1
$\begingroup$

Hello I have tried to prove this result as I know it is true, it is obvious but I dont know how. Ive thought of dividing the interval into two pieces and checking which side the number is on, and then repeating this process indefinately. But i have had trouble convincing myself that it converges to the number itself. Also I have tried to show uniqueness by saying assume there are two expansions which are equal, then some of the corresponding digits are different and getting a contradiction but again have been unable. Can anyone help? Thanks.

Note: The expansion we are considering are non terminating, so for example 1/2= 0.01111111111111.... and not 0.1

  • 1
    What do you mean by unique? Remember that $1$ equals $0.99...$ in decimal. – 2017-02-01
  • 0
    @user251257: He specified non-terminating. The *only* non-terminating binary expansion of $1$ is $0.\overline 1$. Indeed, whenever there are two binary expansions for a number, one of then is terminating. However there is one real number that has no non-terminating binary expansion: $0$. – 2017-02-01

1 Answers 1

2

For existence: Given $0<\alpha<1$ and $n\in \Bbb N_0$, let $c_n=\lceil 2^n\alpha-1\rceil$. Clearly, the $c_n$ are integers with $0\le c_n<2^n$ (in particular, $c_0=0$) and $\lim_{n\to\infty}\frac{c_n}{2^n}=\alpha$. Now for $n\in \Bbb N$, let $a_n=c_n-2c_{n-1}$. Show that $a_n\in\{0,1\}$ and $\frac{c_n}{2^n}=\sum_{k=1}^na_k2^{-k}$. Conclude that $0.a_1a_2a_3\ldots$ is a binary expansion of $\alpha$. Can you see why it is in fact a non-terminating expansion?

For uniqueness: Consider two distinct expansions $\alpha=0.a_1a_2a_3\ldots$ and $\beta=0.b_1b_2b_3\ldots$ for some numbers $\alpha,\beta$. As the expansions are distinct, there are indices $k$ where $a_k\ne b_k$. Let $n$ be the smallest such index. Then $a_i=b_i$ for $1\le in$ with $b_m=1$. Can you show from this that $\beta\ge \alpha+2^{-m}$?