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.
I think answer is (C). Am I correct? How do I prove it rigorously. Please help me.
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.
Best Answer
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.