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LolWolf
Does anyone have any (hysterical) ridiculously-overblown 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.
that needs another complete proof :)
Yeah, of course. Haha, it seems that Fermat's isn't strong enough to prove that...
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
thats interesting too...
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/april-fool/)
I saw another good one for infinite primes but its been a while.
Proof of the Infinity of the Prime Numbers this is simple but i like it http://www.hermetic.ch/pns/proof.htm
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
Oh, yeah, the Euclidean proof. It's nice, and pretty simple, with some extra lemmas to work from the axioms.
http://www.cut-the-knot.org/pythagoras/index.shtml 97 proofs of pythagoreans theorem!
I just want, as they say, to 'nuke a mosquito'. With absolutely absurd proofs for simple statements.
man diophantine equations is my favorite topic :)
well then look up the proof for 2+2 = 4, its about 500 pgaes long.
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...