A community for students.
Here's the question you clicked on:
 0 viewing
 2 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.
 2 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.

This Question is Closed

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.1that needs another complete proof :)

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.1Yeah, of course. Haha, it seems that Fermat's isn't strong enough to prove that...

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.1Another 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

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.1thats interesting too...

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.1Haha, 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
 2 years ago
Best ResponseYou've already chosen the best response.0I saw another good one for infinite primes but its been a while.

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

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.1A 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

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

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

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

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.1man diophantine equations is my favorite topic :)

zzr0ck3r
 2 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.

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.1Haha, 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...
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.