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/sci/ - Science & Math

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File: 195 KB, 1773x837, proof for collatz conjecture.png [View same] [iqdb] [saucenao] [google]
8230652 No.8230652 [Reply] [Original]

welp, I proved the Collatz Conjecture.

>> No.8230659

>>8230652
So go get your $500.

>> No.8230660

>>8230659
I always dreamed of proving something important and instead of taking credit, just posting it anonymously on /sci/

>> No.8230835

>>8230652
>since the sequence has no limit, it can be disregarded
how do you figure that one?

>> No.8230844

brb claiming my prize
thanks op

>> No.8230851

>>8230835
they are just exponents of 2

>> No.8230859

not rigorous enough

>> No.8230862

>>8230859
Being more rigorous would be trivial

>> No.8230867

>>8230862
everything is trivial

>> No.8230868

>A power of 2, minus 1, divided by 3 is composite iff it is an even power of two

So, 5 = (2^4 - 1)/3 is composite. Nice job senpai

>> No.8230870

>>8230868
Read properly

>> No.8230871

Thanks for the free 500

>> No.8230875

>>8230867
and left as an exercise to the reader

>> No.8230887

>>8230870
So you mean: (2^k - 1)/3 is composite implies that it is also an even power of 2?

>> No.8230890

>>8230887
no, it's just a multiple of it
I'm not OP but I understand his proof now. it was all so fucking obvious

>> No.8230895

>>8230890
Is it an actual proof??

>> No.8230896

>>8230895
no

>> No.8230898

>>8230659
OMG if you prove the conjecture those yid give you only 500$ ? is this real

>> No.8230899

>>8230890
What's a multiple of what? (2^6 - 1)/3 = 21 isn't a multiple of any power of 2

>> No.8230900

>>8230895
i don't know what's considered a 'formal' proof, but if it's not a proof it's a much stronger form of the collatz conjecture

still checking it

>> No.8230901

>posted 3 hours and 10 minutes ago
>nobody has yet spotted a flaw
welp, /sci/ solved a millenary problem

>> No.8230902

>>8230901
from Lemma 4 onwards it's all handwavy nonsense. there's no flaw because there's no proof

>> No.8230912
File: 1.06 MB, 2135x1366, 20160726_001717.jpg [View same] [iqdb] [saucenao] [google]
8230912

>>8230899
what he means is (2^(2n))-1 is always divisible by 3. Or rather, 4^n -1 is always divisible by 3.

Another way to think about it is 4^n = (3+1)^n

>> No.8230933

>>8230912
Oh, that's it? That's quite trivial. OP's wording was terrible.

>> No.8230938

>>8230652
no you didnt

>> No.8230940

>>8230660
Don't be perelman fag.
You better have posted this first on arxiv.

If not then someone call Mochizuki or a mod to tell him to save the IP of this bastard and them track him down to give him his prize. Motherfucker.

>> No.8230942

>>8230940
are you being ironic?

>> No.8230945

>>8230942
are you stupid? this is a big deal
i'm angry because i really wanted the 500 dollars, but that won't make me pretend this isn't a big deal

>> No.8230946

>>8230945
again, are you being post-ironic or something? this isn't a proof

>> No.8230949

>>8230660
Good goy. Recognition and money are for other people. You should just do work for free.

>> No.8230953

>>8230942
Fuck no.

If this proof is true then he better receive his prize. Stop being a faggot.

Prove him wrong or call JapMoot to save his bastard's IP.

CALL THE FUCKING FBI AND NSA. WE ARE TRACKING THIS FAGGOT TO GIVE HIM 500 BUCKS.

>> No.8231023

>>8230953
>delusional retard thinks he's funny

no wonder you have no friends anon, nobody can tolerate this kind of shit in real life.

>> No.8231051

Posted 5 hours ago

Nobody has debunked it

Logic holds

This is it gentlemen

>> No.8231061

>>8230652
Nah, it doesn't work I've tried this before. You're basically looking in mod 4 and saying that 0, 1, and 2 all converge under the iteration of the rules.

But numbers congruent to 3 mod 4 don't necessarily converge since after 3n+1 you increase but are not guaranteed to decrease sorry.

>> No.8231063

>>8231061
I already covered that. Read carefully.

>> No.8231075

How does this prove the non-existence of other loops that may exist? It seems you just proved that you can reverse the Collatz on any odd number, but not that all odd numbers lead to one with the Collatz rules, or that all odd numbers can be derived from one using the reverse Collatz. I could just be retarded.

>> No.8231105

>>8230898
Erdős is a greedy guy
>Paul Erdős said about the Collatz conjecture: "Mathematics may not be ready for such problems."[9] He also offered $500 for its solution

>> No.8231115

Replies 35
Unique Posters 17

Something doesn't ADD up here.

>> No.8231128

>>8231115
Posts with inline quotes: 25

It's almost like people exchange two messages

>> No.8231145

>>8230933
Why does he write like it's the 1500s and we haven't developed math notation?

>> No.8231146

I don't know shit about math. Is this legit?

>> No.8231148

>>8231063
no you didn't. everything after "Lemma 4" is nonsense

>> No.8231149

Holy shit, could you define what symbols mean before you use them? This is bordering on unreadable

>> No.8231163

>>8231149
that's the least controversial part... the notation is fucking standard

>> No.8231166

>>8231149
Agreed.

wtf is the main idea of this proof?

>> No.8231169

>>8231163
Why would I be expected to know what fucking [math[\/mu[/math] means? You don't get to say notation is standard in math. You have to define your symbols so that others can read it.

>> No.8231403

I read this interesting answer about why we care about the Collatz conjecture and its generalizations.

First answer here: http://math.stackexchange.com/questions/2694/what-is-the-importance-of-the-collatz-conjecture

>> No.8231426

>>8231403
That's a really interesting take on the ideas underlying the conjecture. I wonder if it's possible to develop some general notions along these sort of "conservation of prime energy" lines that could ultimately lead to a tidy proof.

>> No.8231827

>>8230652
>Zooms in
>looks at a link on the 3rd page
Nice try faggot

>> No.8231844

I don't even get lemma 1 D;

>> No.8231937

>>8231844
You don't get lemma 1? Most of the first page is dedicated to saying that 2^k \equiv 1\pmod{3} iff k is even, and that 2^k+2^{k+1}\equiv 0\pmod{3}, but that is obvious since 2\equiv (-1)\pmod{3} and the most basic understanding of modular arithmetic. The statement of lemma 2 is badly worded (and false as written with counterexample k=4 as already pointed out). The word composite should be replaced with "is an integer."

It is from there that the handwaviness begins with the OP introducing \mu without properly defining it. Even so, it seems he is trying to build the list of what numbers will reach 1. He notes that for those numbers which do reach one, the final set of moves will be repeated divisions by two or will be multiplying by three and adding one followed by repeated divisions by two. Again, nothing new or surprising there.

From there, he tries to use some function f^a which he has not defined or explained what it is. Without having told us what f is, everything from here out is meaningless.

>> No.8231998

>>8231937
this is an accurate and well-articulated summary of the flaws in the OP's proof

>> No.8232057

bump

this is history happen to today in the eye

>> No.8232584

>>8230898
That's quite much for something this irrelevant.

>> No.8233642

>don't believe this
>still posting to be part of history just in case

>> No.8233652
File: 250 KB, 1600x1200, image.jpg [View same] [iqdb] [saucenao] [google]
8233652

I'm here for the memes

>> No.8233656

>>8231937
f is defined right there, dumb shit

it's defined by parts

>> No.8233659

>>8231827
it's a picture from google images

it's unrelated to the proof

>> No.8233769

>>8233652
kek