
In the main result, we check that the decimal expansion is well-defined and that it behaves as ex-
pected. We also give two well-known properties of decimal representations.
Theorem. Let x ∈ R
+
0
have decimal expansion D(x) =
∞
∑
n=0
d
n
10
−n
. Then:
(a) D(x) is a decimal: each d
n
∈ {0, 1, 2, . . . , 9} whenever n ≥ 1.
(b) D(x) converges to x.
(c) The sequence (d
n
) is eventually periodic if and only if x ∈ Q
+
0
.
(d) x equals a unique decimal series, except when D(x) = d
0
.d
1
· · · d
m
terminates (d
m
= 0). In such
a case there is a second decimal representation:
x = D(x) = d
0
.d
1
· · · d
m
= d
0
.d
1
· · · d
m−1
ˆ
d
m
99999 · · ·
where
ˆ
d
m
= d
m
− 1. Otherwise said, we subtract 1 from the final non-zero term and insert an
infinite string of 9’s.
Examples. 1. Part (c) explains why so many people enjoy the challenge of memorizing the digits
of π: since π is irrational, the pattern never repeats.
2. Also referencing part (c), we explicitly evaluate a period-three decimal using geometric series:
3.1279279279279 · · · =
31
10
+
279
10000
∞
∑
n=0
1000
−n
=
31
10
+
279
10000
·
1
1 −
1
1000
=
31
10
+
279
9990
=
1736
555
3. Here are two examples of part (d):
1 = 0.99999 · · · 27.164 = 27.1639999 · · ·
Exercises 1. Compute the decimal expansion of
32
13
.
2. Prove all parts of the Theorem. Here are some hints:
(a) Let E
n
= x −
n
∑
k=0
d
k
10
−k
. Prove by induction that R
n
= 10
n
E
n
and conclude lim E
n
= 0.
(c) A decimal is eventually periodic with period r if
d
0
.d
1
· · · d
m
d
m+1
· · · d
m+r
d
m+1
· · · d
m+r
· · · =
m
∑
k=0
d
k
10
−k
+
r
∑
j=1
d
m+j
10
−m−j
!
∞
∑
l=0
10
−rl
Convince yourself that this is rational. For the converse, is x =
p
q
is rational number
observe that there are only finitely many possible values for the remainders R
n
=
a
q
.
(d) If d
0
.d
1
d
2
· · · = c
0
.c
1
c
2
· · · , let m be minimal such that c
m
< d
m
. . .