Thread: math induction

Prove that n! > 2^n for every integer n greather than or equal to 4.

This is what I have so far.

Proof: Since 4! > 2^4 the statement is true for n greater than or equal to 4. Assume k!>2^k for some positive integer k. Observe: (k+1)! > 2^(k+1).

I am doing okay with math induction with equal signs and not greater or less than signs. Can any one help?

Assume that K>4 and that we know that (K!)>2^K.
Look at (K+1)!=(K+1)(K!) !)>(K+1)2^K>(4)2^K>2^(K+1)

Originally Posted by Plato

Assume that K>4 and that we know that (K!)>2^K.
Look at (K+1)!=(K+1)(K!) !)>(K+1)2^K

I follow you up to this point. How did you come up with (k+1)2^k?

We have:
k + 1 > 4
k! > 2^k

We assume:
(k + 1)! > 2^(k + 1)
(k + 1)k! > 2*2^k

From this we get:
(k + 1)k! > 4k! > 4*2^k > 2^(k+1)