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Completely stuck on this exercise! Hints and a starting point would be greatly appreciated. Nor do I see why is $x \in [ 0,1] \setminus D$ do not have $2$ binary expansions.

Trajan
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    Hint: Prove the contra-positive. Assume a number does not have those two binary expansions and prove it is not a dyadic rational. Also, $x\in [0,1)$ clearly from the definition of D, because $\frac{k}{2^n} < 1$. – Matthew Levy Oct 12 '14 at 12:36
  • Sorry i will probably need more than that as i am struggling to see the general direction of the proof – Trajan Oct 12 '14 at 12:41
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    Actually $0$ which seems to be included in your definition of dyadic rationals does not have two expansions, but other than that exception, the result is correct. – paw88789 Oct 12 '14 at 12:47
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    Oh, I think I may have misinterpreted the problem.. – Matthew Levy Oct 12 '14 at 12:49
  • What book is this from? – AB_IM Nov 20 '24 at 18:54

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Note: $\frac{1}{2^i}+\frac{1}{2^{i+1}}+\cdots=\frac1{2^{i-1}}$.

Also dyadic rationals have a terminating binary expansion.

So you can trade in the last $1$ in the terminating representation for a $0$ in that place followed by a tail of all $1$s.

E.g., (binary representations): $.00101=.00100111111111...$

paw88789
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  • Thanks i understand this but writing this as a formal proof is harder than the describing the solution – Trajan Oct 12 '14 at 13:07
  • Yes, it probably is, but it is also good to strengthen your mathematical writing skills! :-) – paw88789 Oct 12 '14 at 13:20