EGYPTIAN FRACTIONS WITH SMALL NUMERATORS R.ANITHA Department of Mathematics, Saradha Gangadharan College, Puducherry Email id : anithamath09@gmail.com Dr. A.M.S. RAMASAMY Department of Mathematics, Vel Tech Dr. RR & Dr. SR Technical University, Chennai Email id : ramasamyams@veltechuniv.edu.in ABSTRACT. An Egyptian fraction is the sum of distinct unit fractions where all the unit fractions are taken to be different. The sum of an expression of this type is a positive rational number. A classical result in number theory is that every positive rational number can be expressed by an Egyptian Fraction. In this paper, we deal with Egyptian fractions with small numerators. That is, in a positive rational number we fix the numerator as 2, 3 or and then we express, and as the sum of distinct unit fractions. Keywords. Egyptian fraction, Unit fraction. 1. Introduction The Egyptians of 3000 BC had an interesting way of representing fractions. Their trade required that they could multiply, divide and deal with fractions. However, the Egyptian number system was not well suited for performing arithmetic calculations. Therefore, by necessity, they devised methods for multiplication and division which involved addition only. In a general system of fractions, any integer may appear in the numerator. However, in the Egyptian system of fractions, with the exception of the fractions and, the only number which may appear in the numerator for computational purpose was 1. That is, they had a notation as,, and so on. Unit fractions A unit fraction is a rational number written as a fraction where the numerator is 1 and the denominator is a positive integer. A unit fraction is therefore the reciprocal of a positive integer n. Some examples of unit fractions are,,,, etc.
Partition of integers Some integers may be partitioned into a sum of distinct integers whose reciprocals total 1. For example we have the following: From the relation 11 = 2 + 3 + 6, we get the relation + 1. The relation 2 = 2 + + 6 + 12 yields the relation + 1. The positive divisors of 95 are 3, 5, 7, 9, 15, 21, 27, 35, 63, 105 and 135 whose reciprocals sum to 1. i.e., = 1. 2. Egyptian fractions with numerators equal to 2 Suppose that the numerator is 2. If the denominator is an even number we have =. We can express as a sum of two unit fractions. We have the expression eg. Beeckmans [1]). Let us consider the case when the denominator is odd. (see for Theorem 1 If n is odd, then denominators. can be expressed as a sum of two unit fractions with different Proof Denote the denominator by 2r +1. By applying the Greedy Algorithm, we see that largest unit fraction.. However, we have is the If we add to both sides we obtain an expression for as a sum of two unit fractions. To complete the proof when the denominator is 2, we need to show that the denominators are distinct. i.e., we need to check that 1 211.
Since 21 3 the right hand side is at least 3 times the left hand side. So the denominators are distinct as required. Example Expression of as a sum of 3 unit fractions Let us consider the unit fraction expansion of with three unit fractions in different ways. We have the following results: (a) (b) (c) (d) (e) (f) (g) (h) (i) = = = = + + = = = = + + = = = = + + = = = = + + = = = = + + = = = = + + = = = = + + = = = = + + = = = = + + 3. Egyptian fractions with numerators equal to 3 Theorem 2 If n is not divisible by 3, then can be expressed as a sum of at most three unit fractions with distinct denominators. Furthermore, there are fractions of this type that cannot be expressed as a sum of two unit fractions with distinct denominators. Proof
There are two cases depending upon whether the denominator has the form 31 for some 1 or 32 for some 0. Suppose first that the denominator has the form 31 for some 0. In this case it follows that 3 32 1 1 1 132 and we obtain an Egyptian fraction expansion of (r+1)(3r+2) are distinct. This follows in view of the fact that 32 2. if and only if the denominators Suppose now that the denominator has the form 31 for some 1. In this case we have. It becomes necessary to consider two cases depending upon whether the expression (r+1) (3r+1) is even or odd. The product is even if is odd and odd if is even. First we prove that 2 131 1 1 for all 1. The latter is equivalent to the inequality 131 21 and the latter is true since 1 31 2. Suppose now that 2 1 is odd. Since 1 we must have 1. Hence we obtain the expression 2 131 2 262 1 62 and by the preceding inequality this is less than. Therefore we have the identity 3 31 1 62 1 1
where the two summands on the right side are distinct. Therefore if r is odd there is an Egyptian fraction expansion denominators. which is expressible as a sum of two unit fractions with different Suppose now that r is even so that 2 for some 2. Then we have and hence our original fraction are distinct. Considering the expression of will be a sum of three unit fractions if all the denominators as a sum of two unit fractions, we know that the denominators 6 1 and 6 112 81 are distinct. We need to show that neither of these is equal to r + 1. However, the previously established inequality implies that Consequently the unit fraction expansion of fractions as required.. is greater than either of the other two summands in the. This implies that the latter has been expressed as a sum of three distinct unit Fractions that cannot be expressed as a sum of two unit fractions To conclude the discussion of the cases where the denominator is equal to 3, we have to show that some fractions of the form cannot be written as a sum of two unit fractions with distinct denominators. Suppose that 1 so that the fraction under consideration is. If we can write 3 7 1 1 where, then either or vice versa. Without loss of generality, let us assume that. Then we have...,. Because a is an integer, the preceding inequalities imply that 3. The proof that does not have an Egyptian fraction expansion with unit fractions thus reduces to the following computations:
3 7 1 3 2 21 and 3 7 1 5 28. Since neither term on the right hand side is a unit fraction, the assertion regarding the Egyptian fraction expansion of follows. Example Expression of as a sum of 3 unit fractions Let us consider the unit fraction expansion of with three unit fractions in different ways. We have the following results: (a) = = = = + + (b) = = = = + + (c) = = = = + + (d) = = = = + + (e) = = = = + + (f) = = = = + + (g) = = = = + +. Egyptian fractions with numerators equal to (Vaughan [5]) If the denominator is divisible by then is a unit fraction and if it is has the form 2, then 2 2 21 and consequently it has an Egyptian fraction expansion as a sum of two distinct unit fractions.
Theorem 3 If n is odd, then can be expressed as a sum of at most four unit fractions with distinct denominators. Furthermore, there are fractions of this type that cannot be expressed as a sum of two unit fractions with distinct denominators. Proof There are two cases depending upon whether the denominator has the form 1 or 3. Since we are working with fractions in the unit interval we must have 1. If the denominator has the form 1 then we have. This yields an Egyptian fraction expansion of 13 are distinct. This follows because 33. Continuing as in the previous cases we can write. if and only if the denominators 1 and There are two cases depending upon whether 3 divides the denominator or not. Suppose 3 divides the denominator. Then the above equation will define an Egyptian fraction expansion as a sum of two unit fractions if and only if the right hand side is not equal to. However, since 15 we have 1 1 3 51 3 11 and thus the desired inequality is obtained. Now the same inequality also holds irrespective of whether 3 divides 11 or not. In the case where 3 does not divide this number, the fraction 3 11 has an Egyptian fraction expansion with three distinct terms. Now the fraction itself is less than and therefore we see that 1 1 1 3 11
is a sum of with three other unit fractions such that (i) the three unit fractions are distinct and (ii) each is strictly less than the term. It follows that the right hand side yields an Egyptian fraction expansion of with four distinct terms. Fractions that cannot be expressed as a sum of two units fractions To conclude the discussion of this case, we need to show that some fractions of the form cannot be written as a sum of two unit fractions with distinct denominators. Suppose that 1 so that the fraction under consideration is. Suppose we have 5 1 1 where. These equations and the condition 0 imply that 2 5 1 and since a is an integer, the only possibility is a = 2. The proof that does not have an Egyptian fraction expansion with two terms thus reduces to the following computation:. Since the term on the right hand side is not a unit fraction, the assertion regarding the expansion of follows. terms. Similarly we check that does not have an Egyptian fraction expansion with two distinct On expanding into into three Egyptian fractions A notable conjecture of P. Erd s and E. Straus states that the equation 1 1 1 is solvable in positive integers x, y, z for n {, 5, 6, }. According to Guy [3], the problem has been verified by N. Franceschine for 10. Also, Guy [3] has indicated that A. Schinzel has relaxed the condition that,, be positive.
It is known that can be expressed as the sum of Egyptian fractions where the algebraic sign of at least one of the Egyptian fractions is negative. Specifically, one can introduce a single formula that allows a proper fraction of the form to be expressed as, where x, y, z are positive integers and 3. Without loss of generality, we can assume that n is a prime number. Otherwise, if 2 for 2, choose 1/2 and, any positive integer. Then is equivalent to the Egyptian fraction 1/2. Furthermore, if, where p is an odd prime, then. and a three term Egyptian fraction decomposition for can be obtained by expanding in terms of Egyptian fractions and then multiplying. Theorem Let n be an odd prime. Then Proof. 1 1 1 1 1 1 1 1
Examples 1. n = 109. In this case we have 2. n = 1001. Now we have 109 1 1 1 109 2 109 2 109 109. 2 109 2 References : 1001 1 711 13 1 77 1 1 1 12 2 12 2 13 12. 2 12 [1] Beeckmans, L., The splitting algorithm for Egyptian fraction, J. Number Th., 3 (1993), 173-185. [2] Golomb, S., An algebraic algorithm for the representation problems of the Ahmes Papyrus, American Mathematical Monthly, 69 (1962), 785-786. [3] Guy, R. K., Unsolved Problems in Number Theory, 3 rd Edition, Springer-Verlag, Berlin, New York, 200.. [] Stewart, B. W., Sums of distinct divisors, Amer. J. Math., 76 (195), 779-785. [5] Vaughan, R. C., On a problem of Erdos, Straus and Schinzel, Mathemtica, 17 (1970), 193-198. 2