Ex3-1

Chapter 3 Question 1

Question: There is an analogue for bases different from 10 of the usual decimal expansion of a number. If $$b$$ is an integer larger than 1 and $$0\leq x\leq 1$$, show that there exist integral coefficients $$c_k, 0\leq c_k<b$$, such that $$x=\sum_{k=1}^\infty c_kb^{-k}$$. Show that this expansion is unique unless $$x=cb^{-k}, c=1,\ldots,b^k-1$$. in which case there are two expansions.

Proof: We choose $$c_1$$so that $$c_1/b$$ is as large as possible but isn't larger than $$x$$, hence $$c_1/b\leq x\leq (c_1+1)/b$$ (The second inequality can be equality, because $$c_1$$cannot be $$b$$). If we let $$x_1=bx-c_1, $$then we have $$0\leq x_1\leq 1$$. Given $$x_n\in[0,1]$$, we can always choose $$c_{n+1}$$ as large as possible such that $$c_{n+1}/b\leq x_n\leq (c_{n+1}+1)/b$$, then define $$x_{n+1}=bx_n-c_{n+1}$$ we will obtain $$0\leq x_{n+1}\leq 1$$. Choosing by this way, we obtain a sequence $$\{c_1,c_2,\ldots\}$$. One can also verify inductively that

$$x_n=b^nx-b^{n-1}c_1-b^{n-2}c_2-\dots -bc_{n-1}-c_n.$$

Indeed, this is true for $$n=1$$. Suppose this is true for $$n-1$$, then

$$x_n=bx_{n-1}-c_n=b(b^{n-1}x-b^{n-2}c_1-\dots-c_{n-1})-c_n$$

and the relation follows at once. Now we prove $$x=\sum_{k=1}^\infty c_kb^{-k}$$. For each positive integer $$n$$, we have

$$\begin{align}\left|x-\sum_{k=1}^n \dfrac{c_k}{b^k}\right|&=\frac1{b^n}|b^nx-b^{n-1}c_1-\dots-c_n|\\ &=\frac{x_n}{b^n}\\ &\leq \dfrac1{b^n}.\end{align}$$

hence the series converges to $$x$$.

Now we have proved every real number in $$[0,1]$$ has a $$b$$-base expansion, we will observe in which case the expansion is unique. Suppose $$x$$ has two different expansions: $$x=\sum_{k=1}^\infty \dfrac{a_k}{b^k}=\sum_{k=1}^\infty \dfrac{c_k}{b^k}.$$

Then, there exists a smallest index $$n$$for which $$a_n\neq c_n$$, we could assume without loss of generality that $$a_n>c_n$$. After simplification we have

$$a_n-c_n=\sum_{k=1}^\infty \dfrac{c_{n+k}-a_{n+k}}{b^k}.$$

Note that $$c_{n+k}-a_{n+k}\leq b-1$$ by assumption, hence

$$0<a_n-c_n\leq \sum_{k=1}^\infty \dfrac{b-1}{b^k}=1$$

where the last summation is calculated by telescoping. Since $$a_n,c_n$$ are integers, this forces $$a_n-c_n=1$$, and since the equality holds, we must have $$c_{n+k}-a_{n+k}=b-1$$ for $$k\geq 1$$. This means $$a_{n+k}=0, c_{n+k}=b-1$$ for all $$k$$. We now conclude $$x=\sum_{k=1}^\infty a_kb^{-k}=\sum_{k=1}^n a_kb^{-k}=cb^{-n}$$ where $$0\leq c=b^{n-1}a_1+b^{n-2}a_2+\dots +a_n\leq (b-1)(b^{n-1}+b^{n-2}+\dots+1)=b^n-1$$. Using the same logic, if one assumes $$x=\sum a_kb^{-k}=\sum c_kb^{-k}=\sum d_kb^{-k}$$, then two of the sequences from $$\{a_k\}, \{c_k\}, \{d_k\}$$ must be entirely equal. Thus we have proved $$x$$ cannot have more than 2 different expansions. One direction has been proven, that is: If $$x$$ has two different expansion, then $$x=cb^{-k}$$ for some $$0\leq c\leq b^k-1$$. Now we prove the converse.

Suppose $$x=cb^{-n}$$ for some integer $$c\in[0,b^n-1]$$ which is not a multiple of $$b$$ (meaning we are choosing $$n$$ as small as possible). Then we can choose positive integers $$a_1,a_2,\dots,a_n$$ less than $$b$$such that $$c=b^{n-1}a_1+b^{n-2}a_2+\dots+a_n$$with successive division followed by taking remainders. Thus $$x=\sum_{k=1}^n a_kb^{-k}$$is an expansion as described in the problem. For the second expansion, since $$c$$is not a multiple of $$b$$, we find that $$a_n$$is the nonzero positive remainder when $$c$$ is divided by $$b$$. Take another sequence $$\{c_k\}$$ defined as

$$c_k=\begin{cases}a_k&\text{ if }kn,\end{cases}$$

Then $$x=\sum_{k=1}^\infty c_kb^{-k}$$ is another expansion.