Power of two open upper bound q < 2^z and Qabs q < 2^z
Compute a z such that q<2^z.
z shall be close to as small as possible, but we need a compromise between
the tighness of the bound and the computation speed and proof complexity.
Looking just at the log2 of the numerator and denominator, this is a tight bound
except when the numerator is a power of 2 and the denomintor is not.
E.g. this return 4/5 < 2^1 instead of 4/5< 2^0.
If q==0, we return -1000, because as binary integer this has just 10 bits but
2^-1000 should be smaller than almost any number in practice.
If numbers are much smaller, computations might be inefficient.