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

{\displaystyle b}

 is an integer larger than 1 and

0
≤
x
≤
1

{\displaystyle 0\leq x\leq 1}

, show that there exist integral coefficients

c

k

,
0
≤

c

k

<
b

{\displaystyle c\_{k},0\leq c\_{k}<b}

, such that

# k
1

∞

c

k

b

−
k

{\displaystyle x=\sum \_{k=1}^{\infty }c\_{k}b^{-k}}

. Show that this expansion is unique unless

# x
c

b

−
k

,
# c
1
,
…
,

b

k

−
1

{\displaystyle x=cb^{-k},c=1,\ldots ,b^{k}-1}

. in which case there are two expansions.

Proof: We choose

c

1

{\displaystyle c\_{1}}

so that

c

1

/

b

{\displaystyle c\_{1}/b}

 is as large as possible but isn't larger than

x

{\displaystyle x}

, hence

c

1

/

b
≤
x
≤
(

c

1

1
)

/

b

{\displaystyle c\_{1}/b\leq x\leq (c\_{1}+1)/b}

 (The second inequality can be equality, because

c

1

{\displaystyle c\_{1}}

cannot be

b

{\displaystyle b}

). If we let

x

1

=
b
x
−

c

1

,

{\displaystyle x\_{1}=bx-c\_{1},}

then we have

0
≤

x

1

≤
1

{\displaystyle 0\leq x\_{1}\leq 1}

. Given

x

n

∈
[
0
,
1
]

{\displaystyle x\_{n}\in [0,1]}

, we can always choose

c

n
1

{\displaystyle c\_{n+1}}

 as large as possible such that

c

n
1

/

b
≤

x

n

≤
(

c

n
1

1
)

/

b

{\displaystyle c\_{n+1}/b\leq x\_{n}\leq (c\_{n+1}+1)/b}

, then define

x

n
1

=
b

x

n

−

c

n
1

{\displaystyle x\_{n+1}=bx\_{n}-c\_{n+1}}

 we will obtain

0
≤

x

n
1

≤
1

{\displaystyle 0\leq x\_{n+1}\leq 1}

. Choosing by this way, we obtain a sequence

{

c

1

,

c

2

,
…
}

{\displaystyle \{c\_{1},c\_{2},\ldots \}}

. One can also verify inductively that

x

n

=

b

n

x
−

b

n
−
1

c

1

−

b

n
−
2

c

2

−
⋯
−
b

c

n
−
1

−

c

n

.

{\displaystyle x\_{n}=b^{n}x-b^{n-1}c\_{1}-b^{n-2}c\_{2}-\dots -bc\_{n-1}-c\_{n}.}

Indeed, this is true for

# n
1

{\displaystyle n=1}

. Suppose this is true for

n
−
1

{\displaystyle n-1}

, then

x

n

=
b

x

n
−
1

−

c

n

=
b
(

b

n
−
1

x
−

b

n
−
2

c

1

−
⋯
−

c

n
−
1

)
−

c

n

{\displaystyle 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

# k
1

∞

c

k

b

−
k

{\displaystyle x=\sum \_{k=1}^{\infty }c\_{k}b^{-k}}

. For each positive integer

n

{\displaystyle n}

, we have

x
−

∑

# k
1

n

c

k

b

k

=

1

b

n

b

n

x
−

b

n
−
1

c

1

−
⋯
−

c

n

=

x

n

b

n

≤

1

b

n

.

hence the series converges to

x

{\displaystyle x}

.

Now we have proved every real number in

[
0
,
1
]

{\displaystyle [0,1]}

 has a

b

{\displaystyle b}

-base expansion, we will observe in which case the expansion is unique. Suppose

x

{\displaystyle x}

 has two different expansions:

# k
1

∞

a

k

b

k

=

∑

# k
1

∞

c

k

b

k

.

{\displaystyle 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

{\displaystyle n}

for which

a

n

≠

c

n

{\displaystyle a\_{n}\neq c\_{n}}

, we could assume without loss of generality that

a

n

>

c

n

{\displaystyle a\_{n}>c\_{n}}

. After simplification we have

a

n

−

c

n

=

∑

# k
1

∞

c

n
k

−

a

n
k

b

k

.

{\displaystyle 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

≤
b
−
1

{\displaystyle c\_{n+k}-a\_{n+k}\leq b-1}

 by assumption, hence

0
<

a

n

−

c

n

≤

∑

# k
1

∞

b
−
1

b

k

=
1

{\displaystyle 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

{\displaystyle a\_{n},c\_{n}}

 are integers, this forces

a

n

−

c

n

=
1

{\displaystyle a\_{n}-c\_{n}=1}

, and since the equality holds, we must have

c

n
k

−

a

n
k

=
b
−
1

{\displaystyle c\_{n+k}-a\_{n+k}=b-1}

 for

k
≥
1

{\displaystyle k\geq 1}

. This means

a

n
k

=
0
,

c

n
k

=
b
−
1

{\displaystyle a\_{n+k}=0,c\_{n+k}=b-1}

 for all

k

{\displaystyle k}

. We now conclude

# k
1

∞

a

k

b

−
k

=

∑

# k
1

n

a

k

b

−
k

=
c

b

−
n

{\displaystyle x=\sum \_{k=1}^{\infty }a\_{k}b^{-k}=\sum \_{k=1}^{n}a\_{k}b^{-k}=cb^{-n}}

 where

0
≤
# c

b

n
−
1

a

1

b

n
−
2

a

2

⋯

a

n

≤
(
b
−
1
)
(

b

n
−
1

b

n
−
2

⋯
1
# )

b

n

−
1

{\displaystyle 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
∑

a

k

b

−
k

=
∑

c

k

b

−
k

=
∑

d

k

b

−
k

{\displaystyle x=\sum a\_{k}b^{-k}=\sum c\_{k}b^{-k}=\sum d\_{k}b^{-k}}

, then two of the sequences from

{

a

k

}
,
{

c

k

}
,
{

d

k

}

{\displaystyle \{a\_{k}\},\{c\_{k}\},\{d\_{k}\}}

 must be entirely equal. Thus we have proved

x

{\displaystyle x}

 cannot have more than 2 different expansions. One direction has been proven, that is: If

x

{\displaystyle x}

 has two different expansion, then

# x
c

b

−
k

{\displaystyle x=cb^{-k}}

 for some

0
≤
c
≤

b

k

−
1

{\displaystyle 0\leq c\leq b^{k}-1}

. Now we prove the converse.

Suppose

# x
c

b

−
n

{\displaystyle x=cb^{-n}}

 for some integer

c
∈
[
0
,

b

n

−
1
]

{\displaystyle c\in [0,b^{n}-1]}

 which is not a multiple of

b

{\displaystyle b}

 (meaning we are choosing

n

{\displaystyle n}

 as small as possible). Then we can choose positive integers

a

1

,

a

2

,
…
,

a

n

{\displaystyle a\_{1},a\_{2},\dots ,a\_{n}}

 less than

b

{\displaystyle b}

such that

# c

b

n
−
1

a

1

b

n
−
2

a

2

⋯

a

n

{\displaystyle c=b^{n-1}a\_{1}+b^{n-2}a\_{2}+\dots +a\_{n}}

with successive division followed by taking remainders. Thus

# k
1

n

a

k

b

−
k

{\displaystyle x=\sum \_{k=1}^{n}a\_{k}b^{-k}}

is an expansion as described in the problem. For the second expansion, since

c

{\displaystyle c}

is not a multiple of

b

{\displaystyle b}

, we find that

a

n

{\displaystyle a\_{n}}

is the nonzero positive remainder when

c

{\displaystyle c}

 is divided by

b

{\displaystyle b}

. Take another sequence

{

c

k

}

{\displaystyle \{c\_{k}\}}

 defined as

c

k

=

{

a

k

 if

k
<
n
,

a

n

−
1

 if

# k
n
,

b
−
1

 if

k
>
n
,

{\displaystyle 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

# k
1

∞

c

k

b

−
k

{\displaystyle x=\sum \_{k=1}^{\infty }c\_{k}b^{-k}}

 is another expansion.