Problem 31 - Algebra and Trigonometry

Question

Use mathematical induction to prove that each statement is true for every positive integer n. \(n+2>n\)

Step-by-Step Solution

Answer

True. The principle of mathematical induction proves the statement \(n + 2 > n\). First, by checking the base case (n=1), it was found that adding 2 to 1 results in a number greater than 1. Then, assuming the statement is true for an arbitrary positive integer \(k\), it was demonstrated it is also true for \(k + 1\). Therefore, the statement \(n + 2 > n\) is true for every positive integer \(n\).

Step 1: The Base Case

The base case tests the statement for the smallest case, which is n = 1. So, perform the test: is \(1 + 2 > 1\)? The answer is yes, because \(3 > 1\). Therefore the statement holds for the base case.

Step 2: The Induction Hypothesis

Now, let's move onto the induction step. It consists of two parts: the induction hypothesis and the induction consequence. For the Induction Hypothesis, assume that the statement is true for an arbitrary positive integer k, i.e., \(k + 2 > k\). Set this as the induction hypothesis.

Step 3: The Induction Step

The Induction Consequence part of the induction step tries to prove that under the assumption that the statement is true for k, it must also be true for \(k + 1\). Substitute \(k + 1\) instead of \(n\) in the equation and demonstrate it is true: \(k + 1 + 2 > k + 1\). Simplifying the left side gives \(k + 3 > k + 1\). Since 3 is greater than 1, it's clear that \(k + 3 > k + 1\) proving the statement also holds for \(k + 1\) when it holds for \(k\).

Key Concepts

Base Case

Mathematical induction always starts with the base case. This initial step checks whether the given statement holds true for the smallest possible value, typically where the variable is equal to 1. In this particular proof, the statement is examined for when \( n = 1 \).
To do this, we plug \( n = 1 \) into our inequality \( n + 2 > n \) and find:

• \( 1 + 2 > 1 \)
• \( 3 > 1 \)

The result \( 3 > 1 \) is obviously true, confirming that the base case is valid. This first step sets the stage for succeeding parts of the induction process by establishing the truth of the statement at its simplest level.
Confirming the base case verifies that there is a foundation on which to build our induction hypothesis and induction step. Without this first affirmation, the entire process of induction would be incomplete.

Induction Hypothesis

Once the base case has been verified, the next step in mathematical induction is to formulate the induction hypothesis. This part requires you to assume that the statement holds true for some arbitrary positive integer \( k \). In other words, we assume:

• \( k + 2 > k \)

This assumption is crucial because it acts as the bridge that connects the base case to the induction step.
By setting up the induction hypothesis, you're essentially saying, "Let's pretend our statement is true for \( n = k \)."
It allows us to test if this assumption leads to the condition being satisfied for the next integer, \( k + 1 \). This step is where we draw upon the power of induction, laying the groundwork to demonstrate that if it works for one case, it should also work for the next.

Induction Step

The induction step is the climax of mathematical induction. This is where we demonstrate that, assuming our statement holds true for \( n = k \) (our induction hypothesis), it must also hold true for \( n = k + 1 \). For this proof, it involves substituting \( k + 1 \) into the inequality to check its validity.
Our goal is to show:

• \( (k + 1) + 2 > k + 1 \)

Simplifying gives us:

• \( k + 3 > k + 1 \)

We know that \( 3 > 1 \), so naturally, \( k + 3 > k + 1 \) holds true.
By confirming this step, we've established that if \( n + 2 > n \) is true for any integer \( n = k \), it must also be true for \( n = k + 1 \). This successful demonstration brings the inductive proof full circle, solidifying the statement's truth for all positive integers, starting from the base case and moving upwards unerringly.