Problem 9 - Contemporary Precalculus

Question

Use mathematical induction to prove that each of the given statements is true for every positive integer \(n .\) $$n+2>n$$

Step-by-Step Solution

Answer

Question: Use mathematical induction to prove the inequality \(n+2 > n\) for every positive integer \(n\). Answer: The inequality holds true for the base case of \(n=1\) and through the process of mathematical induction, we have demonstrated that if the inequality holds for an arbitrary positive integer \(n=k\), it also holds for \(n=k+1\). Therefore, the inequality \(n+2 > n\) is true for all positive integers \(n\).

Step 1: The base case

First, we will prove that the inequality \(n+2>n\) is true for the smallest positive integer, \(n=1\). $$1 + 2 > 1$$ Which simplifies to: $$3 > 1$$ This statement is true, so the inequality holds for the base case.

Step 2: Inductive step

Now, we will assume the inequality is true for an arbitrary positive integer \(n=k\) for \(k \in \mathbb{Z^+}\): $$(a)~k + 2 > k$$ The goal is to prove that the inequality is also true for \(n= k+1\): $$(b)~(k + 1) + 2 > k + 1$$ To do this, we will add 1 to both sides of our assumption, equation \((a)\): $$k + 2 + 1 > k + 1$$ This simplifies to: $$k + 3 > k + 1$$ Now, substitute \((k+1)\) for \(n\) in the inequality: $$n+2=k+1+2 > k+1+1 > k+1$$ Since the inequality is true for both the base case and the inductive step, we can conclude that the inequality \(n+2>n\) will hold for all positive integers n according to the principles of mathematical induction.

Key Concepts

Base Case

In the realm of mathematical induction, the base case is the starting point of every proof. It is crucial to establish this foundation, as it confirms the validity of the statement for the smallest positive integer. Typically, that is for \(n = 1\).
The process involves substituting the smallest value into the inequality or equation being proved. The purpose is to show that the statement is true at this level. For our problem, the inequality is \(n+2>n\). Substituting \(n = 1\), we check:

• Calculate \(1 + 2\) which equals 3.
• Check if 3 is greater than 1.

Observing that \(3 > 1\) indeed holds true, we confirm our base case. Once established, it sets the stage for proceeding to the next part: the inductive step.

Inductive Step

Once the base case is established, the next crucial element of mathematical induction is the inductive step. It's like climbing a ladder, step by step. Here, the idea is to assume true for some arbitrary positive integer \(n = k\) and then prove that it holds for the next integer, \(n = k+1\).
In our case, we assume the inequality is true for \(n = k\), which means \(k + 2 > k\). This is our inductive hypothesis. Next, we aim to prove it for \(n = k+1\):

• Substitute \((k + 1)\) into the inequality: \((k + 1) + 2 > (k + 1)\).
• Add 1 to both sides of the hypothesis \(k + 2 > k\) to get \(k + 3 > k + 1\).

This deduction solidifies our proof, as it demonstrates that if the statement holds for one integer, it will hold for the next. Therefore, thanks to the principles of mathematical induction, the inequality holds for every positive integer.

Positive Integer

In mathematical induction, the concept of a positive integer is fundamental as it refers to any whole number greater than zero: \(1, 2, 3,\ldots\). These numbers are essential because induction typically seeks to prove statements applicable to all such integers.
When we set up an induction proof, we often start by examining the smallest positive integer, usually \(n = 1\), to establish the base case. This is crucial because 1 is the first step in the sequence of positive integers.
Then, we progress to proving that if the statement is true for any arbitrary integer \(k\), it must also be true for the next integer \(k + 1\). Broadly, positive integers provide the structured sequence we require to apply the induction process methodically. This ensures that our proof covers not just a few cases but all positive integers, comprehensively confirming the validity of the statement in question.