Answer:
Step-by-step explanation:
How did you determine that? One way is to carry out the division. However, that's laborious, especially with larger numbers. A better method is to look at the denominator of the simplified fraction. From here on, we will only be working with simplified fractions of the form
a
b
, where
gcd
(
a
,
b
)
=
1
and
b
>
1
. A well known fact is that if a prime
p
|
b
where
p
∉
{
2
,
5
}
, then the fraction is repeating. Think about why this is.
Here's a quick proof. If some fraction
a
b
has a terminating decimal representation, then there exists a positive integer
k
such that
10
k
⋅
a
b
=
n
, where
n
is a positive integer. Let
p
be a prime factor of
b
, and rewrite this as
10
k
a
=
n
b
. Then
p
|
10
k
a
, but since
gcd
(
a
,
b
)
=
1
,
p
∤
a
, so
p
|
10
k
(ie:
p
=
2
or
5
).
We will now proceed to the main part of this lesson: repeating decimals. We will define the period of a decimal
0.
d
1
d
2
d
3
d
4
…
to be the least positive integer
n
such that for some integer
k
,
d
i
=
d
i
+
n
for all
i
≥
k
. For example, the period of
1
90
=
0.0
¯¯¯
1
is
1
because for all
i
≥
2
,
d
i
=
d
i
+
1
=
1
. Similarly,
1
7
=
0.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
142857
has a period of
6
because
d
i
=
d
i
+
6
for all
i
≥
1
. An intuitive way of thinking about period is that it is the number of digits in the repetend when written optimally. ("Written optimally" informally means that you should write
0.0
¯¯¯
1
instead of
0.0
¯¯¯¯¯¯
11
.)
