Problem 4 - A Gentle Introduction to the Art of Mathematics

Question

Give inductive proofs of the following $$ \forall n \in \mathbb{N}, 3 \mid 4^{n}-1 $$

Step-by-Step Solution

Answer

\(3\) divides \(4^n - 1\) for all \(n \in \mathbb{N}\) by mathematical induction.

Step 1 - Base Case

First, prove the statement for the base case. Let's choose the smallest natural number, which is 0.Substitute \(n = 0\):\[4^0 - 1 = 1 - 1 = 0\]Since 3 divides 0, the base case holds true.

Step 2 - Inductive Hypothesis

Assume that the statement is true for some arbitrary natural number \(k\). This means:\[3 \big| (4^k - 1)\]In other words, there exists some integer \(m\) such that:\[4^k - 1 = 3m\]

Step 3 - Inductive Step

Now, prove that the statement is true for \(k+1\). Consider the expression for \(n = k + 1\):\[4^{k+1} - 1 = 4 \cdot 4^k - 1\]Substitute \(4^k\) from the inductive hypothesis, where \(4^k = 3m + 1\):\[4 \times (3m + 1) - 1 = 4 \times 3m + 4 - 1 = 12m + 3\]Since \(12m + 3\) is divisible by 3, the induction step holds.

Step 4 - Conclusion

By the principle of mathematical induction, since the statement holds for the base case and the inductive step has been proven, the statement is true for all natural numbers \(n\). Thus,\[\forall n \in \mathbb{N}, 3 \big| 4^n - 1\]

Key Concepts

base case

In mathematical induction, the base case is the starting point. It involves proving that the statement is true for the smallest natural number, typically 0 or 1. For our exercise, we start with 0.
When substituting 0 in the given equation, we get:
\[ 4^0 - 1 = 1 - 1 = 0 \]
Since 0 is divisible by 3, the base case holds true. Every induction proof relies on a valid base case to build the argument for all following numbers.

inductive hypothesis

The inductive hypothesis is the assumption that the statement holds for some arbitrary natural number, typically denoted as k. For our problem, the hypothesis is:
\[ 3 \big| (4^k - 1) \]
This means there exists an integer m such that:
\[ 4^k - 1 = 3m \]
This step is crucial because it serves as the foundation for the next part, the inductive step. It is like assuming a ladder step holds so we can confidently move to the next one.

inductive step

In the inductive step, we use the inductive hypothesis to prove that the statement also holds for \( k + 1 \). This shows the statement's validity moves from one natural number to the next.

For our exercise, we need to prove:
\[ 4^{k+1} - 1 \]
We know from our hypothesis:
\[ 4^k = 3m + 1 \]
Substituting this into our equation:
\[ 4 \times (3m + 1) - 1 = 12m + 3 \]
Since \( 12m + 3 \) is divisible by 3, the statement is true for \( k + 1 \). This completes the inductive step.

natural numbers

Natural numbers are the basic building blocks in induction. They start from 0 or 1 and increase by one unit step at a time.

Mathematical induction is a method to prove properties or statements for all natural numbers by relying on this predictable, ordered progression. Using inductive reasoning, we show that if the statement holds for a number k, it must hold for \( k + 1 \) as well. This process repeats indefinitely, covering all natural numbers.

divisibility

Divisibility refers to one number being divisible by another without a remainder. In this exercise, we proved that \( 4^n - 1 \) is divisible by 3 for all natural numbers n.

We demonstrated this through induction, showing:

• Base case: It holds for the smallest number.
• Inductive hypothesis: Assuming it holds for any number k.
• Inductive step: Proving it holds for \( k + 1 \).

The key was transforming the expression under the hypothesis to prove divisibility in the next step. This method ensures the property remains true for an infinite sequence of numbers.