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 January 28th, 2015, 06:33 PM #1 Member   Joined: Aug 2012 Posts: 41 Thanks: 0 Prove that 2n^2 > (n+1)^2 Hello, In the proof, of this, there's a step I don't understand. First they have (N+1) > sqrt root of 2 And then they say, n > (sqrt root of 2) -1 without any further explanations???? How does that help me figuring out the proof? Thanks
 January 28th, 2015, 08:38 PM #2 Senior Member   Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics Assuming N and n are the same, I think they just moved the +1 to the other side, didn't they? BTW, did they say anything else about n? The inequality doesn't always hold (e.g. 0, 0.5...) 2n^2 > (n+1)^2 2n^2 > n^2 + 2(n)(1) + 1 n^2 - 2n - 1 > 0 Thus the inequality only holds for x > 2.41 or x < -0.414. Thanks from Spook Last edited by 123qwerty; January 28th, 2015 at 08:40 PM. Reason: I meant INequality
January 29th, 2015, 03:29 AM   #3
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Quote:
 Originally Posted by Spook . . . First they have . . . (N+1) > sqrt root of 2
I very much doubt that "they" gave that first. It wouldn't make sense to produce such an assertion out of thin air.

 January 29th, 2015, 09:15 AM #4 Member   Joined: Aug 2012 Posts: 41 Thanks: 0 Sorry, 123qwerty nailed the missing step I failed to include. Mea culpa. It's really the jump from n > (sqrt root of 2) -1 to "then it's certainly true that n^2 > (n+1)^2 for all n>= 5" that I don't understand.
 January 29th, 2015, 10:55 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,281 Thanks: 1965 Can you post the entire proof that "they" provided? It's difficult to make sense of tiny snippets of it in isolation.
January 29th, 2015, 11:41 AM   #6
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Quote:
 Originally Posted by Spook It's really the jump from n > (sqrt root of 2) -1 to "then it's certainly true that n^2 > (n+1)^2 for all n>= 5" that I don't understand.
Well it's not true that $n^2 \gt (n+1)^2$ for any non-negative integer $n$, so I suspect you have made an error in what you typed.

I suggest you write out the entire question/proof that you want explained.

 January 29th, 2015, 06:04 PM #7 Member   Joined: Aug 2012 Posts: 41 Thanks: 0 Well, it's the problem #10 of this that I'm doing. http://www.aduni.org/courses/discret..._Solutions.pdf I understand everything up until that step I posted earlier Last edited by Spook; January 29th, 2015 at 06:12 PM.
 January 29th, 2015, 06:15 PM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra They have shown that $2n^2 \gt (n+1)^2$ for all $n \gt \sqrt2 - 1$. Since $\sqrt2 - 1 < 1$, it must be that $2n^2 \gt (n+1)^2$ for all $n \ge 5$, which was the result required. Thanks from Spook
January 30th, 2015, 12:56 AM   #9
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Quote:
 Originally Posted by v8archie They have shown that $2n^2 \gt (n+1)^2$ for all $n \gt \sqrt2 - 1$.
No, they haven't. Their "proof" contains a mistake.

In considering the inequality $n^2 - 2n + 1 > 2$, it states
"In other words, $(n + 1)^2 > 2$
But this is true if and only if $(n + 1) > \sqrt2$
or equivalently $n > \sqrt2 - 1 \approx 0.4$".

That's obviously incorrect. A corrected version could be
"In other words, $(n - 1)^2 > 2$
But this is true (for non-negative $n - 1$) if and only if $(n - 1) > \sqrt2$
or equivalently $n > \sqrt2 + 1 \approx 2.4$".

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