Decimal form of irrational numbers In the decimal form of an irrational number like:
$$\pi=3.141592653589\ldots$$
Do we have all the numbers from $0$ to $9$. I verified $\pi$ and all the numbers are there. Is this true in general for irrational numbers ?
In other words, for an irrational number
$$x=\sum_{n\in \mathbb{Z}} a_n 10^n$$
Does $a_n$ takes all the numbers between $0$ and $9$ ?

This gives a nice opportunity to use cardinality arguments to show that the answer is negative.
Pick two digits $n,k$ such that $\{n,k\}\neq\{0,9\}$. There is a bijection between the numbers in $[0,1]$ whose decimal form includes only $n$ and $k$, and the set of infinite binary sequences.
Therefore the set $\{x\in[0,1]\mid x\text{ has only }n,k\text{ as decimal digits}\}$ is uncountable. So it must include at least one irrational number, and in fact almost the entire set is made of irrational numbers.
The same can be done with three, four, five, six, seven, eight or nine digits. The argument is just the same. Alternatively one can use diagonalization to show that the set is uncountable, it's the same method as you would use otherwise.
