 1 \) is divisible by \( 4 \) for all natural numbers \( n \).

Key Concepts

Base Case

The base case is the starting point of any mathematical induction proof. It's where you verify that the given statement is true for the initial value of your domain, often the smallest natural number, which is typically 1. In our problem, we need to show that \(5^n - 1\) is divisible by 4 when \(n = 1\).

So, let's substitute 1 for \(n\):

• Calculate \(5^1 - 1\), which equals \(4\).
• Clearly, 4 divided by 4 gives us a whole number without any remainder, confirming it's divisible.

This ensures that the base case is true, providing a solid foundation to move forward with the induction process.

Inductive Hypothesis

The inductive hypothesis is a vital concept in mathematical induction. It involves assuming the statement is true for some specific, arbitrary natural number \(k\). This might sound a bit like magic, but it's entirely logical and crucial for the proof.

In our case, assume that \(5^k - 1\) is divisible by 4. This means there exists an integer \(m\) such that:

• \(5^k - 1 = 4m\)

This assumption does not prove anything by itself, but it sets a scenario that allows us to check if the statement holds for the next number in the sequence (\(k+1\)). Think of it as a stepping stone to link the base case with all subsequent cases.

Inductive Step

The inductive step connects the inductive hypothesis to the next natural number in the sequence, \(k+1\). The goal here is to prove that if the hypothesis holds, then the assertion must also be true for \(k+1\).

Let's put this into practice:

• Start with \(5^{k+1} - 1\) and aim to express it so that it shows divisibility by 4.
• Notice that \(5^{k+1} = 5 \times 5^k\).
• Substitute the assumption \(5^k = 4m + 1\) into this expression: \(5 \times (4m + 1) - 1\).
• When simplified, this becomes \(20m + 4\), which is clearly divisible by 4 as it can be rewritten as \(4(5m + 1)\).

By confirming that \(5^{k+1} - 1\) maintains divisibility by 4, we effectively show the statement holds for \(k+1\). Bridging this step with the base case and hypothesis ensures the statement is true for all natural numbers \(n\). With all parts completed, the proof by induction is robust and valid.