Egyptian Fractions This Is My Solution to Problem 2583 in Journal of Recreational Mathematics, Vol 31, #1, Set by Anonymous
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Egyptian Fractions This is my solution to problem 2583 in Journal of Recreational Mathematics, vol 31, #1, set by Anonymous. (a) 4. 50 5×10 (5×2)×(10×9) (9+1)×(89+1) 9×89+89+9+1 = = = = 89 89 89×2×9 2×9×89 2×9×89 1 1 1 1 1 1 1 1 = + + + = + + + 2 18 2×89 18×89 2 18 178 1602 (b) (1) 89 if the fractions can be equal, since 50/89 = 50 x 1/89 and 89 is a prime. (b) (2) 712 if the fractions must be different: 50 1 299 13×23 13×92 (1+12)×(3+89) − = = = = 89 6×89 6×89 6×89 6×4×89 4×6×89 3+3×12+89+12×89 1 1 1 3 = = + + + 4×6×89 8×89 24 2 2×89 1 1 1 1 2 = + + + + 8×89 24 2 2×89 2×89 50 1 1 1 1 1 1 1 1 1 1 1 1 = + + + + + = + + + + + 89 2 24 89 2×89 6×89 8×89 2 24 89 178 534 712 These were found by trial noting the following constraints. To represent a fraction as the sum of two unit fractions, it must be of the general form: p 1 1 (a+b)/d = + = q a⋅c⋅d b⋅c⋅d a⋅b⋅c where (a,b) = 1, and (a+b, c) = 1 and d | (a+b). A solution will exist if the denominator q can be represented as the product a . b . c such that p | (a+b). This can be tested directly by examining all possible factorings of q into three factors. To represent it as the sum of three unit fractions, then the largest must be greater than one third of the fraction, and this limits the values to be examined. For four fractions, then just keep trying to remove one unit fraction to see whether the remainder can be represented as the sum of three; or alternatively, look for solutions of the form: 1 1 1 1 (n+1)⋅( p+1) + + + = m m⋅n m⋅p m⋅n⋅p m⋅n⋅p where n | p+1 and m | n+1. To find the smallest denominator, trial and error was used noting that there was an upper limit from the answer to part (a), and that 89 is a prime, so at least one denominator had to be a multiple of 89. Egyptian Fractions, Farey Sequences and Continued Fractions There is a way of finding an expansion of a proper fraction, p/q, into a sum of unit fractions that guarantees a maximum value for the largest denominator, and that is by the use of Farey sequences. Farey sequences are named after a geologist John Farey (1766 – 1826), although he was not the first to examine them. But he does seem to have made them popular, and so his name became associated with them. His letter containing a conjecture about them to the Philosophical Magazine in 1816 was read by Cauchy who provided a proof. However, Charles Haros had already given similar results in 1802. Neither Cauchy nor Farey knew of that work at the time. A Farey Sequence of order n, is the sorted list of non-negative proper fractions (i.e. in their lowest terms) less than or equal to 1, whose denominators are not greater than n. For example, the sequence of order 6 is: 0 1 1 1 1 2 1 3 2 3 4 5 1 1 6 5 4 3 5 2 5 3 4 5 6 1 These sequences obviously have a finite length, and have a number of peculiar properties. One is that if p/q lies between a/b and c/d, then p/q = (a+c)/(b+d). This can be useful to derive a sequence of order (n+1) from that of order n. Another is that if a/b < c/d are two adjacent terms in the sequence, then (bc – ad) = 1. This can be employed to split a fraction into a sum of unit fractions. Obviously from this fact, each term in a Farey sequence is the sum of all the differences between adjacent terms, which are all unit fractions, with a maximum denominator less than the square of the order of the sequence. But we can do better, and remove the fraction with the largest denominator, leaving the remainder as the preceding term, which necessarily has a smaller denominator and so belongs to a shorter Farey sequence. So by switching to this new sequence, we can remove another term, and reduce it still further. The total number of terms can get quite large, though. The fraction n/(n+1) will reduce n 1 (n−1) to = + which reduces each numerator and denominator by only 1. So the n+1 n(n+1) n total number of unit fractions would be n, and certainly not optimal in any sense of the word. For the example above, this method gives us the following set of predecessors for each term in turn: 50 41 32 23 14 5 1 89 73 57 41 25 9 2 which provides the breakdown into unit fractions: 50 1 1 1 1 1 1 1 = + + + + + + 89 6497 4161 2337 1025 225 18 2 You do not have to generate all Farey sequences of order q to determine the predecessor of the term for p/q. By judicious use of the expansion of p/q as a continued fraction, the final term can be adjusted to form those for the fractions either side of p/q. Repeating this for each reduction in turn will provide a reasonably quick way of getting an expression as a sum of unit fractions. Andy Pepperdine 2020-06-15.