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, 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_1so 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_1cannot 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 nfor 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 bsuch that c=b^{n-1}a_1+b^{n-2}a_2+\dots+a_nwith 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 cis not a multiple of b, we find that a_nis 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 }k<n,\\ a_n-1&\text{ if }k=n,\\ b-1&\text{ if }k>n,\end{cases}
Then x=\sum_{k=1}^\infty c_kb^{-k} is another expansion.