Prove a > 0 is a terminating decimal if and only if it has more than one decimal representation

Question

I want to prove that if we choose a positive real $a$ that has a terminating decimal representation, it has more than one decimal representation, and if a number has more than one decimal representation, one of them is terminating, using the theorem of Completeness of Numbers.

NOTE: There is a similar question asked previously (Which real numbers have two representations?), but the accepted answer there doesn’t provide any proof to the problem.

Answer 1

The first statement is easy to prove.

Suppose $$a=\sum_{k=1}^{n}a_k\cdot 10^{-k},$$ then also $$\sum_{k=1}^{n-1}a_k\cdot 10^{-k} + (a_{n}-1)\cdot 10^{-n} + \sum_{k=n+1}^{+\infty}9 \cdot 10^{-k}$$ converges to $a$, since the last term is a geometric series converging to $10^{-n}$.

---

Let us now suppose that $$a = \sum_{k=1}^{+\infty}a_k\cdot 10^{-k}=\sum_{k=1}^{+\infty}b_k \cdot 10^{-k},$$ with $a_j\neq b_j$ for some $j\in \Bbb Z^+$, and $a_k=b_k$ for $1 \leq k < j$.

Suppose, e.g., that $a_j> b_j$, and consider the series \begin{eqnarray} S &=& \sum_{k=1}^{+\infty}a_k\cdot 10^{-k}-\sum_{k=1}^{+\infty}b_k \cdot 10^{-k}\\ &=&\underbrace{\sum_{k=j}^{+\infty}a_k\cdot 10^{-k}}_A-\underbrace{\sum_{k=j}^{+\infty}b_k\cdot 10^{-k}}_{B}, \end{eqnarray} where we took advantage of absolute convergences of both terms, in order to use the associative property. With our hypothesis, $S\to 0$.

Let $h$ the position of the first non-$9$ digit following $b_j$. Then $$A>\sum_{k=j}^h a_k\cdot 10^{-k}=A^-$$ and $$B < \sum_{k=j}^{h-1}b_k\cdot 10^{-k} + (b_h+1)\cdot 10^{-h}=B^+.$$

Thus $$S > A^--B^+>0$$ cannot converge to $0$, as required. We have a contradicion, and $b_j$ therefore must be followed by an infinite sequence of $9$'s (of course this also implies $b_j=a_j-1$ and $a_k = 0$ for $k>j$).

Answer 2

I assume you're familiar with the proof for the converse and are asking about a proof of the conditional, i.e if a number has more than one decimal representation it must be terminating. We can prove this by contradiction.

Suppose there exists a non-terminating decimal with two decimal representations, call them $\alpha$ and $\beta$. Let the $k$th decimal place be the first place they differ, and say from the $k$th place onward, the decimal representations $\alpha$ and $\beta$ are $\;\overline{a_0a_1a_2a_3...}\;$ and $\;\overline{b_0b_1b_2b_3...}\;$ respectively; WLOG assume $a_0 \gt b_0$.

Introduce two terminating decimals, $\alpha_0$ and $\beta_0$, that agree with $\alpha$ and $\beta$ up to the $k$th decimal place and then terminate. Since $a_0 \gt b_0$, clearly $\alpha_0 \gt \beta_0$. Now consider $\beta_0 + 10^{-k}$ (essentially adding 1 to the last decimal place of $\beta_0$.) Then $\alpha_0 \ge \beta_0 + 10^{-k}$. But this leads to

$$\alpha \gt \alpha_0 \ge \beta_0 + 10^{-k} \ge \beta$$

which contradicts the assumption that $\alpha$ and $\beta$ represent the same non-terminating decimal.

[The first inequality in the above statement comes from the fact that at least one of the $a_{j\gt0}$ must be non-zero, else $\alpha$ would be terminating; the last inequality from the fact that even if we maximize $\beta$ by setting each $b_{j\gt0}$ to 9, $\beta$ would still only be equal to $\beta_0 + 10^{-k}$.]