TITLE: Irrational Numbers Containing Other Irrational Numbers
QUESTION [6 upvotes]: 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?

REPLY [5 votes]: 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.