Prove k<2^k for all positive integers k

Answered question

2021-12-14

Prove k<2^k for all positive integers k through the proces of mathematical induction.

Answer & Explanation

alenahelenash

alenahelenash

Expert2023-04-21Added 556 answers

To prove that k<2k for all positive integers k, we will use mathematical induction.

First, we will prove the base case for k = 1:

k<2k
1<21
1<2

This is true, so the base case holds.

Next, we will assume that k<2k for some positive integer k and use this assumption to prove that k+1<2k+1.

Assuming that k<2k, we can multiply both sides by 2 to get:

2k<2⋅2k

Simplifying the right-hand side gives:

2k<2k+1

Adding k to both sides gives:

k+2k<k+2k+1

Simplifying the left-hand side gives:

3k<k+2k+1

We know that k is a positive integer, so k≥1. Therefore, we can say that:

k+2k+1≥1+2k+1>2k+1

Substituting this inequality back into the previous inequality, we get:

3k<2k+1

Finally, dividing both sides by 2 and simplifying gives:

k+1<2k

So we have shown that if k<2k, then k+1<2k+1. Therefore, by mathematical induction, we can conclude that k<2k for all positive integers k.

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