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So here is what I have to prove by induction:

$2^n\gt(n+1)^2$ for $n\ge6$

So, first lets say $n=6$

$$2^6\gt(6+1)^2$$ $$64\ge49$$ Now, assume $n=k$ $$2^k\gt(k+1)^2\text{ for } k\ge6$$

Prove $n=k+1$ is true $$2^{k+1}>(k+2)^2$$ $$2^k *2> k^2+4x+4$$ We can replace $2^k$ with $(k+1)^2$ $$(k+1)^2*2>k^2+4x+4$$ $$k^2+2>4$$ $$k^2>2$$ Now, since $k\ge6$, $k^2$ must be greater than $2$ $$$$ So, my question is, did I do the proof correctly, or is there a more rigorous method to get this done?

Daniel W. Farlow
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user1843145
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2 Answers2

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If I were you, this is how I would structure the meat of your proof: \begin{align} 2^{k+1} &= 2^k\cdot 2\tag{definition}\\[0.5em] &> 2\cdot(k+1)^2\tag{inductive hypothesis}\\[0.5em] &= 2(k^2+2k+1)\tag{expand}\\[0.5em] &= 2k^2+4k+2\tag{expand again}\\[0.5em] &> k^2+4k+4\tag{since $k\geq 6$}\\[0.5em] &= (k+2)^2.\tag{factor} \end{align} That is really all you need to show (of course, in your formal write up, you should include the base case, the inductive assumption, etc.).

Daniel W. Farlow
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  • I (sort of) understand your solution (working on deciphering it), but if I was to put my answer on a test, would it be correct? – user1843145 Feb 28 '15 at 03:20
  • @Anonymous Which part do you not understand? – Daniel W. Farlow Feb 28 '15 at 03:21
  • So, on the left hand side, $2^{k+1}$ would be there throughout the proof? – user1843145 Feb 28 '15 at 03:23
  • @Anonymous If you were to write it all out in one line, it would look like this: $$ 2^{k+1}=2^k\cdot2>2\cdot(k+1)^2=2k^2+4k+2>k^2+4k+4=(k+2)^2.\tag{1} $$ The idea behind structuring it the way I have is to clearly indicate what is going on at each step. As evident in $(1)$ above, by the chain of relations, you have $2^{k+1} > \cdots > (k+2)^2$, which is what you wanted to show. – Daniel W. Farlow Feb 28 '15 at 03:28
  • Oh that makes SOO much more sense, thanks! – user1843145 Feb 28 '15 at 03:33
  • Just curious, does my method work? Maybe not for all inequalities, but for some? – user1843145 Feb 28 '15 at 03:35
  • @Anonymous I'm not sure what you meant by "your method." The reason I posted my answer was ultimately to clean everything up. Don't take offense at this, but your attempted proof is not very well structured. In fact, you may want to check out an answer I posted yesterday that shows how one might reasonably structure a very clear proof. You may find it of interest. Part of the struggle is in actually typesetting though. So beef up on $\rm\LaTeX$ and you'll be fine! – Daniel W. Farlow Feb 28 '15 at 03:40
  • @Anonymous You may also want to check out this post which explains why induction really works. :) – Daniel W. Farlow Feb 28 '15 at 03:42
  • Will do, thanks! – user1843145 Feb 28 '15 at 04:36
  • Wait, @crash , couldn't you have just said $(k+1)^2>(k+2)^2$ ? – user1843145 Feb 28 '15 at 22:31
  • @user1843145 When would that inequality you mentioned be true? You can't use that. – Daniel W. Farlow Feb 28 '15 at 22:40
  • Oh crap, I meant the other way around, but I see that I messed up. But how do you know $2k^2+4k+2>k^2+4k+4$? Is it because you end up with $k^2>2$? – user1843145 Feb 28 '15 at 22:54
  • @user1843145 That's correct. The inequality $k^2>2$ is true for all $k\geq 2$; since you're considering a proposition for all $k\geq6$, then certainly $k^2>2$. – Daniel W. Farlow Feb 28 '15 at 22:55
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Hint: $2 \geq ((n+1)/n)^2$ for your assumption $n \geq 6$.

user2566092
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