Problem 36 - A Graphical Approach to Precalculus with Limits

Question

Prove each statement by mathematical induction. \(4 n<2^{n},\) if \(n \geq 5\)

Step-by-Step Solution

Answer

The statement \( 4n < 2^n \) is true for all \( n \geq 5 \) by induction.

Step 1: Base Case

To begin a proof by induction, we first verify the base case, which is the simplest instance of the statement. Here, set \( n = 5 \) and check if \( 4n < 2^n \):\[4 \times 5 = 20 < 2^5 = 32\]Since \( 20 < 32 \), the statement is true for \( n = 5 \). This confirms the base case is correct.

Step 2: Inductive Hypothesis

Assume that the statement holds true for some arbitrary integer \( k \geq 5 \), i.e., \( 4k < 2^k \). This assumption is called the inductive hypothesis.

Step 3: Inductive Step

We need to prove that if the inductive hypothesis holds, then it must also hold for \( k + 1 \), i.e., prove \( 4(k + 1) < 2^{k+1} \).Start by expressing \( 4(k + 1) \) as:\[4(k + 1) = 4k + 4\]

Step 4: Using the Inductive Hypothesis

Since \( 4k < 2^k \) by the inductive hypothesis, we add 4 to both sides:\[4k + 4 < 2^k + 4\]Now, to prove the inductive step, we need to show that:\[2^k + 4 \leq 2^{k+1}\]Simplify the right side as \( 2^{k+1} = 2^k + 2^k \), so the inequality becomes:\[2^k + 4 \leq 2^k + 2^k\]

Step 5: Verify the Inequality

From the previous step, verify that:\[4 \leq 2^k\]Since \( k \geq 5 \), we know that \( 2^k \geq 2^5 = 32 \), which is indeed greater than 4. Therefore, the inequality holds.

Step 6: Concluding Inductive Proof

Since the base case is true and the inductive step has been verified, by mathematical induction, the statement \( 4n < 2^n \) is true for all \( n \geq 5 \).

Key Concepts

Base Case

The base case is the foundation of any proof by mathematical induction. It serves as the starting point of the proof, confirming that the statement we want to prove is valid for the smallest value of the variable. In our exercise, the stated inequality is to be verified for the smallest integer, which is 5. Plugging in, we calculate

• For base case: \( 4 \times 5 = 20 \) and \( 2^5 = 32 \).
• Comparing these, we find \( 20 < 32 \).

Since 20 is indeed less than 32, the base case holds true for \( n = 5 \). When the base case is verified, it means our statement has a valid beginning and is ready for the next step in induction.

Inductive Hypothesis

The inductive hypothesis is a pivotal part of mathematical induction. It involves assuming the statement is true for some arbitrary integer \( k \), which holds the place for any number after the base case.Let's assume that for some \( k \geq 5 \), the inequality \( 4k < 2^k \) is true. This supposition acts as a stepping stone because it provides a base from which we use logical reasoning to prove the statement for the next number, \( k+1 \).In essence, the inductive hypothesis sets up the conditions to test the continuity of the statement beyond the initial instance verified in the base case.

Inductive Step

The inductive step is the crux of the induction process. It's where we show that if the statement holds for \( n = k \), it also holds for \( n = k + 1 \).Here, we must demonstrate \( 4(k+1) < 2^{k+1} \). We start by rewriting:

• \( 4(k+1) = 4k + 4 \).
• Using the inductive hypothesis \( 4k < 2^k \), add 4 to both sides, giving \( 4k + 4 < 2^k + 4 \).

Now, the goal is to prove \( 2^k + 4 \leq 2^{k+1} \). Recognizing that \( 2^{k+1} = 2^k + 2^k \), we assert \( 2^k + 4 \leq 2^k + 2^k \) as valid.The essential aim here is to show the truth of the inequality carries forward to the next integer, effectively linking each case to the subsequent one, thereby proving the generality.

Inequality Proof

Inequality Proof in this context focuses specifically on establishing the validity of the inequality \( 2^k + 4 \leq 2^k + 2^k \).From the inductive step, we recognize we need to confirm \( 4 \leq 2^k \). Since our problem defines \( k \geq 5 \), the inequality \( 2^k \geq 2^5 = 32 \) holds true. Therefore, it is evident that:

• \( 4 \) is less than \( 32 \).
• Hence, \( 4 \leq 2^k \) is always satisfied for \( k \geq 5 \).

Conclusively, this proofs that the original inequality \( 4n < 2^n \) is valid through mathematical induction for all \( n \geq 5 \). The inequality steps ensure the foundation and logic align to complete the proof accurately and thoroughly.