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anonymous
 3 years ago
Does anyone have any (hysterical) ridiculouslyoverblown proofs of simple statements? I'll give an example:
\(\sqrt[n]{2}\) is irrational for all \(n>2\in \mathbb{Z}\)
Proof:
Suppose \(\sqrt[n]{2}=\frac{p}{q}\) for some \(p,q\in \mathbb{Z}\), then
\[
2=\frac{p^n}{q^n} \implies\\
2q^n=q^n+q^n=p^n
\]Contradicting Fermat's Last Theorem.
I'd love to see some more of these, haha.
anonymous
 3 years ago
Does anyone have any (hysterical) ridiculouslyoverblown proofs of simple statements? I'll give an example: \(\sqrt[n]{2}\) is irrational for all \(n>2\in \mathbb{Z}\) Proof: Suppose \(\sqrt[n]{2}=\frac{p}{q}\) for some \(p,q\in \mathbb{Z}\), then \[ 2=\frac{p^n}{q^n} \implies\\ 2q^n=q^n+q^n=p^n \]Contradicting Fermat's Last Theorem. I'd love to see some more of these, haha.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that needs another complete proof :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, of course. Haha, it seems that Fermat's isn't strong enough to prove that...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Another interesting proof: There is an infinitude of primes: We begin by stating some large number \(n\) over which there exists no primes. If \(n>1\), there must exist a prime \(p\) such that \(n<p<2n\), thus our original statement contradicts Bertrand's postulate. Proof for the postulate here: http://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulate

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thats interesting too...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Haha, it's sort of absolutely ridiculous... I'm trying to come up with these as I finish up some homework... but I'm having a hard time. (Although the first one is from http://rjlipton.wordpress.com/2010/03/31/aprilfool/)

zzr0ck3r
 3 years ago
Best ResponseYou've already chosen the best response.0I saw another good one for infinite primes but its been a while.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Proof of the Infinity of the Prime Numbers this is simple but i like it http://www.hermetic.ch/pns/proof.htm

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0A proof that \(\nexists x, y \in \mathbb{Z} \text{ s.t. }x^3+113y^3=1\) http://books.google.com/books?id=YXDYKJvZY0QC&pg=PA111&lpg=PA111&dq=R+Finkelstein+and+H+London,+On+D.+J.+Lewis%27s+equation&source=bl&ots=YiyOhNRaLi&sig=JmtUPqakWZ5Mf4tqab8UMzQ_ecE&hl=en#v=onepage&q=R%20Finkelstein%20and%20H%20London%2C%20On%20D.%20J.%20Lewis%27s%20equation&f=false

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, yeah, the Euclidean proof. It's nice, and pretty simple, with some extra lemmas to work from the axioms.

zzr0ck3r
 3 years ago
Best ResponseYou've already chosen the best response.0http://www.cuttheknot.org/pythagoras/index.shtml 97 proofs of pythagoreans theorem!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I just want, as they say, to 'nuke a mosquito'. With absolutely absurd proofs for simple statements.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0man diophantine equations is my favorite topic :)

zzr0ck3r
 3 years ago
Best ResponseYou've already chosen the best response.0well then look up the proof for 2+2 = 4, its about 500 pgaes long.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Haha, it actually IS. To prove that like \(0\cdot m=0\) takes quite a bit of work, from axioms. Oh, and that \(1 \in \mathbb{N}\), also does...
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