Cardinality of the set of multiple-representation decimals

The set of real numbers in the open interval $(0,1)$ which have more than one decimal expansion is

(A) Empty

(B)non-empty but finite

(C)Countably infinite

(D)uncountable

I know that, $$\frac{1}{10}=.10000...=.0999999...$$ $$\frac{1}{10}=.010000...=.099999...$$ $$\frac{1}{100}=.0010000...=.0099999...$$

$$...$$

$$\frac{1}{10^n}$$ has also the two binary representation.

When I was doing the above question, this question came in to mind.

I have question that The set of real numbers in the open interval $(0,1)$ which have more than one binary expansion is

(A) Empty

(B)non-empty but finite

(C)Countably infinite

(D)uncountable

I think the elements of the form $\frac{1}{2^n}$ has two binary expansion in $(0,1)$ . but I couldn't give the rigorous proof. Please help me.

• By the way ,Maneesh, I heard this question from somebody else yesterday. I get the feeling it has appeared in a competitive exam recently. – астон вілла олоф мэллбэрг Dec 11 '17 at 4:46
• yes. TIFR GS 2018 – Unknown x Dec 11 '17 at 5:23
• Exam held yesterday. I feel to mark (C). But due to lack of confidence, i didn't mark ;'( – Unknown x Dec 11 '17 at 5:24
• That is unfortunate, now you cannot do much about it. However, if there are any other questions from that exam, you may post them as well. But of course , +1. – астон вілла олоф мэллбэрг Dec 11 '17 at 5:25
• math.stackexchange.com/questions/2560250/… – Unknown x Dec 11 '17 at 5:29

In any base $b$, using $h$ to denote the highest possible digit, all the numbers in $(0,1)$ that have more than one representation have the form $$0.xxx\dots p000\ldots=0.xxx\dots qhhh\dots$$ where $p\ne0$ and $q=p-1$. In other words, they have a terminating expansion.

Thus the set of all such numbers is countably infinite by the following bijection between the terminating expansions and the positive integers: $$(0.x_1x_2\dots x_k)_b\to(x_k\dots x_2x_1)_b$$ (C) is correct for both questions.

A number has more than one decimal expansion if and only if it has terminating decimal expansion.

To see this, if it does terminate, then if it goes like $\overline{a_1\cdots a_n . b_1\cdots b_n}$ then an alternating decimal expansion is also $\overline{a_1\cdots a_n.b_1\cdots b_{n-1} 999...}$

On the other hand, any non-terminating decimal expansion is determined uniquely, for example by repeated long division, which does not terminate and therefore does not give us room to introduce the chain of $9$s, as done in the terminating case.

Hence, the set of such real numbers actually is countable.

A similar argument can be given for the binary case, with the same conclusion.