Every real number has a decimal expansion

Statement

Let $x$ be a real number with $x \ge 0$. Then $x$ has a decimal expansion: there are a natural number $a_0$ and digits $d_k \in \{0, \dots, 9\}$ for $k \ge 1$ with $x = a_0 + \sum_{k=1}^{\infty} d_k 10^{-k}$.

Remarks

One choice is $a_0 = \lfloor x \rfloor$ and $d_k = \lfloor 10^k x \rfloor - 10 \lfloor 10^{k-1} x \rfloor$, with the floor function. With this choice, the truncation $a_0.d_1 \dots d_k$ satisfies $0 \le x - a_0.d_1 \dots d_k < 10^{-k}$. The expansion is unique, except for the numbers with a finite expansion. They also have a second expansion that ends in infinitely many $9$.

Explanations

L1 human

You can find the digits of a number one at a time. At each step, you choose the largest digit that keeps the truncated decimal at or below the number. The truncations then get closer to the number by a factor of ten at each step.