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I suspect the answer to the title question is "yes," but can it be proved? One would expect digits match in one tenth of cases.

David G. Stork
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Bernardo Recamán Santos
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    Neither number is known to be normal. – Mason May 26 '20 at 18:56
  • A reminder that irrational is not the same as normal. The number $0.10110111011110111110111111\dots$ is irrational, but clearly the digit $2$ never occurs... much less a tenth of the time. – JMoravitz May 26 '20 at 19:02
  • I would approach this problem by considering the question in bases other than $10$. (What's so special about $10$, after all?) What about base $2$? – David G. Stork May 26 '20 at 19:18
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    Base 2 is too special because not sharing a base-2 digit forces the digits to be 0 and 1; therefore if two real numbers fail to have coinciding base-2 digits infinitely often, then their sum must be a rational number whose denominator is a power of 2. – Greg Martin May 26 '20 at 19:26
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    @Integrand, not quite equivalent, because of the potential for borrowing. – Barry Cipra May 26 '20 at 19:30

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I'm sure your expectation is correct in reality; but as commenters have mentioned, this is the type of problem that is currently hopeless given what we know about proving such statements.

Just to point out how far away we seem to be from being able to address this problem, let me make the following provocative and surely false assertion:

After finitely many digits, the decimal expansion of $\pi$ consists only of $3$s and $8$s and the decimal expansion of $\sqrt2$ consists only of $1$s and $5$s.

That statement is ridiculous—but we can't even disprove that, as far as I'm aware! And the conjecture in the OP is far stronger than simply disproving the above provocative statement.

Greg Martin
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