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Chapter 1: Problem 15

Prove that \(2 n-3 \leq 2^{n-2}\) for all \(n \geq 5, n \in \mathbb{N}\)

Short Answer

Expert verified
The inequality \(2 n-3 \leq 2^{n-2}\) is proven to be true for all natural numbers \(n \geq 5\) using mathematical induction.

Step by step solution

01

Base Case

We first need to verify the base case, which is \(n = 5\). We plug \(n = 5\) into both sides of the inequality to get \(2(5) - 3\) and \(2^{(5-2)}\), respectively. Simplified, these equal 7 and 8. So, 7 is less than or equal to 8 which verifies our base case.
02

Inductive Hypothesis

We assume that the inequality holds for a particular \(n = k\), hence we have that \(2k - 3 \leq 2^{k-2}\). This forms our induction hypothesis, which will be used to prove the statement for \(k + 1\).
03

Inductive Step

In this step, we try to prove the statement for \(n = k + 1\). So, we need to prove that \(2(k + 1) - 3 \leq 2^{(k + 1) - 2}\). Simplifying the left side gives us \(2k - 1\). The right side can be written as \(2 \cdot 2^{k-2}\). We know from our induction hypothesis that \(2k - 3 \leq 2^{k-2}\), and since \(2k - 1 > 2k - 3\), we can say that \(2k - 1 \leq 2 \cdot 2^{k-2}\), which proves the inductive step.

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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 first step in a proof by induction. It establishes that the given statement is true for the initial value. In our exercise, the base case is set at \(n = 5\). This means we check the inequality \(2n - 3 \leq 2^{n-2}\) when \(n = 5\).

Here's how it works:
  • Substitute 5 into the inequality: \(2(5) - 3\) and \(2^{(5-2)}\).
  • Simplify both sides resulting in: 7 and 8 respectively.
  • Since 7 is indeed less than or equal to 8, the base case holds true.
This step confirms that our inequality starts off on the right foot.
Inductive Hypothesis
An Inductive Hypothesis is an assumption made for a particular instance \(n = k\). It assumes that the inequality holds true for this specific \(k\) value. In our case, we assume:
  • \(2k - 3 \leq 2^{k-2}\).
This forms the basis for using induction to extend the proof to the next integer, \(k + 1\).

Imagine this is like a domino effect. You are assuming that if one statement is true, the next will also be true. Here, the inductive hypothesis is the vital assumption that helps bridge from the known base case to a broader conclusion.
Inductive Step
The Inductive Step is critical as it uses the inductive hypothesis to prove the next case, \(n = k + 1\). Our goal here is to verify:
  • \(2(k + 1) - 3 \leq 2^{(k+1)-2}\) holds.
Let's go through the steps:
  • Simplify the left side: \(2(k+1) - 3\), which becomes \(2k - 1\).
  • The right side becomes \(2 \cdot 2^{k-2}\), by breaking it into a product.
  • Using the hypothesis \(2k - 3 \leq 2^{k-2}\), and since \(2k - 1\) is larger, it follows \(2k - 1 \leq 2 \cdot 2^{k-2}\).
Completing this step proves that if the statement is true for \(n=k\), then it is true for \(n=k+1\). This confirms the domino effect in full swing, ensuring the inequality holds for all \(n \geq 5\).

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Most popular questions from this chapter

Conjecture a formula for the sum of the first \(n\) odd natural numbers \(1+3+\cdots+(2 n-1)\), and prove your formula by using Mathematical Induction.

For \(a, b \in \mathbb{R}\) with \(a

Let \(f(x):=x^{2}\) for \(x \in \mathbb{R}\), and let \(E:=\\{x \in \mathbb{R}:-1 \leq x \leq 0)\) and \(F:=(x \in \mathbb{R}: 0 \leq x \leq 1\\}\). Show that \(E \cap F=\\{0\\}\) and \(f(E \cap F)=\\{0\\}\), while \(f(E)=f(F)=\\{y \in \mathbb{R}: 0 \leq y \leq 1\\}\) Hence \(f(E \cap F)\) is a proper subset of \(f(E) \cap f(F)\). What happens if 0 is deleted from the sets \(E\) and \(F ?\)

Prove that \(5^{2 n}-1\) is divisible by 8 for all \(n \in \mathbb{N}\).

If \(A\) and \(B\) are sets, show that \(A \subseteq B\) if and only if \(A \cap B=A\).
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