g The ancient Egyptians expressed rational numbers as the finite sum of distinct unit fractions. Were Theorem 1. Suppose x is a positive integer and n = lcm(x, x 1,...,3, 2) + 1. Then x will have − n the Egyptians limited by this notation? In fact, they were not as every rational number can be written a length x Engel expansion. Through this project we have shown a method for choosing a denominator that will always produce as a finite sum of distinct unit fractions. Moreover, these expansions are not unique. There exist an Engel expansion whose length is equal to its numerator. We have also demonstrated that while this is not the only possible way to produce a length x expansion, n = lcm(x, x 1,...,3, 2) + 1 is several different algorithms for computing Egyptian fraction expansions, all of which produce different − conjectured to be the least denominator that does so for some fixed x. Tables 1 and 2 demonstrate representations of the same rational number. Example 2. Let x =4. Then by Theorem 1, n = lcm(4, 3, 2) + 1 = 13. In order clarify why One such algorithm, called an Engel series, produces a finite increasing sequence of integers for every this for small values of x. There are other values for n which produce a length x Engel expansion this n produces the desired expansion, 13 will be written as lcm(4, 3, 2) + 1 for the majority of this x rational number. This sequence is then used to obtain an Egyptian fraction expansion. Motivated by 4 for n. example. To find the Engel expansion, first let u1 = . The first term of the Engel series the work of M. Mays, this project aims to investigate properties of natural number denominators n lcm(4,3,2)+1 The paper includes more extensive tables for small numer- x will be 3 ators that demonstrate that n = lcm(x, x 1,...,3, 2)+1 is that produce length x Egyptian fraction expansions using Engel series for n between 0 and 1. 1 lcm(4, 3, 2) + 1 lcm(4, 3, 2) 1 Engel expansions for n − a1 = = = + . the least n that will produce an a length x Engel expansion While computing Engel expansions using Mathematica, a helpful pattern emerged: rational numbers "u # " 4 # " 4 4# x k 1 x 1 x which produce length x Egyptian fraction expansions were those whose denominator n = lcm(x, x n i=1 a1 a2 ai for . We are open to the possibility that a counter exam- n lcm(4,3,2) · ··· n − 3 1 1 x 1,...,3, 2) was divisible by every natural number between (and including) x and 2. We conjecture By the lemma, since 4 =3is an integer,a1 can be rewritten as: + ple may be found. That is, some n$ <nsuch that both n 4 ! 2 2 2 x $ · and produce x length Egyptian fraction expansions using and prove that this will always hold for length x Engel series of non-integer rational numbers between 3 1 1 n x lcm(4, 3, 2) 1 lcm(4, 3, 2) lcm(4, 3, 2) + 4 5 2 + 2 5 Engel series. None has been found yet. 0and1.Furthermore,weconjecturethatanalgorithmforfindingthesmallestn such that n produces a1 = + = +1= =4. · alengthx Engel series is to add one to the lowest common multiple of x, x 1, ...,3,and2.Aproof 4 "4# 4 4 3 1 1 1 7 3 + 3 4 + 3 4 7 − · · · that such an n works is included along with an investigation of whether or not this is the least such n. Next find u2 by subtracting one from the product of u1 and a1 to get: Table 1 In the future, we would like to determine and prove an al- 4 lcm(4, 3, 2) + 4 lcm(4, 3, 2) + 4 lcm(4, 3, 2) 1 gorithm for obtaining families of length x expansions for u2 = 1= − − x lcm(4, 3, 2) + 1 · 4 − lcm(4, 3, 2) + 1 n. That is, we are interested in defining and proving a 3 3 Engel expansions for 4 pattern that will produce each length x expansion for any = = . n lcm(4, 3, 2) + 1 13 x k 1 given x. In addition, we are interested in the ”best case” n i=1 a a a 1· 2··· i 4 1 1 1 scenarios (Engel expansions of length two). We would like The Rhind Papyrus (see Figure 1) is the best example of Egyptian 5 !2 + 2 2 + 2 2 5 The second term of the Engel series is: · · · 4 1 1 to employ methods similar to those used in this project to mathematics. It begins with expansion tables for rational numbers of 7 2 + 2 7 2 · obtain a pattern for these length 2 expansions then prove the form .ThesetablesdemonstratetheunusualwaytheEgyptians 1 lcm(4, 3, 2) + 1 lcm(4, 3, 2) 1 4 1 1 n a2 = = = + 9 3 + 3 3 that this pattern holds. · wrote out rational numbers. Instead of expressing rational numbers "u2# " 3 # " 3 3# 4 1 1 11 3 + 3 11 as proper or improper fractions, they would write them as a sum of · lcm(4,3,2) 4 1 1 1 1 1 1 2 Again, by the lemma, since =4is an integer,a can be rewritten as: 13 4 + 4 5 + 4 5 7 + 4 5 7 13 distinct unit fractions (e.g. 4 + 28 instead of 7). These expansions 3 2 · · · · · · make certain real-life problems much easier to solve. One such real- lcm(4, 3, 2) 1 lcm(4, 3, 2) lcm(4, 3, 2) + 3 Table 2 life application focuses on sharing whole items (like loaves of bread) a2 = + = +1= =5. with multiple people. For example, how can five loaves be split evenly 3 "3# 3 3 amongst eight people? The solution can be written as the proper frac- The next u term is: tion, 5.Thismeanseachpersongetsfiveeightsofaloaf(seeFigure 8 3 lcm(4, 3, 2) + 3 lcm(4, 3, 2) + 3 lcm(4, 3, 2) 1 2). This is not a very practical solution. It makes distribution of the u3 = 1= − − Figure 1: Rhind Papyrus loaves tedious (lots of slicing) and inconvenient (since no one receives lcm(4, 3, 2) + 1 · 3 − lcm(4, 3, 2) + 1 2 2 References any single larger piece of bread only small pieces). = = . lcm(4, 3, 2) + 1 13 [1] Erdos, Shallit. “New bounds on the length of finite Pierce and Engel Series.” S´eminaire de Th´eorie Luckily, this solution is not the only possible solution. Egyp- The third term in the Engel series is: des Nombres Bordeaux 3 (1991), 43-53. tian fraction expansions can be used to find a more practi- [2] Graham, Knuth, Patashink. Concrete Mathematics. Addison Wesley, Massachusetts, 1st Edition cal solution. instead of giving each person five small pieces 1 lcm(4, 3, 2) + 1 lcm(4, 3, 2) 1 (1989), Chapter 3. 5 1 1 a3 = = = + of bread, Egyptian fraction expansions show that 8 = 2 + 8. "u3# " 2 # " 2 2# [3] V. Laohakosol, T. Chaichana, J. Rattanamoong, and N.R. Kanasri. “Engel Series and Cohen-Egyptian Thus everyone will receive half a loaf first, then one eighth of lcm(4, 3, 2) 1 lcm(4, 3, 2) lcm(4, 3, 2) + 2 Fraction Expansions.” Hindawi Publishing Corporation. International Journal of Mathematics and = + = +1= =7. aloaf. ThisisnottheonlypossiblesolutionusingEgyptian 2 "2# 2 2 Mathematical Sciences. Article ID 865705 (2009), 15 pages. fractions. Another expression of the same sharing problem is [4] Mays, Michael E. “A Worst Case of the Fibonacci-Sylvester Expansion.” The Journal of Combinato- 5 1 1 1 Continue to get: rial Mathematics and Combinatorial Computing. 1 (1987), 141-148. 8 = 2 + 10 + 40. Although, this is again, an impractical solu- tion. 2 lcm(4, 3, 2) + 3 lcm(4, 3, 2) + 2 lcm(4, 3, 2) 1 [5] O’Reilly, Declan “Creating Egyptian Fractions” Mathematics in School. Vol. 21, No. 5 (Nov., 1992), u = 1= − − Page 41. 4 lcm(4, 3, 2) + 1 · 2 − lcm(4, 3, 2) + 1 [6] Renyi, A. “A New Approach to the Theory of Engel’s Series.” Annales Universitatis Scientiarum The algorithm that this poster will focus on produces an Egyptian Figure 2: Impractical Solution 1 1 = = . Budapestinensis de Rolando E¨otv¨os Nominatae. Volume 5 (1962), 25-32. fraction expansion using an Engel series. An Engel series (for rational lcm(4, 3, 2) + 1 13 numbers) is a sequence of increasing integers, a ,a ,...a [6]. The [7] Weisstein, Eric W. “Egyptian Fraction.” From MathWorld–A Wolfram Web Resource. 1 2 k http://mathworld.wolfram.com/EgyptianFraction.html sequence is infinite for irrational numbers. These integers can be used to The fourth (and hopefully last) term of the Engel series is: form an Egyptian fraction expansion by using the following summation: 1 lcm(4, 3, 2) + 1 k 1 . Any real number can be derived using an Engel series in i=1 a1 a2 ai a4 = = = lcm(4, 3, 2) + 1 = 13. auniqueway.Irrationalnumberscanberepresentedasaninfinitesum· ··· "u4# " 1 # ! [3] and rational numbers can be represented as a finite sum. An Engel Finally, series for x can be computed by defining u = x and a = 1 . From n 1 n 1 ui 1 lcm(4, 3, 2) + 1 " # u5 = 1=0 here, each subsequent ui+1 and ai+1 is obtained by first computing lcm(4, 3, 2) + 1 · 1 − Figure 3: Practical Solution 1 and the process stops. The corresponding Egyptian fraction expansion is ui+1 = ui ai 1 and ai+1 = . · − "ui+1# 4 4 1 1 1 1 I would like to thank Julia Bergner for all of her guidance.