Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
LolWolf
Group Title
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
 2 years ago
LolWolf Group Title
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
 2 years ago

This Question is Closed

mukushla Group TitleBest ResponseYou've already chosen the best response.1
but for n=2 ?
 2 years ago

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

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

LolWolf Group TitleBest 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
 2 years ago

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

LolWolf Group TitleBest 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/)
 2 years ago

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

mukushla Group TitleBest 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
 2 years ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.0
thats the one.
 2 years ago

LolWolf Group TitleBest 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
 2 years ago

LolWolf Group TitleBest 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.
 2 years ago

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

LolWolf Group TitleBest 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.
 2 years ago

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

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

LolWolf Group TitleBest 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...
 2 years ago
See more questions >>>
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.