In our
monetary system, different coins represent different fractions of a dollar, and
most of us have memorized a few basic fraction to decimal conversions. For example, one quarter is worth ¼ of a
dollar or $0.25, so we know that ¼ = 0.25.
Similarly, one dime is worth 1/10 of a dollar or
$0.10, one nickel is worth 1/20 of a dollar or $0.05, and
one penny is worth 1/100 of a dollar or $0.01. Therefore, we know that ¼ = 0.25, 1/10
= 0.1, 1/20 = 0.05, and 1/100 =
0.01.
One of the
reasons these particular values were picked for coins was because the decimals
for these fractions terminate, despite the fact that most fractions do
not. For example, a quarter can be
represented with long division as the following:
0.
|
2
|
5
|
|
4
|
1.
|
0
|
0
|
–
|
8
|
↓
|
|
2
|
0
|
||
–
|
2
|
0
|
|
0
|
The
remainder reaches zero and so the decimal terminates, which makes for easier
calculations for adding and subtracting amounts in cents. But imagine a coin that was worth 1/3
of a dollar. Using long division, we
would get:
0.
|
3
|
3
|
3
|
…
|
|
3
|
1.
|
0
|
0
|
0
|
…
|
–
|
9
|
↓
|
|||
1
|
0
|
||||
–
|
9
|
↓
|
|||
1
|
0
|
||||
–
|
9
|
||||
1
|
The
remainder never reaches zero, and so the threes carry on infinitely, which is
why we don’t have a coin worth 1/3 of a dollar. In mathematical terms, the three is repeated
and the decimal can be expressed as 1/3 = 0.333…
The
following table shows the decimal notation of the first twenty integers as
denominators with a numerator of one:
Fraction
|
Decimal
|
Repeated Part
|
Period
|
1/1
|
1
|
(terminates)
|
0
|
1/2
|
0.5
|
(terminates)
|
0
|
1/3
|
0.333…
|
3
|
1
|
1/4
|
0.25
|
(terminates)
|
0
|
1/5
|
0.2
|
(terminates)
|
0
|
1/6
|
0.1666…
|
6
|
1
|
1/7
|
0.142857…
|
142857
|
6
|
1/8
|
0.125
|
(terminates)
|
0
|
1/9
|
0.111…
|
1
|
1
|
1/10
|
0.1
|
(terminates)
|
0
|
1/11
|
0.090909…
|
09
|
2
|
1/12
|
0.08333…
|
3
|
1
|
1/13
|
0.076923…
|
076923
|
6
|
1/14
|
0.0714285…
|
714285
|
6
|
1/15
|
0.0666…
|
6
|
1
|
1/16
|
0.0625
|
(terminates)
|
0
|
1/17
|
0.0588235294117647…
|
0588235294117647
|
16
|
1/18
|
0.0555…
|
5
|
1
|
1/19
|
0.052631578947368421…
|
052631578947368421
|
18
|
1/20
|
0.05
|
(terminates)
|
0
|
There does
not seem to be any pattern to the length of the repeated part of the decimal,
and as we will see later that is because there is a relationship between it and
prime numbers, which has no known pattern.
However, there are certain rules that can be applied.
First of
all, all denominators comprised of only
factors of 2 or 5 terminate. For
example, 2, 4 (22), 5, 8 (23), 10 (2∙5), 16 (24),
20 (225), and so on all terminate.
This is due to our numbering system being in base 10, which is 2 times
5.
Secondly, if the denominator is a prime number (other
than 2 or 5), then the period is a factor of one less than the denominator. For example, 7 is a prime number, and the
period for 1/7 is 6, which is a factor of 7 – 1 = 6. Also, 11 is a prime number, and the period
for 1/11 is 2, which is a factor of 11 – 1 = 10. The reasoning behind this is somewhat
involved. You may recall from your high
school math class that the sum of an infinite geometric series is t1/1
– r (when |r| < 1), where t1 is the first term and r is the
ratio. This means that any repeating
decimal (such as 0.090909…) can be expressed as a fraction by first rewriting
it as an infinite geometric series (0.090909… = 0.09 + 0.0009 + 0.000009 + …)
and then using the infinite geometric sum formula (t1/1 – r
= 0.09/1 – 0.01 = 0.09/0.99 = 9/99
= 1/11). In
general, the numerator is the repeated part of the decimal and the denominator must
contain a string of nines as long as the period of the decimal (before
simplifying). Since 7 has a period of 6,
7 must divide evenly into a string of 6 nines (999999 / 7 = 142857). Since 11 has a period of 2, 11 must divide
evenly into a string of 2 nines (99 / 11 = 9).
Mathematically, the string of nines n digits long can be represented as
10n – 1. Now, Fermat’s Little
Theorem states that “if p is prime and a is coprime to p,
then p divides ap−1 – 1.” So if a = 10 and p is a prime number (other
than 2 or 5 so that a and p are coprime) and n = p – 1, then p must divide into
10n – 1, which matches the string of nines needed in the denominator
for a repeating decimal, meaning that all prime numbers must have a period that
repeats itself every p – 1 digits. That
is why the prime number 7 has a period of 6.
So why doesn’t the prime number 11 have a period of 10 instead of
2? Technically, it does have a period of
10, but only if you include five sets of two (1/11 =
0.0909090909…). Because of the repetition
of numbers in the period of 10 itself, the final period is short-circuited to a
factor of 10, in this case 2.
Third, if the denominator d has a period of one
less than itself, then d is prime. Such
prime numbers are called long primes. For
example, 7 must be a long prime because it has a period of 7 – 1 = 6, 17 must
be a long prime because it has a period of 17 – 1 = 16, and 19 must be a long prime
because it has a period of 19 – 1 = 18.
The reason behind this is due to Lehmer’s Theorem, which states that “if
there exists an a where p divides into ap - 1 – 1, but p does not
divide into ap – 1/q – 1 for all primes q dividing p – 1, then p is
prime.” Because of our argument in the
second point dealing with infinite geometric sums, since 7 has a period of 6, 7
cannot evenly divide into 9, 99, 999, 9999, or 99999, but finally must evenly divide
into 999999. In other words, if p = 7
and a = 10, p divides into ap - 1 – 1, but p does not divide into ap
– 1/q – 1 for all primes q dividing p – 1, so by Lehmer’s Theorem p = 7
must be prime. With the help of a
computer, other long primes include 7, 17, 19, 23, 29, 47, 59, 61, 97, 109,
113, 131, 149, 167, 179, 181, 193, 223, 229, 233, 257, 263, 269, 313, 337, 367,
379, 383, 389, 419, 433, 461, 487, 491, 499, 503, 509, 541, 571, 577, 593, 619,
647, 659, 701, 709, 727, 743, 811, 821, 823, 857, 863, 887, 937, 941, 953, 971,
977, 983, and so on.
It’s pretty
amazing that long primes even exist, especially with larger numbers. If you think about it, using long division on
a long prime must result in every single remainder between one and the period
before being repeated (otherwise the period would be too short to be a long
prime). For example, using long division
on the long prime 7 into 1 looks like the following:
0.
|
1
|
4
|
2
|
8
|
5
|
7
|
…
|
|
7
|
1.
|
0
|
0
|
0
|
0
|
0
|
0
|
…
|
–
|
7
|
↓
|
||||||
3
|
0
|
|||||||
–
|
2
|
8
|
↓
|
|||||
2
|
0
|
|||||||
–
|
1
|
4
|
↓
|
|||||
6
|
0
|
|||||||
–
|
5
|
6
|
↓
|
|||||
4
|
0
|
|||||||
–
|
3
|
5
|
↓
|
|||||
5
|
0
|
|||||||
–
|
4
|
9
|
||||||
1
|
The
remainders are 3, 2, 6, 4, 5, and 1, which are all the numbers between 1 and 6
exactly once, before cycling through again.
That means that dividing a larger long prime number such as 983 would
result in every single number between 1 and 983 appearing as a remainder
exactly once before the cycle is repeated again!
The repeated
part of the quotient of long primes, called cyclic numbers, also have
fascinating properties. First of all, multiplying a cyclic number by any integer
less than its long prime number results in a number with the same digits in the
same order. For example, the cyclic
number of the long prime 7 is 142857.
Multiplying 142857 by any integer less than 7 results in a number with
the same digits in the same order:
142857 x 3 =
428571
142857 x 2 =
285714
142857 x 6 =
857142
142857 x 4 =
571428
142857 x 5 =
714285
142857 x 1 =
142857
Note that to
make the diagonal number pattern, the multipliers are in the same order as the
remainders of dividing 1/7 by long division. Changing the numerator shifts the numbers of
the remainders and the digits in the quotient.
So if you divide 3/7 by long division, your first
remainder is 2, then 6, then 4, then 5, then 1, then 3, which is the same order
of the remainders in 1/7, therefore resulting in the same
order of digits in the quotient in 1/7.
0.
|
4
|
2
|
8
|
5
|
7
|
1
|
…
|
|
7
|
3.
|
0
|
0
|
0
|
0
|
0
|
0
|
…
|
–
|
2.
|
8
|
↓
|
|||||
2
|
0
|
|||||||
–
|
1
|
4
|
↓
|
|||||
6
|
0
|
|||||||
–
|
5
|
6
|
↓
|
|||||
4
|
0
|
|||||||
–
|
3
|
5
|
↓
|
|||||
5
|
0
|
|||||||
–
|
4
|
9
|
↓
|
|||||
1
|
0
|
|||||||
–
|
7
|
|||||||
3
|
Since 1/7
x 3 = 3/7, 0.142857… x 3 = 0.428571… which means 142857 x
3 = 428571.
A second
fascinating property is that the first
half of the digits of a cyclic number added to the second half of the digits
results in a string of nines. For
example, for the cyclic number 142857, the first half of this number is 142 and
the second half is 857. Adding these two
numbers results in:
142
|
+ 857
|
999
|
For another
example, the cyclic number of the long prime 17 is 0588235294117647 and
05882352
|
+ 94117647
|
99999999
|
The reason
for this also quite involved. From our
rules of repeating decimals, 1/7 = 142857/999999,
and since 999999 = 106 – 1 = (103 – 1)(103 +
1) = (999)(1001), 1/7 = 142857/999999
can be rearranged to 1001/7 = 142857/999. As mentioned above, since 7 has a period of
6, 7 cannot evenly divide into 9, 99, 999, 9999, or 99999, but finally must evenly
divide into 999999. Applying the
difference of squares, 999999 = 106 – 1 = (103 – 1)(103
+ 1), so 7 must evenly divide into (103 – 1)(103 + 1). However, we already said that 7 cannot evenly
divide into 999 or 103 – 1, so 7 must divide into 103 + 1,
which means that 1001 divided 7 must be an integer. Therefore, since 1001/7 =
142857/999, 142857/999 must be an
integer. Now, 142857/999
= 142000/999 + 857/999 or
142.142142142… + 0.857857857… Since this
sum must be an integer, each digit after the decimal must be 9, because 0.999…
= 1, meaning the first half of the digits of this cyclic number added to the second
half of the digits of this cyclic number must result in a string of nines. (A quick proof that 0.999… = 1 is that 0.999…
= (0.333…)3 = (1/3)3 = 1.) A similar
argument can be made for any cyclic number, since the period will always be even,
which means the difference of squares can always be applied.
Therefore,
using long division to turn fractions into decimals leads to finding
fascinating properties of terminating and repeating decimals. The period of repeating decimals do not seem
to follow any sort of pattern, but certain rules can be applied, especially
concerning prime numbers. Investigating a
little deeper leads to the discovery of long primes and cyclic numbers, which
also have some amazing properties.
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