Irrational Numbers Containing Other Irrational Numbers

Question

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?

Answer 1

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.