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Prove that $2^{n} < n!$ $\forall$ n > 4

$n=5:$ $$2^{5}<5!$$

$$32 < 120$$

This is true.

Now, after knowing it worked for $n$ we need to show it works for every other, so $n+1$:

$$2^{n+1} < (n+1)!$$

$$2^{n} < \frac{(n+1)!}{2}$$

We know from beginning that $2^{n} < n!$

So we replace it with it here and then show:

$$n! < \frac{(n+1)!}{2}$$

$$n! < \frac{n!\cdot(n+1)}{2}$$

$$1<\frac{n+1}{2}$$

Task say for all $n > 4$ so the thing on the right side will really be greater than $1$.

I hope everything is ok?

Edit: The possible-duplicate-link didn't help me because I'm not really looking for a solution to the task. I'm rather interested in knowing if MY proof is correct.

Anne Bauval
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cnmesr
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    Correct. Everything is fine. – Sarvesh Ravichandran Iyer Sep 20 '16 at 12:34
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    Be careful on direction. What is your scratch work and what is your proof? It seems you start off with what you want to show. –  Sep 20 '16 at 12:37
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    What's with the close vote? This is a perfectly fine question. Even for my high standards. –  Sep 20 '16 at 12:39
  • I think in my language (not English) I will take care if it. There are several names for the parts in induction proof I have used here but I don't know them in English that's why I have skipped them here. Or you didn't even refer to the names? Is there something else important I have forgotten, maybe at the end? – cnmesr Sep 20 '16 at 12:39
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    You start off your induction step with $2^{n+1}<(n+1)!$ which is what you are trying to prove! –  Sep 20 '16 at 12:41
  • It's not done like that? We firstly show it works for $n$, then if it worked, we show it will work for $n+1$. Take the inequality from the beginning, the thing to show, and replace every $n$ with $n+1$. – cnmesr Sep 20 '16 at 12:43
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    First you show it holds for $n=5$. Then you show IF it holds for $n$, then it holds for $n+1$. –  Sep 20 '16 at 12:51
  • Ahhh I understand now! Thank you from now I will always write that in my proofs after shown for $n$ :-) In maths language would be it be enough to write: $n \rightarrow n+1$ ? – cnmesr Sep 20 '16 at 12:52
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    "Now, after knowing it worked for n=5 we need to show it works for every other, – Airymouse Sep 20 '16 at 12:54
  • See also http://math.stackexchange.com/questions/111146/prove-by-induction-that-n2n and the other four appearnaces of this question that are linked at that page. – Gerry Myerson Sep 20 '16 at 13:03
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    Sorry I hit enter in place of ". That's what you want to know, but what you need to prove is that if the statement holds for n it holds for n+1. My granddaughter is being asked: 3x=15 Find x. She has not been told that what she is doing is backwards: she is thinking if 3x=15 then I can divide by 3 and so x must be 5. That's right, the only number that could work is 5, but she doesn't yet know that 5 works. Her method makes sense only after she has found out that ax=b has a solution whenever a is not 0. – Airymouse Sep 20 '16 at 13:08
  • this is true for n=4 also,$2^4<4!$ – Sathasivam K Sep 20 '16 at 14:42
  • At least two answers there do exactly what you did. Does this not confirm what you do is correct? Plus several answers here seem near identical to those on the target. Do they not answer your question either? If so please let the posters know via a comment. – quid Sep 20 '16 at 15:18
  • See here for a more general and conceptual method. – Bill Dubuque Oct 21 '16 at 03:14

6 Answers6

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I would simply note that for sufficiently large $n$, $2$ < $n$ + $1$. Then by induction, when $2^n$ < $n$!, $2^n$ ${\times}$ $2$ < $n$! ${\times}$ $2$ and $n$! ${\times}$ $2$ < $n$! ${\times}$ $n + 1$, so by transitivity $2^n$ ${\times}$ $2$ < $n$! ${\times}$ $n + 1$.

PMar
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Your results are indeed correct. The only thing is, whatever steps you wrote down, follow them in reverse order, i.e write the last step first and so on. Then, the proof will be a well-written inductive proof.(And it is a pretty good strategy to prove results, isn't it)?

codetalker
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    It does not matter whether the steps are in implicative or necessitary order. What is much more important is that there should be arrows ($\Rightarrow$ or $\Leftarrow$) indicating the direction of implication. – Anon Sep 20 '16 at 14:32
  • I guess the crux of my answer was based on what you just said.. – codetalker Sep 20 '16 at 16:25
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Hint:

$$4<n\implies2^{n+1}=2^n\cdot2<n!\cdot2<(n+1)!$$

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A precise proof is as follows:

For 4 ≤ n we have: $2$ < $n$ + $1$. Now using this and by induction, assuming $2^n$< $n$! we may simply get:

$2{\times}$$2^n$< $(n + 1)$ ${\times}$ $n$! or $2^{n+1}$$< (n+1)$!

The above argument is just based on this basic fact that if a>b and c>d then a×c>b×d , for all positive integers. Note that this is true for 4 ≤ n.

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In addition to my solution, I should note that there is a serious basic error in your proof. Note that you can NEVER use this formula $2^{n+1} < (n+1)!$ in any step in your proof procedure (by induction), as it should be merely gotten as the final consequence. As I've shown in the solution, we should necessary start with the initial assumptions which are: $2^4$< $4$! and $2$ < $n$ + $1$ (which is true for n ≥ 4).

Now as you see in the solution, by induction supposing $2^n$< $n$!, we can get strightforwardly:

$2^{n+1} < (n+1)!$

as the main requirement. In general, there is a similar procedure in any proof by induction.

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Are you stating that if $a<b$ and $a<c$ then $b<c?$ But you have the correct answer. I'd do it in this way.

Assume its true that it's true that $2^n<n!=>\frac{2^n}{n!}<1$ then for $n+1$ we have to prove that:

$2^{n+1}<(n+1)!$

$2^n<n!(n+1) * 1/2$

$\frac{2^n}{n!}<(n+1)*1/2$

And then state that as $\frac{2^n}{n!}<1$ and $\frac{n+1}{2}>1$ for any $n>4$ Therefore the inequality is true.

Sargsyan Grigor
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