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Advanced Math/Mathematical Induction


I am practicing this question for a test.

Prove that n!>2^n for every positive integar n.

This is the proof I have written, but I am not sure I did it right and there is probably a more elegant way to to this.

Proof: Induction on n
Step: Let n=4. Then
Assume that k!>2^k for integars greater or equal to 4.
Now replacing k with k+1,
we have (k+1)!>2^(k+1)
Since k is greater or equal to 4, k+1>2 and
k!>2^k can be expressed as
kx(k-1)x...x4x3x2x1 > 2x2x2x2x(2^k-4).
Since 4x3x2x1=24 > 2x2x2x2=16,
and (k)(k-1) > 2^(k-4), then k!>2^k.
Now we observe that k!>2^k and
k+1 > 2, so the product of k!(k+1) is
greater than the product of (2)(2^k).

I'm not sure if this even shows induction or if it even proves this so if you could help me out that would be great! Thanks!

First, you say "for every positive integer n" when you probably mean "for ever integer n≥4." This statement is not true for n=1,2,3.

You show the base case 4, but you don't apply induction correctly -- you're making it too complicated. All you need to do is this:

Assume it is true for n=k, so that:

k! > 2^k

Now consider that:

(k+1)! = (k+1) k!

By the induction assumption:

(k+1)! = (k+1) k! > (k+1) 2^k

And because k > 4, k+1 is definitely greater than 2, so:

(k+1)! = (k+1) k! > (k+1) 2^k > 2 2^k = 2^(k+1)

That one line sums up the entire induction step of the proof.

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Clyde Oliver


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