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MHB Crankfest

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
ths thrd is devpted to teh discusion of proofs of unsovled/hard prblems and connjecturs in math.

teh rules r :

1. strt with any hard/unprved theomr/conjectur as a subejct and presnt teh crnkiest prof posssibb;e
3. one cn post a crnk-prof only once and 1 at time. other thn taht, any oter comemt/rply is not restricted.
4. try t be as much as a crnk possible!

Lt teh cntest beginn!

No idea what's going on? See here :

This thread is devoted to the discussion of (crank) proofs of unsolved problems and conjectures or relatively well-known solved theorems in mathematics (i.e., Riemann Hypothesis, Collatz conjecture, Fermats Last Theorem, Fermats little Theorem (yes!), Trisection problem, Goldbach conjecture and many others, everything you like!)

The rules are :

[1] Start with any hard/unproved theorem/conjecture as a subject and present the crankiest proof possible.

[2] One can post a crank proof only once and one at a time. Other than that, any other comments/replies are welcome!

[3] Please try to be as much as cranky as possible!!

Let the contest begin!

PS : This thread is partially inspired by this topic on MMF
 
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mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
I start the initial example by trying to prove FLT via elementary techniques by maximal crank possible : (NOTE* : I am not one of the competitors!)

[HR][/HR]
flt was an unslved prblem for a long time utnill i prvd it. it is believd that femrta has a proof of ths an it was essntlally same as mine. recently wiles tried t prve it but a flw has been ponted out in hs wrk. i hve been de only one to derve such a result. you cn laugh at me, bt cn you laugh at fermat tooo?

if someon can find a flaw in me proof, i will gv him $100.

teh prof cn be done by applyng balarka charectrisitcs. it basically says fr n<2, we hav teh pythagoras identity and sin^2 +cos^2= 1 whc shows it is true fr n<2. For n>2, it imlies that tereh must've a f nd g s.t., f^n +g^nn = 1 whc is impssible since it mst be n = 2 fr f,g = sine, cos as esily cn be seen form godell's prop.

tis prf also imples beal's conj hence is great> i dnt knw why no one found it bt i prvd id by elemntry apprch.

i psted in in another forum, but teh mods tere deletd it.

if anyne hlps me publsh it, i wll give him 1000$ f teh prize money.

Note : any persn using tis prove witou permiss. of teh autor would be dealt in court. (C : Copyrighted) COPYRIGHTED-COPYRIGHTED-COPYRIGHTED-COPYRIGHTED-COPYRIGHTED-COPYRIGHTED-COPYRIGHTED-COPYRIGHTED

[HR][/HR]

Appropriate measuring of Caldwell-Gruenberger-Aaranson score gives 179C, 30G and 6A. This can be improved largely of course, though a maximal score of 511C, 100G and 10A is hard to obtain.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
My favorite theorem:

$2 = 0$.

We have, from elementary algebra, that:

$x^2 = x \cdot x = x + x + x + \cdots + x\ (x \text{ times})$.

Differentiating both sides, we obtain:

$2x = 1 + 1 + 1 + \cdots + 1\ (x \text{ times})$.

Differentiating again, we have:

$2 = 0 + 0 + 0 + \cdots + 0\ (x \text{ times}) = 0x = 0$.

Corollary:

All fields are $\Bbb Z_2$.

Thx 4 ur 5u|>|>0r7 :p
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Let \(\displaystyle x=red\), \(\displaystyle y=green\), and \(\displaystyle z=blue\), then:


\(\displaystyle x + y = yellow\)

\(\displaystyle y + z = silver\)

\(\displaystyle x + z = carmine\)


Can someone pls check ma results? I'm a tad colour blind, see... (Drunk)
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Let \(\displaystyle x=red\), \(\displaystyle y=green\), and \(\displaystyle z=blue\), then:


\(\displaystyle x + y = yellow\)

\(\displaystyle y + z = silver\)

\(\displaystyle x + z = carmine\)


Can someone pls check ma results? I'm a tad colour blind, see... (Drunk)
I read your theorem with great interest, but I must say it conflicts with another theorem I've read on these forums claiming everything is white....
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
I must say it conflicts with another theorem I've read on these forums claiming everything is white....
Is it the one that follows from the lemma that all horses are the same color? That's one of my favorites.
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
Special Relativity is wrong.

Proof: It's too complicated.

(Can I just say how many times I've heard this (Swearing) argument?)

-Dan
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Our winner is decided! I give Dhananjay P. Mehendale the Fence medal (*) for proving twin prime conjecture, Goldbach conjecture, Sierpinski conjecture, prime k-tuple conjecture, infinitude of Sophie Germain prime, Mersenne prime, Fermat prime and nominate his name in professor Dudley's next publication (+) as a reputed mathematician for researching on primes and twin primes in arithmetic progression and proving further that there is always a twin prime between n and 2n (which is obvious from his ingenious generalization).

(*) For the guards of the field.
(+) Dudley, Underwood. "The man who made history among the cranks"

PS : I knew this name from secret agents of mine. (Bandit)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
I read your theorem with great interest, but I must say it conflicts with another theorem I've read on these forums claiming everything is white....
That's OK, Deveno. My theorem - not that it deserves such a lofty title - is just random verbiage. Like a dog-walker minus any realistic chance of acquiring a dog, if you will.

Actually, that makes it the Oxfam of mathematical hypotheses, I think. (Hug)
 

eddybob123

Active member
Aug 18, 2013
76
Let P be a priem number an dlet NP be a number that is not priem. By Fermat's lastt hereom, NP is a compossite number. So P =\= NP and this prbolem was hard fro more than 3000 yares!

Edit: I think I may have overdone it a bit here. (Smirk)
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Let P be a priem number an dlet NP be a number that is not priem. By Fermat's lastt hereom, NP is a compossite number. So P =\= NP and this prbolem was hard fro more than 3000 yares!

Edit: I think I may have overdone it a bit here. (Smirk)
I think if you work a bit on that proof, it just may fit in the margin....