TITLE: Proving that the Calkin-Wilf tree enumerates the rationals.
QUESTION [8 upvotes]: The Calkin-Wilf tree is an infinite undirected graph (tree) which is constructed as follows: starting from the root at $\frac{1}{1}$, each node $\frac{a}{b}$ has two children:

a left child $\frac{a}{a+b}$
a right child $\frac{a+b}{b}$


This tree has the property that every rational appears in it exactly once, in lowest terms. I'm interested in ways to intuitively understand this fact.
Most of what I know on this topic comes from this wonderful blog, which gives a proof [*] of the above at the link. He points out that every child uniquely defines a parent, and that every parent has a either a smaller numerator or a smaller denominator than its child. Therefore, if you start from any fraction $\frac{p}{q}$ in lowest terms, you can always trace a path back to $\frac{1}{1}$, the root.
This is a really nice proof, but it feels a bit "backwards" to me, in that we visualize walking the tree from the bottom up. Does anyone know of alternate proofs of this fact? I don't need rigor, just intuition.
Thoughts:
Clearly, all children must have either a greater numerator or a greater denominator than any of their ancestors, so they can't be repeats of an ancestor. (We also need "lowest terms" for this, but that follows by a separate argument -- see footnote).
So I'm only worried about "cousins". Perhaps there is some property that all the left children of a node share, which the right children do not? That would solve the problem, I believe.
*My summary only covers the part of his argument that proves "every rational appears in it exactly once." The "in lowest terms" part involves Euclid's Algorithm, and is covered in the next post.

REPLY [2 votes]: I know this is old thread but I can't help to mention that it may not need to be this complicated. For any positive rational number $\frac{m}{n}$ in its simplest form, if $m>n$ we know it is a right-side child node and the parent should be $\frac{m-n}{n}$; otherwise it is a left-side child node and the parent is $\frac{m}{n-m}$. The path from $\frac{m}{n}$ to the root 1 is thus unique and can be found easily by repeatedly applying this strategy. This is exactly the algorithm to find $\gcd(m,n)$ by Euclidean Division, and since $m$ and $n$ are coprime, it is guaranteed to end up with the root (ie, $m=n=1$).