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?
 A: 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.
A: 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.
