Problem 41 - College Algebra

Question

Some statements are false for the first few positive integers, but true for some positive integer \(m\) on. In these instances, you can prove \(S_{n}\) for \(n \geq m\) by showing that \(S_{m}\) is true and that \(S_{k}\) implies \(S_{k+1}\) when \(k \geq m .\) Use this extended principle of mathematical induction to prove that each statement in Exercises \(41-42\) is true. Prove that \(n^{2}>2 n+1\) for \(n \geq 3\). Show that the formula is true for \(n=3\) and then use step 2 of mathematical induction.

Step-by-Step Solution

Answer

The inequality \(n^{2}>2n+1\) was validated through mathematical induction. The base case was verified for \(n=3\). Assuming it holds for arbitrary \(n=k\) (Induction Hypothesis), we showed it also holds for \(n=k+1\) (Inductive Step), thus proving the inequality for \(n \geq 3\).

Step 1: Establish the Base Case

To start, validate if the inequality holds for the base case where \(n=3\). Substitute \(n=3\) into \(n^{2}>2n+1\) equation, and check if the resulting inequality is true.

Step 2: Assumption of Induction Hypothesis

Assume the inequality \(n^{2}>2n+1\) is true for some arbitrary \(n=k\). Hence, when replaced with \(k\), the inequality becomes \(k^{2}>2k+1\). This is called the induction hypothesis.

Step 3: Induction Step

In this step, we need to prove that if the inequality is true for \(n=k\), it must also be true for \(n=k+1\). First, substitute \(n=k+1\) into the inequality \(n^{2}>2n+1\) to obtain \((k+1)^{2}>2(k+1)+1\). Next, simplify this inequality and express it in terms of the expression for the induction hypothesis obtained in Step 2. If the inequality holds, then we've proven that the statement is true for \(n=k+1\).

Step 4: Conclusion

After executing the above steps, if all parts have been successfully implemented, then we can conclude that the inequality \(n^{2}>2n+1\) is true for all \(n=3\), \(n=4\), ..., and up to \(n=k\), and also true for \(n=k+1\). This completes the proof via mathematical induction.

Key Concepts

Base Case

In mathematical induction, the base case serves as the foundation of your proof. It involves validating whether the statement is true for the initial value, often denoted as the smallest integer in the set involved. For our problem, proving the inequality \( n^2 > 2n + 1 \) starts with checking it for \( n = 3 \). To do this:

• Substitute \( n = 3 \) into the inequality, resulting in \( 3^2 > 2 \times 3 + 1 \).
• Simplify to get \( 9 > 7 \), which is true.

With this verification, the base case holds, and we can move ahead to ensure the rest of the proof is based on a true initial step.

Induction Hypothesis

The induction hypothesis assumes the statement is true for some arbitrary positive integer \( k \). In our context, we suppose that \( k^2 > 2k + 1 \).This assumption forms the backbone of the proof because:

• It sets a proving launchpad for the subsequent step.
• Everything else builds upon it, so it's crucial to state clearly.

Think of the induction hypothesis as your climbing rope – it secures your growth to newer heights in reasoning. It is this assumption that we will later use to show the truth of the statement for \( n = k + 1 \).

Induction Step

In the induction step, demonstrate that if the statement is true for \( n = k \), then it must also be true for \( n = k + 1 \). This verifies the transitional progress in our proof by induction.Here's how to initiate:

• Substitute \( n = k + 1 \) in the inequality to get \( (k+1)^2 > 2(k+1) + 1 \).
• Simplify to \( k^2 + 2k + 1 > 2k + 2 + 1 \).
• Further reduce and rearrange to \( k^2 > 2k + 1 \), which corresponds to our induction hypothesis.

By proving this implication, you've secured the truth of the statement for \( n = k + 1 \), effectively linking it back to the hypothesis.

Inequalities

Inequalities are mathematical expressions involving the symbols \( >, <, \geq, \leq \). In mathematical proofs, especially those using induction, inequalities often assess relationships between numbers.Here are some key points:

• Inequalities allow us to express conditions of size, boundary, and value comparison.
• During the induction step, they guide simplification and logical transformations.
• Precision is vital – mishandling inequality signs can derail entire arguments.

In the current exercise, handling inequalities carefully builds up steps of the proof: from establishing initial truth with a base case to carrying it through with the induction hypothesis and induction step, reinforcing the importance of direction and magnitude.