Here&#x27;s a (slighty) less hand-wavey proof.<p>Note that for a base 10 normal irrational by the definition of &quot;convergence&quot; in this case means || (number of digits of value a) &#x2F; (number of digits of value b) ||_\infty &lt; ε (sup norm over all a, b from 0 to 9, a ≠ b) for some ε &gt; 0 such that for some fixed n (the number of digits in the decimal expansion, halted), we have that whenever N &gt; n this inequality holds.<p>In the case where there are a fixed number m of leading &quot;a&quot;s (whatever digit a is), we have ||(m + dig_a)&#x2F;(dig_b)||_\infty = ||m&#x2F;dig_b + dig_a&#x2F;dig_b||_\infty -- we only need this to be less than SOME fixed ε (not necessarily the same ε as above) for every N &gt; n for some n we choose.<p>By assumption ||dig_a &#x2F; dig_b||_\infty &lt; ε whenever n &gt; N. Choose n&#x27; &gt; n such that dig_b satisfies m&#x2F;dig_b &lt; ε<p>(Note: this is possible as dig_b increases as n&#x27; grows: since m is constant and if dig_b were to be constant for n &gt; N&#x27; then ||dig_a&#x2F;dig_b||_\infty would be larger than ε for some n).<p>Then we have ||m&#x2F;dig_b||_\infty &lt; ε and ||dig_a&#x2F;dig_b||_\infty &lt; ε so ||m&#x2F;dig_b + dig_a&#x2F;dig_b||_\infty &lt; 2ε.<p>Thus it converges -- satisfying the same definition as above chooseing n = n&#x27; and ε = 2ε.