I want to prove the collatz conjecture and part of my proof is

File: 1415654830376.jpg (851KB, 1958x1980px) Image search: [Google]

851KB, 1958x1980px

You're expecting the virgins of /sci/ to teach you how to solve a millennium problem and get yourself million dollars.

>>8710687
Just carry the ones bro

(2x+1)*2^y is always even. Dude.

>>8710708
/thread

>>8710687
https://en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect

>y is the greatest power of 2 that appears in the decimposition of a given number n
>n/2^y is odd so set it equal to 2x+1
>this only works for x, y nonnegative integers and f is a surjection onto N*, not N
Bruh if you're this retarded kys
Stop making maths look so bad
>>8710708
N stands for non-negative integers in some places. But I agree, no matter how you look at it it's wrong

>>8710687
You won't prove the Collatz conjecture and if you can't prove that you're a literal retard.

>>8710708
If we assume naturals are [math]\mathbb{N}\cup 0[/math] then there's an obvious surjection, but even if not there still is because the space of functions is super large.

The real problem is OP thinks he has solved the Collatz conjecture.

>>8710708
No it's not f.e.
(2*0+1)*2^0 = 1
(2*1+1)*2^0 = 3
(2*2+1)*2^0 = 5
...

all odd numbers

>>8710741
I was expecting that everyone would call me retarded

>>8710742
>>no matter how you look at it it's wrong
but why?

(2*0+1)*2^0 = 1
(2*0+1)*2^1 = 2
(2*1+1)*2^0 = 3
(2*0+1)*2^2 = 4
(2*2+1)*2^0 = 5
(2*1+1)*2^1 = 6
(2*3+1)*2^0 = 7
(2*0+1)*2^3 = 8
...

Whats wrong?

>>8710762
Literally nothing. It's quite obvious that every number's prime decomposition can be written uniquely in the form 2^n(2k+1). There's nothing to prove.

>>8710756
>The real problem is OP thinks he has solved the Collatz conjecture.
Nigga I havent solved anything I just have an idea I want to checkout

>>8710764
THANKS

the next surjection is this:

1/6((2*y+3)3^x-3), x> 0 and y >=0 and x,y eof N

It should also map onto N hows that proven?

>>8710772
Not a surjection

Algebraically you can rewrite it as (2^(y-1)+3/2)(x-1). If we set this equal to 1, and look at our domains, we see that (x-1)>=0 and (2^(y-1)+3/2)={2, 5/2, 7/2, 11/2, ..., 4k+1/2,...}. But each of (2^(y-1)+3/2)>1, and for no value of x is the following true: 0<(x-1)<1 so 1 is not in our image. Thus the mapping is not surjective.

>>8710791
I dont get how its wrong,
1 is in our image -> 1/6((2*0+3)3^1-3 = 1

Can you give me a counter example using 1/6((2*y+3)3^x-3 where x> 0 and y >=0 and x,y eof N ?

>>8710840
Sorry i read the operation on y incorrectly

>>8710843
No problem. Any Idea how to prove the function is a surjection?

>>8710870
Yo, I'm that guy:
>>8710741
I guess I was a little too harsh on you, so here is how I would go about proving your last conjecture:
If you set x = 1 you can simplify your term to read y + 1. So for every a eof N you found a y (which is a -1) and a x (which is 1) so that the equation a = 1/6((2*y+3)3^x-3) is true.

>>8710924
sry was a bad Idea to call the variable "a" since its also a word in the english language.
What I was trying to say: Pick x = 1 and you are left with f(x,y) = y + 1, which is obviously surjective.

>>8710938
I dont really understand the reasoning, I want to hit every number in N with this function. How does setting x to 1 guarantee that?

File: 1487308567421.jpg (99KB, 640x640px) Image search: [Google]

99KB, 640x640px

>>8710705
>collatz conjecture
>millennium problem
???
it's only worth 500$

>>8710981
For every b eof N you are looking for a y and a x so that b = 1/6((2*y+3)3^x-3) right?
I am saying you can even find x and y for every b if you limit yourself to x = 1, so that only y is variable.
Lets try it:
b = 1/6((2*y+3)3^1-3)
6b = (2 * y + 3)*3 -3
6b + 3 = (2 * y + 3) * 3
2b + 1 = 2 * y + 3
2b - 2 = 2 * y
b - 1 = y

This means, that for every b eof N, we found a x (which is 1) and y (which is b - 1) so that b = 1/6((2*y+3)3^x-3).
Which is what surjection means.
There exist x and y for every b so that f(x,y) = b.

>>8711024
ok thanks that was a good explanation

>>8710687
Guilherme, você é muito velho pro 4chan.