Repeating Decimals

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 a^p−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 a^{p - 1} – 1, but p does not divide into a^{p – 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 a^{p - 1} – 1, but p does not divide into a^{p – 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.

Pascal's Pyramids

As mentioned in a previous blog, Pascal’s Triangle is a famous triangular array of numbers that has applications for the expansion of binomials, powers of two, sequences of numbers, statistics, and properties of simplexes.  Each line of the triangle starts and ends with the number 1, and the numbers in between can be obtained by finding the sum of the two numbers above it.  The triangle can then be continued indefinitely.

A similar method can be extended in three dimensions to make an array of numbers in the shape of a pyramid with either a triangular base or a square base.  Both of these pyramids also have amazing properties.

Triangular-Based Pascal’s Pyramid

The triangular-based Pascal’s Pyramid can be formed by layers of expanding equilateral triangles.  This means the first layer has one number (a triangle with one number per side), the second layer has three numbers (a triangle with two numbers per side), the third layer has six numbers (a triangle with three numbers per side), and so on. 

Expanding a Pyramid by Adding Equilateral Triangle Layers

The numbers in this pyramid can be obtained by finding the sum of the three numbers in the layer above it.  For example, the center number 6 in the fourth layer was obtained by adding the three 2’s in the third layer.  The pyramid can also be continued indefinitely.

Steps for Finding a New Triangular Layer

Layers in a Triangular-Based Pascal’s Pyramid

The triangular-based Pascal’s Pyramid has several interesting properties.  The numbers along the edges are the same numbers as Pascal’s Triangle, and therefore all of the interesting properties of Pascal’s Triangle also apply to Pascal’s Pyramid.  In addition to this, the coefficients for the expanded form of the trinomial (x + y + z)ⁿ match up with the numbers in the triangular layers of Pascal’s Pyramid.  For example, (x + y + z)² = x² + y² + z² + 2xy + 2xz + 2yz (1, 1, 1, 2, 2, 2), and (x + y + z)³ = x³ + y³ + z³ + 3x²y + 3x²z + 3xy² + 3xz² + 3y²z + 3yz² + 6xyz (1, 1, 1, 3, 3, 3, 3, 3, 3, 6).  (Recall that the coefficients for the expanded form of the binomial (x + y)ⁿ match up with the numbers in Pascal’s Triangle.)  Finally, the numbers of each triangular layer of Pascal’s Pyramid add up to a power of 3.  For example, the numbers in the third layer (1, 2, 1, 2, 2, 1) add up to 9 which is 3², and the numbers in the fourth layer (1, 3, 3, 1, 3, 6, 3, 3, 3, 1) add up to 27 which is 3³. (Recall that the numbers of each row of Pascal’s Triangle adds up to a power of 2.)

Square-Based Pascal’s Pyramid

The square-based Pascal’s Pyramid can be formed by layers of expanding squares instead of triangles.  This means the first layer has one number (a square with one number per side), the second layer has four numbers (a square with two numbers per side), the third layer has nine numbers (a square with three numbers per side), and so on. 

Expanding a Pyramid by Adding Square Layers

The numbers in this pyramid can be obtained by finding the sum of the four numbers in the layer above it.  For example, one of the 9’s in the fourth layer was obtained by adding the 1, 2, 2, and 4 in the third layer.  This pyramid can also be continued indefinitely.

Steps for Finding a New Square Layer

Layers in a Square-Based Pascal’s Pyramid

The square-based Pascal’s Pyramid also has several interesting properties.  Like the triangular-based Pascal’s Triangle, the numbers along the edges of the square-based Pascal’s Pyramid are the same numbers as Pascal’s Triangle, and therefore all of the interesting properties of Pascal’s Triangle also apply.  In addition, just as the numbers of each row of Pascal’s Triangle adds up to a power of 2, and just as the numbers of each triangular layer of the triangular-based Pascal’s Pyramid add up to a power of 3, the numbers of each square layer of Pascal’s Pyramid add up to a power of 4.  For example, the numbers in the third layer (1, 2, 1, 2, 4, 2, 1, 2, 1) add up to 16 which is 4², and the numbers in the fourth layer (1, 3, 3, 1, 3, 9, 9, 3, 3, 9, 9, 3, 1, 3, 3, 1) add up to 64 which is 4³. Finally, each square layer of Pascal’s Pyramid represents a multiplication table, in which any number in the square is the product of the row header and column header.  For example, any of the 9’s in the fourth layer is can be found by multiplying the row header of 3 by the column header of 3.  Because the numbers in each square layer are symmetrical, all numbers in the diagonals are square numbers.

Variations of Pascal’s Pyramid

There are a few other variations of Pascal’s Pyramid.  In the triangular-based pyramid, the next layer can be obtained by adding a triangle of numbers that is upside-down from the usual triangle of numbers.  The result is a new hexagonal layer that still holds the property that the numbers along the edges are the same as the numbers in Pascal’s Triangle, and also the property that the sum of the numbers in each layer is a power of 3.

Usual Way of Obtaining a New Layer

Alternate Way of Obtaining a New Layer

A hexagonal-based pyramid can also be made by numbers in a honeycomb pattern.  Each new number can be obtained by finding the sum of the three numbers in the layer above it.  Once again, this pyramid still holds the property that the numbers along the edges are the same as the numbers in Pascal’s Triangle, but also has the additional property that the sum of the numbers in each layer is a power of 6.

Steps for Finding a New Hexagonal Layer

Layers in a Hexagonal-Based Pascal’s Pyramid

Conclusion

Extending Pascal’s Triangle into three dimensions reveals even more unique properties.  The edges of each pyramid have the same numbers as Pascal’s Triangle, and the sum of the numbers in each layer can be represented as a power between two integers.  Each layer in a triangular-based Pascal’s Pyramid has application in expanding trinomials, and each layer in a square-based Pascal’s Pyramid is a multiplication table.  All of these properties are truly amazing!