I want to prove the collatz conjecture and part of my proof is that I need to show that (2x+1)*2^y is a surjection onto...

I want to prove the collatz conjecture and part of my proof is that I need to show that (2x+1)*2^y is a surjection onto N (y,x eof N) - actually I need to prove it for a bunch of functions (they all are surjections onto N) but this is the easiest one and I need to know the pattern of the proof.

How I do that? Please help bros.

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

Just carry the ones bro

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

/thread

en.wikipedia.org/wiki/Dunning–Kruger_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
N stands for non-negative integers in some places. But I agree, no matter how you look at it it's wrong

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

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.

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

I was expecting that everyone would call me retarded

>>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?

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.

>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

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?

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

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 ?

Sorry i read the operation on y incorrectly

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

Yo, I'm that guy:
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.

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.

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?

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

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.

ok thanks that was a good explanation

Guilherme, você é muito velho pro Veeky Forums.