Here's a (slighty) less hand-wavey proof. Note that for a base 10 normal irrational by the definition of "convergence" in this case means || (number of digits

Here's a (slighty) less hand-wavey proof. Note that for a base 10 normal irrational by the definition of "convergence" in this case means || (number of digits of value a) / (number of digits of value b) ||_\infty < ε (sup norm over all a, b from 0 to 9, a ≠ b) for some ε > 0 such that for some fixed n (the number of digits in the decimal expansion, halted), we have that whenever N > n this inequality holds. In the case where there are a fixed number m of leading "a"s (whatever digit a is), we have ||(m + dig_a)/(dig_b)||_\infty = ||m/dig_b + dig_a/dig_b||_\infty -- we only need this to be less than SOME fixed ε (not necessarily the same ε as above) for every N > n for some n we choose. By assumption ||dig_a / dig_b||_\infty < ε whenever n > N. Choose n' > n such that dig_b satisfies m/dig_b < ε (Note: this is possible as dig_b increases as n' grows: since m is constant and if dig_b were to be constant for n > N' then ||dig_a/dig_b||_\infty would be larger than ε for some n). Then we have ||m/dig_b||_\infty < ε and ||dig_a/dig_b||_\infty < ε so ||m/dig_b + dig_a/dig_b||_\infty < 2ε. Thus it converges -- satisfying the same definition as above chooseing n = n' and ε = 2ε.