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Blah I accidentally deleted this the first time I posted… - Mathematicus Ridiculus [entries|archive|friends|userinfo]
Mathematicus Ridiculus

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[Nov. 13th, 2006|04:44 pm]
Mathematicus Ridiculus
mathphiles
[anagrammed]
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[mood |sillyabsurd]
[music |I Hate Everything About You - Three Days Grace]

Blah I accidentally deleted this the first time I posted it><  And it's not even worth typing out again but I shall!

Okay I got this problem from my calc book.  It's easy, but the proof is quite nice.  Simple.  Yeah so here goes.

Let  √2 > a/b, where a, b > 0

Prove (a + 2b)/(a + b) > √2


(a + 2b)/(a + b) > √2

↔ 1 + 1/(a/b + 1) > √2



As a/b > √2,

1 + 1/(a/b + 1) > 1 + 1/(√2 + 1)


(√2 - 1)(√2 + 1) = 1

√2 = 1 + 1/(√2 + 1)

Thus 1 + 1/(a/b + 1) > √2

and a+2b/a+b > √2 > a/b 

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Comments:
[User Picture]From: stupidpot
2006-11-14 03:30 am (UTC)
that's quite nice! how does it have anything to do with calculus though? haha i should contribute more to mathphiles too x)
(Reply) (Thread)
From: anagrammed
2006-11-14 06:30 am (UTC)
the first chapter is on set theory. it's annoying.
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