How to prove that for all digits there exists a transcendental number which contains it infinite number of times in its representation?

Question

Soon I was told a statement that for all digits a transcendental number can be found, containing the digit infinitely many times. It seems obvious, but cannot find good enough argument for it. Can anyone show it to me or at least some idea why this must be true (I think the main argument is hidden in some (un)countability comparisons)?

Answer 1

Let $x = \sum_{i=1}^{\infty} a_i 10^{-i}$ with $a_i \in \{0,...,9\}$ be a transcendental number. Let $k \in \{0,...,9\}$ be arbitrary.

Let $l \in \{0,...,9\}$ such that $l = a_i$ for $i \in I$ with $I \subseteq \mathbb{N}$ infinite.

Let $y := \sum_{i \in I} 10^{-i}$.

First case: $y$ is transcendental. If $k = 0$, take $y$, since it has infinitely many digits $0$. If $k\neq0$, then $ky$ is also transcendental and has infinitely many digits $k$.

Second case: $y$ is algebraic. Then $x - (l-k)y$ is transcendental and has infinitely many digits $k$.

Answer 2

Liouville's idea can be used here: given any finite string of digits, form a decimal expansion with those digits occurring over-and-over, separated by strings of $0$'s which grow factorially in length. Then the Liouville idea that numbers that can be very-well approximated by rationals can be invoked: if $\alpha$ is a real number such that, for any exponent $d$, for all sufficiently large integer $q$, there is integer $p$ such that $|\alpha-p/q|<1/q^d$, then $\alpha$ is transcendental. (Hilariously, this follows from the Mean Value Theorem.)

Answer 3

Others have given algebraic solutions, I'll give a simple cardinality argument for this.

Consider $\mathcal P(\mathbb N)$, the set of subsets of $\mathbb N$. For any natural $k$, there's at most $|\mathbb N^k|=\aleph_0$ subsets of size $k$. Thus, there's $\aleph_0$ finite subsets, and uncountably many infinite subsets.

We can define the function $f:\mathcal P(\mathbb N)\to\mathbb R$ given by $$f(I)=\sum_{i\in I} 100^{-i}.$$ This function is injective, so there's uncountably many numbers we can get as the output of an infinite set. There's only countably many algebraic numbers, so one of these outputs must be transcendental. This number $x$ will have infinitely many $0$s and $1$s. Furthermore, for any digit $d\ne 0$, the number $dx$ will also be transcendental and have infinitely many digits $d$.