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I have been asked to prove the following:

For n ≥ 4, n$^2$ ≤ 2$^n$

I will argue by induction the statement P(k): for n ≥ 4, n$^2$ ≤ 2$^n$.

First, consider the base case P(4) = 16 ≤ 16 which is true, so we assume P(n) holds and consider P(n+1).

2$^{n+1}$ = 2$^n$2

We know that 2$^n$2 ≥ 2$^n$ and from our induction hypothesis we also know that n$^2$ ≤ 2$^n$, so we know that 2$^n$2 ≥ 2$^n$ ≥ n$^2$.

This is the part that makes me a bit uncomfortable and I think is wrong:

(n+1)$^2$ ≥ 2$^n$ by our induction hypothesis and definition of inequality.

Then, also by definition of inequality:

2$^n$2 ≥ (n+1)$^2$ ≥ 2$^n$ ≥ n$^2$

Thus, we have completed our induction step and shown that 2$^{n+1}$ ≥ (n+1)$^2$.

I am worried about my jump that (n+1)$^2$ ≥ 2$^n$. Do I need to some kind of subproof here to prove this relation? Any help would be appreciated!

jmoore00
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  • $2^n2 \neq 4^n$. – A. Thomas Yerger Mar 08 '17 at 02:55
  • Ahh, that's an embarrassing mistake that I have corrected. Thanks :) – jmoore00 Mar 08 '17 at 02:57
  • What is your induction hypothesis? – Eclipse Sun Mar 08 '17 at 03:02
  • My induction hypothesis is just that for n ≥ 4, n^2 ≤ 2^n – jmoore00 Mar 08 '17 at 03:07
  • Going back to your proof, what do you mean by $2^n2\geq 2^{n*}$ – Juniven Acapulco Mar 08 '17 at 03:25
  • Oops! That was the result of some shoddy html on my part. Just fixed. :) – jmoore00 Mar 08 '17 at 03:29
  • Remember that our objective is to show that $$2^{n+1}\geq (n+1)^2.$$ In your proof, you did it correctly that $$2^{n+1}\geq n^2.$$ What is next then? So, to attain the objective, you need to show that $$n^2\geq (n+1)^2$$ and use transitive property. But the latter inequality is false. Hope this help. That is why, I present to you my answer which I think the logical one. – Juniven Acapulco Mar 08 '17 at 03:36
  • One thing more I noticed in your proof and I quote."Then, also by definition of inequality:

    2$^n$2 ≥ (n+1)$^2$ ≥ 2$^n$ ≥ n$^2$". You already made a mistake in here. Look at the first inequality. That is supposed to be the thing that you are going to prove. What you did, you assumed it. Did you see?

     – Juniven Acapulco Mar 08 '17 at 03:45
  • Sigh. Yes I did and I knew I was jumping the gun but for lack of any other ideas went ahead and did it anyway. – jmoore00 Mar 08 '17 at 03:47

3 Answers3

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Observe the following:

$$\begin{align} n\geq 4&\implies n^2\geq 4n\\ &\implies n^2\geq 2n+2n\\ &\implies n^2> 2n+1. \end{align}$$ $P(n)$ is assumed to be true and so, $$2^n\geq n^2$$ and so multiplying by $2$ to both sides, we get $$2^{n+1}\geq 2n^2.$$ Thus, if $n\geq 4$ we get

$$\begin{align} 2^{n+1}&\geq 2n^2\\ &=n^2+n^2\\ &>n^2+(2n+1)\\ &=(n+1)^2 \end{align}$$ This means that $P(n+1)$ is also true.

Juniven Acapulco
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  • These proofs can be so clever! I guess my question is next whether or not what I have done so far in my proof is correct, and if not, if I can alter it slightly to make it so. I am trying to develop an intuition in which to approach problems like this. – jmoore00 Mar 08 '17 at 03:20
  • @agra94 Okay I understand. So this is just a formal proof that I know. Let me analyze your proof. wait for a moment.--) – Juniven Acapulco Mar 08 '17 at 03:23
  • Thanks again! I guess my question is now whether or not you have any tips for approaching induction. I feel like I always get kind of lost and don't realize the subtleties. I'm not sure if that's something you can really answer or if the answer is really just practice. – jmoore00 Mar 08 '17 at 03:43
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Hint For $n\ge 4$ you know that $(n+1)/n = 1 + 1/n \le 5/4$ so $(n+1)^2/n^2 \le 25/16 \le 2.$

spaceisdarkgreen
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This is more a comment than an answer, but it does provide an answer.

In my answer to my own question, I prove this:

If $n$ and $k$ are integers and $k \ge 2$ and $n \ge k^2+1$, then $2^n > n^k$.

Here it is:

Prove that $n^k < 2^n$ for all large enough $n$

marty cohen
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