Unit fractions are the reciprocals of the positive integers where the numerator is always one. They are also called Egyptian fractions, and were used exclusively by the Egyptians to represent all forms of fractions. The two fractions that they used that did not have a unit fraction was 2/3 and 3/4. The only other fractions that they seemed to have a strong interest in were those of the form 2/n where n was any positive odd number. The Rhind papyrus contains a list of unit fractions representing a series of 2/n for odd n's from 5 to 501. It is unclear as to why they found these 2/n fractions so important. These and other 2/n fractions may be derived from 2/n = 1/[n(n+1)/2] + 1/[n+1)/2].
In the year 1202, Leonardo Fibonacci proved that any ordinary fraction could be expressed as the sum of a series of unit fractions in an infinite number of ways. He used the then named greedy method for deriving basic unit fraction expansions. He described the greedy method in his Liber Abaci as simply subtracting the largest unit fraction less than the given non unit fraction and repeating the process until only unit fractions remainded. It was later shown that the greedy method, when applied to any fraction m/n, results in a series of no more than "m" unit fractions. An example will best illustrate the process.
Reduce the fraction 13/17 to a sum of unit fractions. Dividing the fraction yields .7647. Of the unit fractions 1/2, 1/3, 1/4, 1/5, etc., 1/2 is the largest that is smaller than 13/17 so we compute 13/17 - 1/2 = 9/34 making 13/17 = 1/2 + 9/34. Repeating with 9/34, we have 9/34 - 1/4 = 2/136 = 1/68 making 13/17 = 1/2 + 1/4 + 1/68. Alternatively, divide the numerator into the denominator and use the next highest integer as the new denominator. 17/13 = 1.307 making 2 the denominator of the fraction to be subtracted from 13/17.
Reduce the fraction 23/37 to a sum of unit fractions. 23/37 - 1/2 = 9/74 - 1/16 = 70/1184 - 1/17 = 6/20,128 - 1/3355 = 1/33,764,720 making 23/37 = 1/2 + 1/16 + 1/17 + 1/3355 + 1/33,764,720.
There are many methods or algorithms that derive the unit fractions for any fraction m/n. Much more information regarding unit fractions can be found at
Unit Fractions Search at
http://mathpages.com/cgi-local/AT-mathpakbsearch.cgi
Egyptian Fractions at
http://www1.ics.uci.edu/~eppstein/numth/egypt/
Unit Fraction from Math World at
http://mathworld.wolfram.com/UnitFraction.html
Algorithms for Egyptian Fractions at
http://www1.ics.uci.edu/~eppstein/numth ... force.html
Creating Unit Fractions at
http://www.mathpages.com/home/kmath150.html
The unit fractions derived by means of the method shown, or any other method, can be further broken down into other unit fractions by means of the identity 1/a = 1/(a+1) + 1/a(a+1), also known to Fibonacci. For example, 1/2 = 1/(2+1) + 1/2(2+1) = 1/3 + 1/6. Further still, 1/3 = 1/(3+1) + 1/3(3+1) = 1/4 + 1/12 and 1/6 = 1/(6+1) + 1/6(6+1) = 1/7 + 1/42 yielding 1/2 = 1/4 + 1/7 + 1/12 + 1/42. While the number of unit fractions derivable for any given fraction is therefore infinite, there is apparantly no known procedure for deriving a series with the least number of unit fractions or the smallest largest denominator.
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There are other ways of breaking down a unit fraction into other unit fractions. Positive integer solutions to 1/x + 1/y = 1/z can be derived from x = (z + k) and y = (z + z^2/k) where k = any of the factors/divisors of z^2. Therefore, the total number of solutions is N = (a + 1)(b + 1)(c + 1)......where a, b, c, etc., are the prime exponents of the prime factorization of z^2 = (p1^a)(p2^b)(p3^c)........etc.
Example:
1/x + 1/y = 1/z = 1/11
z^2 = 121
Prime factorization of 121 = 11^2
Total factors/divisors = (2+1) = 3
Factors of 11^1 = 1, 11, 121
x = 11 + 1 = 12 and y = 11 + 121/1 = 132
x.= 11 + 11 = 22 and y = 11 + 121/11 = 11
x = 11 + 121 = 132 and y = 11 + 121/121 = 12
Therefore, 1/11 + 1/132 = 1/11
1/x + 1/y = 1/15
z^2 = 225
Prime factorization of 225 = 3^2(5^2)
Total factors/divisors f(N) = (2+1)(2+1) = 9
Factors of 15^2 = 1, 3, 5, 9, 15, 25, 45, 75, 225
x = 15 + 1 = 16 and y = 15 + 225 = 240
x = 15 + 3 = 18 and y = 15 + 75 = 90
x = 15 + 5 = 20 and y = 15 + 45 = 60
x = 15 + 9 = 24 and y = 15 + 25 = 40
x = 15 + 15 = 30 and y = 15 + 15 = 30
x = 15 + 25 = 40 and y = 15 + 9 = 24
x = 15 + 45 = 60 and y = 15 + 5 = 20
x = 15 + 75 = 90 and y = 15 + 3 = 18
x = 15 + 225 = 240 and y = 15 + 1 = 16
After examining the two examples it becomes obvious that there are really only (f(N) - 1)/2 distinct solutions as we do not count the one where both denominators are the same and the solutions on either side of this one are numerically identical.
Alternatively, solutions can be derived from x = x and y = zx/(x-z). For y to be an integer, (x-z) must be a positive divisor of z^2. The number of solutions is the same as derived above. Therefore, letting x = 16, (x - z) becomes 16 - 15 = 1 which divides 225 making y = 15(16)/(16 - 15) = 240. For x = 18, (x - z) becomes 18 - 15 = 3 which divides 225 making y = 15(18)/(18 - 15) = 90. As you can readily see, the same list of solutions will result.
Alternatively, if a value is selected for x, we then have 1/y = 1/z - 1/x = (x - z)/zx resulting in y = zx/(x - z) the solution then becoming (x, y) = (x, zx/(x - z). We must therefore find the positive integers that make (x - z) a positive divisor of zx, where z is a given positive integer. This requires that (x - z) be = or > 1, i.e., x = or > (z + 1). From y = zx/(x - z) we can derive y = [(x - z)z + z^2]/(x - z) = z + z^2/(x - z).
For y to be an integer, (x - z) must be a divisor of the constant z^2, noting that (x - z) must be positive.
If a positive integer k divides n^2, then setting (x - z) = k, we get the solution (x, y) = [(z + k), (z + z^2/k)]. Again, the total number of solutions is N = (a + 1)(b + 1)(c + 1)......where a, b, c, etc., are the prime exponents of the prime factorization of z^2.
Example:
1/x + 1/y = 1/10.
10^2 = 100
Prime factorization = 2^2(5^2)
Total factors/divisors = (2+1)(2+1) = 9
Factors of 100 = 1, 2, 4, 5, 10, 20, 25, 50, 100
For k = 1 x = 11 and y = 110
1/11 + 1/110 = 11/1210 + 110/1210 = 121/1210 = 1/10.
For k = 5, x = 15 and y = 30
1/15 + 1/30 = 15/450 + 30/450 = 45/450 = 1/10.
And so it goes.
Alternative approach -
Since x, y, and z are positive integers, x > z and y > z. Letting x = z + u and y = z + v, with u and v > 0, the equation 1/x + 1/y = 1/z reduces to z^2 = uv. Therefore, for every z, we need only decompose z^2 into the product of two integers u and v. For 1/x + 1/y = 1/15, z^2 = 225 = uv leading to (u,v) = (1,225), (3,75), (5,45), (9,25), and (15,15) which lead to the exact same solutions.
Alternative approach -
For 1/x + 1/y = 1/a, we can write ax + ay = xy or xy - ax - ay + a^2 = a^1 or (x - a)(y - a) = a^2.
The last expression has 2µ - 1 solutions, where µ is the number of divisors of the integer a^2 (including 1 and a^2 as divisors). To define all the solutions, we can write 2µ systems of expressions of the form x - a = k, y - a = a^2/k and x - a = -k, y - a = -a^2/k (where k is a divisor of a^2), and discard the x - a = -a, y - a = -a, which leads to x = y = 0.
For a = 14, we get a^2 = 196, and the divisors "k" become 1, 2, 4, 7, 14, 28, 49, 98, 196.
For k = 1, we have x = 15 and y = 210
Then, 1/15 + 1/210 = 15/3150 + 210/3150 = 225/3150 = 1/14.
For k = 2, we have x = 16 and y = 112
Then, 1/16 + 1/112 = 16/1792 + 112/1792 = 128/1792 = 1/14.
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Other unit fractions of frequent interest are those that sum to unity. For example, determine n different unit fractions whose sum is 1, i.e., 1/a + 1/b + 1/c + 1/d + 1/e + 1/f + 1/g = 1. Lets start with a simpler problem, 1/x + 1/y + 1/z = 1.
If you pick up a typical ruler, it will become obvious that the subdivisions break down in a predictable and/or orderly manner. With pencil and paper in hand, you will note that
...........1 = 1/2 + 1/2 followed by
........1/2 = 1/4 + 1/4
..............= 1/3 + 1/2(1/3) = 1/3 + 1/6
..............= 1/5 + 1.5/5 = 1/5 + 3/10
..............= 1/6 + 2/6 = 3/6
..............= 1/7 + 2.5/7 = 1/7 + 5/14
..............= etc.
This simple exercise immediately leads to the solution of 1/x + 1/y + 1/z = 1 as 1/2 + 1/3 + 1/6 = 1.
After examining these fractions for a while, it becomes obvious that each of them can also be further broken down as
1/3 = 1/4 + 1/3(1/4) = 1/4 + 1/12 = 1/5 + 1/3(2/5) = 1/5 + 2/15
1/6 = 1/7 + 1/6(1/7) = 1/6 + 1/42 = etc.
Being interested only in unit fractions though, and after a little more exploration, it appears that each fraction can be further broken down in accordance with the general expression
...............................1/n = 1/(n + 1) + 1/n(n + 1)
which is identical to the one described earlier. After checking out several other examples, you will be convinced that this is the key to breaking down any expression of the form
...........................1/a + 1/b + 1/c + 1/d + 1/e + 1/f + 1/g + ............1/? = 1
into integer solutions. As for a solution for 1/a + 1/b + 1/c + 1/d + 1/e + 1/f + 1/g = 1
1/2 + 1/2 = 1 from which we can derive
1/2 + 1/3 + 1/6 = 1 from which we can derive
1/2 + 1/4 + 1/12 + 1/6 = 1 from which we derive
1/2 + 1/4 + 1/12 + 1/7 + 1/42 = 1 from which we derive
1/2 + 1/5 + 1/20 + 1/12 + 1/7 + 1/42 = 1 from which we derive
1/2 + 1/5 + 1/20 + 1/12 + 1/8 + 1/56 + 1/42 = 1.
