Does a dyadic number have exactly 2 binary representations?

Question

I came across this question in my mind trying to see the "almost isomorphism" between the outcome set of a one-dimensional random walk, and the interval $\left[0,1\right]$. I do know that in $\left[0,1\right]$,

• a nondyadic number has a unique binary representation,
• $0$ and $1$ have unique fractional binary representations which are $0.000...$ and $0.111...$
• a dyadic number in $\left(0,1\right)$ has at least one finite and one infinite binary representation.

Answer 1

A dyadic number $x$ has the form $\frac{m}{2^n}$, and its two known binary representations are $$\left(a_1,a_2,\ldots,a_n,0,0,0,\ldots\right) \text{where } a_1a_2\ldots a_n=m$$ $$\left(b_1,b_2,\ldots,b_n,1,1,1,\ldots\right) \text{where } b_1b_2\ldots b_n=m-1$$ Any binary representation $0.c_1c_2...\,=\sum_{i=1}^\infty c_i2^{-i}=y$ has the initial part $c_1c_2...c_n = m'$ for some integer $m'$ between $0$ and $2^n-1$, and the latter part $0.00\ldots0c_{n+1}c_{n+2}...\in[0,2^{-n}]$. Then consider the following 4 cases:

• If $m'>m$, then $y>x$.
• If $m'<m-1$, then $y<x$.
• If $m'=m$ and $c_{n+k}=1$ for some $k$, then $y>x$.
• If $m'=m-1$ and $c_{n+k}=0$ for some $k$, then $y<x$.

I figured out this answer a while after troubling myself with my own question.