Can 2^n Always Be Greater Than n for Natural Numbers?

[amiv4]

Homework Statement

Prove that for all the natural numbers n that 2^n > n

Homework Equations

The Attempt at a Solution

Base Case is easy
Then the inductive step you have 2^k > k as the inductive hypothesis
show that p[k+1] holds
2^(k+1) > k+1
on the left side 2^(k+1) = 2^k * 2
but idk what else to do

 

[LCKurtz]

Homework Statement

Prove that for all the natural numbers n that 2^n > n

Homework Equations

The Attempt at a Solution

Base Case is easy
Then the inductive step you have 2^k > k as the inductive hypothesis
show that p[k+1] holds
2^(k+1) > k+1
on the left side 2^(k+1) = 2^k * 2

And what does your induction hypotheses tell you about the 2^k on the right side of that last equation? And after you answer that, what are you wanting thge right side of the inequality to be greater than?

 

[amiv4]

Well 2^k * 2 > 2^k > k. And I am guessing you are talking about k+1 and idk what i want that to be greater than. It is greater than k but idk how that helps

 

[LCKurtz]

Well 2^k * 2 > 2^k > ... And I am guessing you are talking about k+1 and idk what i want that to be greater than. It is greater than k but idk how that helps

But you are giving away something by dropping the 2. You have:

2^k+1 = 2^k*2 so far, and you know 2^k > k

Use that without dropping anything else and see if you can see the result is > k + 1.