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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.
 one year ago
 one year 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.
 one year ago
 one year ago

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mukushlaBest ResponseYou've already chosen the best response.1
that needs another complete proof :)
 one year ago

LolWolfBest ResponseYou've already chosen the best response.1
Yeah, of course. Haha, it seems that Fermat's isn't strong enough to prove that...
 one year ago

LolWolfBest ResponseYou've already chosen the best response.1
Another 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
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
thats interesting too...
 one year ago

LolWolfBest ResponseYou've already chosen the best response.1
Haha, 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/)
 one year ago

zzr0ck3rBest ResponseYou've already chosen the best response.0
I saw another good one for infinite primes but its been a while.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
Proof of the Infinity of the Prime Numbers this is simple but i like it http://www.hermetic.ch/pns/proof.htm
 one year ago

LolWolfBest ResponseYou've already chosen the best response.1
A 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
 one year ago

LolWolfBest ResponseYou've already chosen the best response.1
Oh, yeah, the Euclidean proof. It's nice, and pretty simple, with some extra lemmas to work from the axioms.
 one year ago

zzr0ck3rBest ResponseYou've already chosen the best response.0
http://www.cuttheknot.org/pythagoras/index.shtml 97 proofs of pythagoreans theorem!
 one year ago

LolWolfBest ResponseYou've already chosen the best response.1
I just want, as they say, to 'nuke a mosquito'. With absolutely absurd proofs for simple statements.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
man diophantine equations is my favorite topic :)
 one year ago

zzr0ck3rBest ResponseYou've already chosen the best response.0
well then look up the proof for 2+2 = 4, its about 500 pgaes long.
 one year ago

LolWolfBest ResponseYou've already chosen the best response.1
Haha, 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...
 one year ago
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