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January 28th, 2015, 05:33 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, 07:38 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. Last edited by 123qwerty; January 28th, 2015 at 07:40 PM. Reason: I meant INequality 
January 29th, 2015, 02:29 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,965 Thanks: 2214  
January 29th, 2015, 08:15 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, 09:55 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,965 Thanks: 2214 
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, 10:41 AM  #6  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra  Quote:
I suggest you write out the entire question/proof that you want explained.  
January 29th, 2015, 05:04 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 05:12 PM. 
January 29th, 2015, 05:15 PM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 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. 
January 29th, 2015, 11:56 PM  #9  
Global Moderator Joined: Dec 2006 Posts: 20,965 Thanks: 2214  Quote:
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 nonnegative $n  1$) if and only if $(n  1) > \sqrt2$ or equivalently $n > \sqrt2 + 1 \approx 2.4$".  

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