[1] http://benpaulthurstonblog.blogspot.com/2012/05/estimating-s...
[1] http://benpaulthurstonblog.blogspot.com/2012/05/estimating-s...
The argument is that the continued fraction representation for sqrt(N) grows as O(lg(N) sqrt(N)), making the representation blow up.
[0] https://www.facebook.com/bshlgrs/posts/10215278471769811?com...
[1] https://mathworld.wolfram.com/PeriodicContinuedFraction.html...
Edit: Also found a more accessible website describing how to do arithmetic with continued fractions: https://perl.plover.com/yak/cftalk/. In case anyone wants to try it out.
It's just that a proof of this is considerably harder, since you can't just assume the fractional part of irrational numbers is random.
Also, two continued fractions are not necessarily easy to compare. It's not a positional notation like decimal or binary.
sqrt(1-x) = 1- \sum_n C(n)/2^(2n+1) x^(n+1)
where x is in (0,1) and C(n)=binomial(2n,n)/(n+1) is the n'th catalan number.
[Learned this from http://www.math.chalmers.se/~wastlund/coinFlip.pdf]