Math Problem Statement

 since we already have the base case as n = 0, we can assume k > 0 for our inductive hypothesis.

Prove using induction that 2^n > n for all integers n ≥ 0.

Solution

To prove the inequality 2n>n2^n > n for all integers n0n \geq 0 using mathematical induction, we can follow these steps:

Step 1: Base Case

We check the base case n=0n = 0:

20=1and1>02^0 = 1 \quad \text{and} \quad 1 > 0

The inequality 2n>n2^n > n holds true for n=0n = 0.

Step 2: Inductive Hypothesis

Assume that the inequality 2k>k2^k > k holds for some integer k0k \geq 0. This is our inductive hypothesis.

Step 3: Inductive Step

We need to show that if 2k>k2^k > k, then 2k+1>k+12^{k+1} > k + 1 as well.

Starting with the left side, we use the property of exponents to express 2k+12^{k+1}:

2k+1=22k2^{k+1} = 2 \cdot 2^k

Since we assume 2k>k2^k > k, we can multiply both sides of this inequality by 2:

2k+1=22k>2k2^{k+1} = 2 \cdot 2^k > 2 \cdot k

Thus,

2k+1>2k2^{k+1} > 2k

Now, we need to show that 2kk+12k \geq k + 1, which would imply 2k+1>k+12^{k+1} > k + 1.

Rearrange 2kk+12k \geq k + 1 to k1k \geq 1. Therefore, for ( k \geq 1 \

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Mathematical Induction
Inequalities
Exponents

Formulas

2^n > n

Theorems

Principle of Mathematical Induction

Suitable Grade Level

Grades 10-12