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NeGaTiVe
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PostSubject: one random topic :P   Sun Dec 06, 2009 8:21 am

Every time i see FoG's name.. i always want to pronounce it like " F of G " lol!

only because i will have taken every math course offered at my school when i take calculus next semester.. damn i hate math.
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AvP*FoG
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PostSubject: Re: one random topic :P   Sun Dec 06, 2009 8:32 am

lol@ i hate math's too

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Omega
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PostSubject: Re: one random topic :P   Sun Dec 06, 2009 8:34 am

Statement: If a and b are perfect squares, then a + b - 2 √ab is a perfect square.
Proof Let a and b be arbitrarily chosen perfect squares. We will show, using a direct proof, that a + b - 2 √ab is a perfect square. By the definition of perfect squares, we know that a = k2 and b = j2 for some k,j  Z Then
k2 + j2 - 2√k2j2
k2 + j2-2kj
(k-j)(k-j)
(k-j)2
Note that (k-j)2 Is an integer by closure under multiplication. Since our equation can be simplified as an integer squared, it must be a perfect


Real math right there...

~mega
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Rav3n
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PostSubject: Re: one random topic :P   Sun Dec 06, 2009 8:43 am

Omega wrote:
Statement: If a and b are perfect squares, then a + b - 2 √ab is a perfect square.
Proof Let a and b be arbitrarily chosen perfect squares. We will show, using a direct proof, that a + b - 2 √ab is a perfect square. By the definition of perfect squares, we know that a = k2 and b = j2 for some k,j  Z Then
k2 + j2 - 2√k2j2
k2 + j2-2kj
(k-j)(k-j)
(k-j)2
Note that (k-j)2 Is an integer by closure under multiplication. Since our equation can be simplified as an integer squared, it must be a perfect


Real math right there...

~mega

Where the fuck would any1 fuckin use that..

math has so many pointless shit in it...




bounce bounce bounce

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NeGaTiVe
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PostSubject: Re: one random topic :P   Sun Dec 06, 2009 8:46 am

Omega wrote:
Statement: If a and b are perfect squares, then a + b - 2 √ab is a perfect square.
Proof Let a and b be arbitrarily chosen perfect squares. We will show, using a direct proof, that a + b - 2 √ab is a perfect square. By the definition of perfect squares, we know that a = k2 and b = j2 for some k,j  Z Then
k2 + j2 - 2√k2j2
k2 + j2-2kj
(k-j)(k-j)
(k-j)2
Note that (k-j)2 Is an integer by closure under multiplication. Since our equation can be simplified as an integer squared, it must be a perfect


Real math right there...

~mega

nerd :-P jk

we had to memorize all the trig identities...all 12 of em i think.
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*AVP*SEAN
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PostSubject: Re: one random topic :P   Sun Dec 06, 2009 11:33 am

Rav3n wrote:
Omega wrote:
Statement: If a and b are perfect squares, then a + b - 2 √ab is a perfect square.
Proof Let a and b be arbitrarily chosen perfect squares. We will show, using a direct proof, that a + b - 2 √ab is a perfect square. By the definition of perfect squares, we know that a = k2 and b = j2 for some k,j  Z Then
k2 + j2 - 2√k2j2
k2 + j2-2kj
(k-j)(k-j)
(k-j)2
Note that (k-j)2 Is an integer by closure under multiplication. Since our equation can be simplified as an integer squared, it must be a perfect


Real math right there...

~mega

Where the fuck would any1 fuckin use that..

math has so many pointless shit in it...




bounce bounce bounce

lmao... i kind of like math ... i rather play quake tho!
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Omega
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PostSubject: Re: one random topic :P   Sun Dec 06, 2009 2:38 pm

That is how the math you used is based. I hate this class in college... it makes me want to cry sometimes.

~mega
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