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Chapter 8: Problem 21

Prove the statement by mathematical induction. $$ 2^{n}>2 n \text { if } n \geq 3 $$

Short Answer

Expert verified
The inequality \( 2^n > 2n \) is true for all \( n \geq 3 \) by mathematical induction.

Step by step solution

01

Base Case

First, let's verify the statement for the initial value of \( n = 3 \). Substitute \( n = 3 \) into the inequality:\[ 2^3 > 2 \times 3 \] Simplifying both sides gives \( 8 > 6 \), which is true. Thus, the base case holds.
02

Inductive Hypothesis

Assume that the statement holds for some integer \( k \geq 3 \). That is, assume \( 2^k > 2k \). This assumption is known as the inductive hypothesis.
03

Inductive Step

Prove that if the statement holds for \( n = k \), then it also holds for \( n = k + 1 \). Based on the inductive hypothesis, we want to show:\[ 2^{k+1} > 2(k+1) \] We know that \( 2^{k+1} = 2 \times 2^k \). Using the inductive hypothesis \( 2^k > 2k \), we can substitute to get:\[ 2 \times 2^k > 2 \times 2k \] Simplifying, we have \( 2^{k+1} > 4k \). To prove \( 2^{k+1} > 2(k+1) \), notice that when \( k \geq 3 \), \( 4k \geq 2k + 2 \), which implies \( 2^{k+1} > 2(k+1) \) is true for \( k \geq 3 \). Therefore, the inductive step holds.
04

Conclusion

Since the base case holds and the inductive step proves that if the statement is true for \( n = k \), then it is also true for \( n = k+1 \), by mathematical induction, the statement \( 2^n > 2n \) holds for all \( n \geq 3 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Base Case
The base case is the foundation of a mathematical induction proof. Here, we need to verify that our statement holds for the initial value given. In the exercise, the statement to prove is \( 2^n > 2n \) for \( n \geq 3 \). So, we start by substituting \( n = 3 \) into the inequality.Let's do the math:\[ 2^3 > 2 \times 3 \]This simplifies to \( 8 > 6 \), which is indeed true. This step is crucial as it confirms the statement for the baseline value.
  • If our base case fails, the induction cannot proceed as it relies on the statement being true initially.
In simpler terms, if you can't trust a small domino to fall, the rest can't follow suit. Ensure the base is strong!
Inductive Hypothesis
The inductive hypothesis is a critical step where you assume the statement is true for some arbitrary integer \( k \). This hypothesis acts as a bridge from the known to the unknown values. For our exercise, assume:\[ 2^k > 2k \] This assumption is like a stepping-stone, allowing us to establish validity for the next integer.
  • This hypothesis should be clearly stated. It's what makes the induction "chain" hold together across all values.
  • Remember, you're not proving it for \( k \) at this point; you're just assuming it to help prove it for \( k + 1 \).
While this assumption might seem abstract, it's a logical leap necessary for the method of induction.
Inductive Step
The inductive step is where we prove the domino theory in action: if one domino falls (the statement is true for \( n = k \)), the next will too (it's true for \( n = k+1 \)).In the exercise, we need to prove:\[ 2^{k+1} > 2(k+1) \]We know from our inductive hypothesis that \( 2^k > 2k \). Using this, let's substitute in our proof:\[ 2^{k+1} = 2 \cdot 2^k \]Hence,\[ 2 \cdot 2^k > 2 \cdot 2k \]Simplify this to get \( 2^{k+1} > 4k \). Now, let's ensure it leads to \( 2^{k+1} > 2(k+1) \): observe that:
  • For \( k \geq 3 \), we know \( 4k \geq 2k + 2 \) intuitively when doing the simple arithmetic comparison.
  • This confirms that \( 2^{k+1} > 2(k+1) \) to be true.
This step is vital as it extends the truth of our base case through every subsequent positive integer meeting the criteria.

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