Thread: difficulty following proof by induction

1. difficulty following proof by induction

This is given as an example of proof by induction. Prove that $\displaystyle n! > 2^n$ for all n >= 4

$\displaystyle n_0=4$ and $\displaystyle 4!=24>16=2^4$. Thus $\displaystyle n_0$ is true. Now suppose that k >= 4 and the statement is true for n=k. Thus we suppose $\displaystyle k! > 2^k$ We must prove that the statement is true for n=k+1; that is, we must prove that $\displaystyle (k+1)! > 2^{k+1}$. Now
$\displaystyle (k+1)!=(k+1)k!>(k+1)2^k$
using the induction hpothesis. Since k >= 4 certainly k+1 > 2 so $\displaystyle (k+1)2^k>2 x 2^k = 2^{k+1}$. We conclude that $\displaystyle (k+1)! > 2^{k+1}$ as desired. By the principle of mathematical induction we conclude that $\displaystyle n!>2^n$ for all integers n>=4

I don't get the part after "certainly k+1>2". What happened to the factorial?

2. Re: difficulty following proof by induction

It doesn't say that. It says k+1>2 which it is because k is at least 4. In fact, k+1>4, but all that's needed for the proof is k+1>2.

Also you have a typo. That minus sign towards the end should be an equal sign.

3. Re: difficulty following proof by induction

Ok but how does that show anything about (k+1)!?

4. Re: difficulty following proof by induction

Originally Posted by Jskid
Ok but how does that show anything about (k+1)!?
You know $\displaystyle \displaystyle (k + 1)! = (k + 1)k! > (k + 1)2^k$ and you have shown $\displaystyle \displaystyle (k + 1)2^k > 2^{k + 1}$.

Therefore $\displaystyle \displaystyle (k + 1)! > 2^{k + 1}$.