Most people think that the Collatz conjecture is true, but I think that I can prove the opposite.
Let's make two functions, $f(x)$ and $g(x)$.
$f(x) = $ The amount of numbers that can be solved in x steps.
$g(x) = $ The amount of numbers that can be solved in x or less steps.
What we see is that x of the numbers that can be solved in x steps or less is a power of 2. This means that the percentage of numbers that is a power of 2 and that can be solved in x or less steps is $$\dfrac{x}{g(x)}$$.
If the collatz conjecture would be true, it would mean that as $x \rightarrow \infty$, the distribution of powers of 2 in $g(x)$ would be the same as the distribution of powers of 2 in all numbers and thus:
$$\dfrac{x}{g(x)}=\dfrac{x}{2^x}$$
Now let's take a close look at the relationship between $f(x)$ and $g(x)$.
We see that as $x \rightarrow \infty$: $\dfrac{f(x)}{f(x-1)}=a$, where a is a constant smaller than 2 and greater than 1.
This means that: $$g(x) = \sum\limits_{i=0}^{x-1} a^x = \dfrac{a^x-1}{a-1}$$
Using our earlier equation we get: $$\dfrac{a^x-1}{a-1}=2^x$$
Multiplying both sides by $a-1$: $$a^x-1=(a-1)2^x$$
Since $x \rightarrow \infty$, we can ignore the $-1$: $$a^x=(a-1)2^x$$
Dividing both sides by $2^x$: $$({\dfrac{a}{2}})^x=a-1$$
Since $x \rightarrow \infty$ and $a<2$, $$({\dfrac{a}{2}})^x=0$$ and thus $a=1$
But as we just stated a is greater than 1, a contradiction.
The only thing I didnt really prove was the existance of $a$, but if you take a closer look you don't need a constant as long as the ratio is between 1 and 2, which is easy to prove.
Question. Is this attempted proof valid (probably not) and why or why not?
You say:
This means you claim if the Collatz conjecture is true, then $g(x)=2^x$. So you claim the Collatz conjecture implies that $2^x$ numbers can be solved in $x$ or less steps, but I see no justification for this at all.
Also in latter parts the handling of "limits" is unusual. You seem to try to use "asymptotic equalities" but in ways one cannot use them.