Problem 31 - Contemporary Precalculus

Question

Use the Extended Principle of Mathematical Induction (Exercise 28 ) to prove the given statement. $$n^{2}>n \quad \text { for all } n \geq 2$$

Step-by-Step Solution

Answer

Using the Extended Principle of Mathematical Induction, we proved that the inequality \(n^2 > n\) is true for all integers \(n \geq 2\). We first demonstrated the base case, where \(n=2\), and found that the inequality holds. Next, we assumed that the inequality is true for some integer \(m \geq 2\) and used this assumption to prove the statement for \(n=m+1\). Ultimately, we showed that the original statement is true for all values of \(n \geq 2\).

Step 1: Prove the Base Case

First, we need to prove that the statement is true for \(n=2\). We want to show that \(2^2 > 2\). Plug \(2\) into the inequality: $$2^2 = 4$$ Since \(4>2\), the inequality holds for the base case.

Step 2: Inductive Hypothesis

Now, we assume that the statement is true for \(n=k, k+1, \cdots, m\), where \(m \geq 2\). In other words, we assume that the following inequalities hold: $$k^2 > k, \quad (k+1)^2 > k+1, \quad \cdots, \quad m^2 > m$$

Step 3: Prove the Statement for \(n=m+1\)

Using the inductive hypothesis, we want to show that the statement is true for \(n=m+1\). That is, we want to show that \((m+1)^2 > m+1\). Let's first expand the left side of the inequality: $$(m+1)^2 = m^2 + 2m + 1$$ Now, consider the inequality \(m^2 > m\), which we know to be true from our inductive hypothesis. We can add \(2m+1\) to both sides of this inequality: $$m^2 + 2m + 1 > m + 2m + 1$$ Simplify the right side: $$m^2 + 2m + 1 > 3m + 1$$ Now notice that we have derived the inequality \((m+1)^2 > 3m + 1\). In order to complete the proof, we need to show that \(3m+1 \geq m+1\). We can rearrange this inequality to \(2m \geq 0\), which is true for all \(m \geq 2\) because \(m\) is an integer greater than or equal to \(2\). Thus, we have proved that \((m+1)^2 > m+1\) using the inductive hypothesis. By the Extended Principle of Mathematical Induction, the statement \(n^2 > n\) is true for all \(n \geq 2\).

Key Concepts

Base Case

When working with mathematical induction, precisely understanding the base case is crucial. It is the starting point that ensures our domino effect of logic has a solid foundation to commence from. In regard to the provided exercise, the base case is established by initially testing the given inequality, \(n^2>n\), for \(n=2\).

To confirm the base case, we simply substitute \(2\) into our inequality, resulting in \(2^2=4\), then compare it to \(2\). It's evident that \(4>2\), thus, the base case is valid. This step is akin to placing the first domino in a line—the one you'll nudge to watch the rest fall in succession. Without proving the base case, the entire inductive argument would crumble like a house without a foundation.

Inductive Hypothesis

Following the base case confirmation, we step into the world of the inductive hypothesis. This is where we assume the statement to be true not just for a single value, but for a range of values up to a certain point, creating a sort of 'assumed truth zone'.

In the context of the exercise, we assume the inequality holds for \(n=k, k+1, \text{...}, m\) where \(m \text{is greater than or equal to} 2\). Explicitly, we're claiming that \(k^2 > k\), \((k+1)^2 > k+1\), \text{...,} \(m^2 > m\). The acceptance of these assumed truths forms the second step of our logical domino setup.

Inequality Proofs

The final keystone in the arch of mathematical induction proofs is confirming the inductive step—applying the 'assumed truth zone' to validate the next element, in this case for \(n=m+1\). Here, we explore the proof of the inductive step through inequality.

We begin with expanding \((m+1)^2\) which gives us an expression \(m^2 + 2m + 1\). Using our inductive hypothesis, which tells us \(m^2 > m\), we add \(2m+1\) to both sides yielding the inequality \(m^2 + 2m + 1 > 3m + 1\). The remaining task for completeness is to show \(3m+1\) is always greater than or equal to \(m+1\) for our range, which simplifies to the true statement \(2m \text{is greater than or equal to} 0\) for all integers \(m\) within the range we're studying.

Once we cross this logical chasm, we underpin the statement \((m+1)^2 > m+1\). This allows us to affirm, with mathematical vigor, that our original inequality, \(n^2>n\), stands true for all integers \(n \text{greater than or equal to} 2\). It's the final push that sends the domino of \(m+1\) toppling, continuing the cascade of truth established by our base case and carried by our inductive hypothesis.