>>14353528>>14353498This guy
>>14353502 meant to say "q is between sqrt(2) and p".
If you have p < sqrt(2) (so p is in A), you want to find q with p < q < sqrt(2). So clearly q is a function of p.
You could just take the average of p and sqrt(2), but then that wouldn't be rational.
Up to here everything is probably clear to you, but the rest requires a bit of experience.
There are other ways to create q, and one of the simplest ways is to use Mobius transformations. These are just functions that look like (ax+b)/(cx+d). Notice that in the proof, the function used is q = (2p+2)/(p+2).
Anyways, you have f(x) = (ax+b)/(cx+d), and you want it to satisfy a few conditions:
>a,b,c,d should be integers (if you had them as rational, you could just multiply the numerator and denominator of f by the denominators of these four numbers, so you may as well just assume they're already integers)>f(sqrt2) = sqrt2 (because for p < sqrt2, you will repeatedly apply f to p to get larger and larger values, while still remaining below sqrt2, and for p > sqrt2, you want to stay above sqrt2. So the only place for sqrt2 is itself)>The derivative of f is positive at sqrt2 (from what we said in the previous point, f is increasing)Okay, so using S = sqrt2 for convenience, (aS+b)/(cS+d) = S from the second condition, so aS+b = S(cS+d) = cS^2 + dS = dS + 2c. Rearrange to get (a-d)S = 2c-b. If a =/= d, then the left side is irrational, while the right side is an integer, which is impossible. So you need a = d, and so 0 = 2c-b, and b = 2c.
So f(x) = (ax+2c)/(cx+a). We need the derivative to be positive, and the derivative is (a^2-2c^2)/(cx+a)^2.
The denominator of that is always positive, so we only need the numerator to be positive, and using some algebra, we get (a/c)^2 > 2.
Now we can just take c = 1 and a = 2.
There we have it: f(x) = (2x+2)/(x+2).
