[extremelearning]

Yes.

Consider π. For any S>0, you can construct an infinite number of rational approximations that have a score of less than S.

But for any quadratic irrational (surd), as the depth of the corresponding continued fraction increases, the score will converge (in an alternating manner) to a critical score, S.

This means that for any score S < S, there is only a finite set of rational approximations that have a score of less than S.

For example, in figure 3, for S=0.4 < 1/√5 ≃ 0.447, there is only one fraction that gives a score of less than S=0.4.

Hope that helps!