My Math Forum  

Go Back   My Math Forum > Math Forums > Math

Math General Math Forum - For general math related discussion and news


Thanks Tree3Thanks
  • 1 Post By 123qwerty
  • 1 Post By v8archie
  • 1 Post By skipjack
Reply
 
LinkBack Thread Tools Display Modes
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
Spook is offline  
 
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
123qwerty is offline  
January 29th, 2015, 03:29 AM   #3
Global Moderator
 
Joined: Dec 2006

Posts: 20,281
Thanks: 1965

Quote:
Originally Posted by Spook View Post
. . . 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.
skipjack is online now  
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.
Spook is offline  
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.
skipjack is online now  
January 29th, 2015, 11:41 AM   #6
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,599
Thanks: 2587

Math Focus: Mainly analysis and algebra
Quote:
Originally Posted by Spook View Post
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.
v8archie is offline  
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.
Spook is offline  
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
v8archie is offline  
January 30th, 2015, 12:56 AM   #9
Global Moderator
 
Joined: Dec 2006

Posts: 20,281
Thanks: 1965

Quote:
Originally Posted by v8archie View Post
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 $n > \sqrt2 + 1 \approx 2.4$".
Thanks from Spook
skipjack is online now  
Reply

  My Math Forum > Math Forums > Math

Tags
&gt, >, 2n2, prove



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
how to prove it baxy Applied Math 1 April 6th, 2013 08:56 AM
Prove DP= DC Vasu Algebra 1 March 6th, 2013 07:04 AM
Goldbach's conjecture (to prove or not to prove) octaveous Number Theory 13 September 23rd, 2010 05:36 AM
prove prove prove. currently dont know where to post qweiop90 Algebra 1 July 31st, 2008 07:27 AM
prove prove prove. currently dont know where to post qweiop90 New Users 1 December 31st, 1969 04:00 PM





Copyright © 2019 My Math Forum. All rights reserved.