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#1
03-12-2006, 05:04 PM
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Basic math proof
If there a theorum or proof about the following?
The difference between the squares of two consecutive numbers is the sum of those numbers. For instance: 4^2 - 3^2 = 3 + 4 16 - 9 = 7 7 = 7 10^2 - 9^2 = 9 + 10 100 - 81 = 19 19 = 19 btw - This works for both positive and negative, though it needs to be modified a bit for negative numbers to work correctly because of the negative sign. 
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#3
03-12-2006, 05:28 PM
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Re: Basic math proof
Just a little algebra:
n^2-(n-1)^2=n^2-(n^2-2n+1)=n^2-n^2+2n-1=2n-1= n+n-1=n+(n-1) If n=4, 4^2-(4-1)^2=4+(4-1), or 4^2-3^2=4+3 Edit: Kerth is speedier than I am. [img]/images/graemlins/smile.gif[/img]
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#4
03-12-2006, 08:38 PM
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Re: Basic math proof
Heh, I meant the name of it (if there is one).
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#6
03-12-2006, 11:43 PM
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Re: Basic math proof
[ QUOTE ]
[ QUOTE ] Heh, I meant the name of it (if there is one). [/ QUOTE ] Its just a neat observation, hardly worthy of a name. [/ QUOTE ] I suppose. I thought of it back in the 8th grade (21 years ago) and my math teacher at the time blew it off. I remembered it again a couple of days ago and posted it to see if mattered.
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#8
03-13-2006, 02:49 AM
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Re: Basic math proof
[ QUOTE ]
[ QUOTE ] [ QUOTE ] Heh, I meant the name of it (if there is one). [/ QUOTE ] Its just a neat observation, hardly worthy of a name. [/ QUOTE ] I suppose.... [/ QUOTE ] How about the "Curious-Kerth theorem"? C-K for short.
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#9
03-13-2006, 03:12 AM
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Re: Basic math proof
LOL, every child in the 7th grade knows that (a-b)*(a+b) = a^2-b^2.
Do you really think the special case of a=b+1 should be called a "theorem"? Try this: add the first n odd numbers (1, 1+3, 1+3+5, 1+3+5+7, ...) and see if you can find a pattern. Explain. [img]/images/graemlins/laugh.gif[/img]
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