# Irrational Numbers Containing Other Irrational Numbers

Does $\sqrt{2}$ contain all the digits of $\pi$ in order? Does it contain all the digits of $\pi$ in order an infinite number of times? Does $\pi$ contain all the digits of $\sqrt{2}$ in order?

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What do you mean by 'all the digits'? –  Daniel Littlewood Jan 8 '13 at 19:20
3.1415926535897932384626433832795... somewhere in Root 2 digits. –  user55601 Jan 8 '13 at 19:23
If you mean do they after a while coincide, then no. If as a subsequence, then almost certainly, though I doubt there is a known proof. –  André Nicolas Jan 8 '13 at 19:25
So the complete digits of pi in order are not in Root 2. –  user55601 Jan 8 '13 at 19:30

It is not known if $\pi$ is a normal number in base $10$. If it is, then the answer to your question is ‘yes’, in the sense that infinitely many subsequences of the decimal expansion of $\pi$ will be the decimal expansion of $\sqrt{2}$. What I just said remains valid if we swap the roles of $\pi$ and $\sqrt{2}$.

The funny thing is that almost every real number is normal in base $10$ w.r.t. the Lebesgue measure, but aside from constructions that have been designed on purpose to create normal numbers, almost no examples exist.

On the other hand, the decimal expansion of $\pi$ cannot be exactly the decimal expansion of $\sqrt{2}$ after some point. Otherwise, we would have $$\pi = q + r \sqrt{2}$$ for some $q,r \in \mathbb{Q}$ (I shall leave this as an easy exercise for you), thus implying that $\pi$ is algebraic over $\mathbb{Q}$, which is a contradiction.

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Much less than "$\pi$ is normal" is sufficient for some subsequence of the decimal digits of $\pi$ to be the decimal expansion of $\sqrt{2}.$ You don't even need the digits of $\pi$ to be a disjunctive sequence. In fact, if we only know that the decimal digits of $\pi$ contain infinitely many occurrences of each of the $10$ decimal digits (the validity of this for $\pi$ is not known, BTW), then for each real number the decimal expansion of that real number will appear as a subsequence of the decimal digits of $\pi.$ –  Dave L. Renfro Jan 8 '13 at 19:56
@DaveL.Renfro: Is it known if the decimal digits of $\sqrt{2}$ contain infinitely many occurrences of each of the 10 decimal digits? –  P.. Jan 8 '13 at 20:08
@Pambos: As far as I know, no irrational real number having any reasonable type of explicit form (algebraic, closed form in the sense of Timothy Chow, etc.) is known to have infinitely many occurrences of more than one digit. (Note that infinitely many occurrences of fewer than two digits forces a number to be rational.) –  Dave L. Renfro Jan 8 '13 at 20:15
@Dave: Yes, you are right. Of course, if $\pi$ is normal, then the question automatically has an affirmative answer. If the distribution of a particular digit is sparse but there are still infinitely many of them, then we are fine. –  Haskell Curry Jan 8 '13 at 20:21
Something that occurred to me driving home last night (right after I made the last comment) is that we even have the following: If we knew that the decimal digits of $\pi$ contain infinitely many occurrences of each of the $10$ decimal digits, then for each real number (between $0$ and $1,$ say) the decimal expansion of that real number will appear as a subsequence of the decimal digits of $\pi$ for continuum many subsequences (continuum $= 2^{{\aleph}_{0}}$). To show this, first show that $\pi$ would contain continuum many subsequences each of which contains each of the $10$ decimal digits. –  Dave L. Renfro Jan 9 '13 at 15:04
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Does $\sqrt{2}$ contain all the digits of $\pi$ in order?No, because it would mean that $\sqrt{2} = p + q\cdot\pi$ for some $p,q\in\mathbb{Q}$, so $\pi$ would be an algebraic number. But we know that $\pi$ is transcendental.

Does it contain all the digits of $\pi$ in order an infinite number of times?No, because it does not contain them even once (see above). Another reason is that $\pi$ is irrational and hence contains infinite number of digits and they are aperiodic. So, there is no way how this infinite sequence could appear in another sequence multiple times starting from different positions.

Does $\pi$ contain all the digits of $\sqrt{2}$ in order?No, because it would mean that $\pi = p + q\cdot\sqrt{2}$ for some $p,q\in\mathbb{Q}$, so $\pi$ would be an algebraic number. But we know that $\pi$ is transcendental.

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