What does proving the Collatz Conjecture entail? From the get go: i'm not trying to prove the Collatz Conjecture where hundreds of smarter people have failed. I'm just curious. 
I'm wondering where one would have to start in proving the Collatz Conjecture. That is, based on the nature of the problem, what's the starting point for attempting to prove it? I know that it can be represented in many forms as an equation(that you'd have to recurse over):
$$\begin{align*}
f(x) &= 
  \left\{
    \begin{array}{ll}
      n/2 &\text{if }n \bmod2=0 \\
      3n+1 &\text{if }n \bmod2=1
    \end{array}
  \right.\\
\strut\\
a_i&=
\left\{
\begin{array}{ll}
n &\text{if }n =0\\
f(a_i-1)&\text{if }n>0
\end{array}
\right.\\
\strut\\
a_i&=\frac{1}{2}a_{i-1} - \frac{1}{4}(5a_{i-1} + 2)((-1)^{a_i-1} - 1)
\end{align*}$$
Can you just take the equation and go from there?
Other ways I thought of would be attempting to prove for only odd or even numbers, or trying to find an equation that matches the graph of a number vs. its "Collatz length"
I'm sure there's other ways; but I'm just trying to understand what, essentially, proving this conjecture would entail and where it would begin. 

One approach could be to use the inverse function. Start with unity and apply the inverse function $g(n)$ recursively. Visualize each positive integer as a node and each iteration as a path. To prove the conjecture, one has to show that every node will be reached eventually. To disprove the conjecture, one has to show that there exist at least one node which will never be reached. 
\begin{align*}
g(n) &= 
  \left\{
    \begin{array}{ll}
      2n &\text{if }n \bmod2=0 \\
      (n-1)/3 &\text{if }n \bmod2=0 \hspace{5pt} \& \hspace{5pt} n \bmod3=1  \\
      2n &\text{if }n \bmod2=1
    \end{array}
  \right.\\
\strut\\
\end{align*}
Note that while $f(n)$ is a one to one function, $g(n)$ is not.
See this question.
