## anonymous 3 years ago 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.

1. anonymous

but for n=2 ?

2. anonymous

that needs another complete proof :)

3. anonymous

Yeah, of course. Haha, it seems that Fermat's isn't strong enough to prove that...

4. anonymous

lol

5. anonymous

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

6. anonymous

thats interesting too...

7. anonymous

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/)

8. zzr0ck3r

I saw another good one for infinite primes but its been a while.

9. anonymous

Proof of the Infinity of the Prime Numbers this is simple but i like it http://www.hermetic.ch/pns/proof.htm

10. zzr0ck3r

thats the one.

11. anonymous

Oh, yeah, the Euclidean proof. It's nice, and pretty simple, with some extra lemmas to work from the axioms.

12. zzr0ck3r

http://www.cut-the-knot.org/pythagoras/index.shtml 97 proofs of pythagoreans theorem!

13. zzr0ck3r

lol

14. anonymous

I just want, as they say, to 'nuke a mosquito'. With absolutely absurd proofs for simple statements.

15. anonymous

man diophantine equations is my favorite topic :)

16. zzr0ck3r

well then look up the proof for 2+2 = 4, its about 500 pgaes long.

17. anonymous

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...