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Exploring Unit Fractions Page 36 Exploring Unit Fractions Alan Christison [email protected] INTRODUCTION In their article “The Beauty of Cyclic Numbers”, Duncan Samson and Peter Breetzke (2014) explored, among other things, unit fractions and their decimal expansions6. They showed that for a unit fraction to terminate, its denominator must be expressible in the form 2푛. 5푚 where 푛 and 푚 are non-negative integers, both not simultaneously equal to zero. In this article I take the above observation as a starting point and explore it further. FURTHER OBSERVATIONS 1 For any unit fraction of the form where 푛 and 푚 are non-negative integers, both not 2푛.5푚 simultaneously equal to zero, the number of digits in the terminating decimal will be equal to the larger 1 1 of 푛 or 푚. By way of example, has 4 terminating decimals (0,0625), has 5 terminating decimals 24 55 1 (0,00032), and has 6 terminating decimals (0,000125). The reason for this can readily be explained by 26.53 1 converting the denominators into a power of 10. Consider for example which has six terminating 26.53 decimals. In order to convert the denominator into a power of 10 we need to multiply both numerator and denominator by 53. Thus: 1 1 53 53 53 125 = × = = = = 0,000125 26. 53 26. 53 53 26. 56 106 1000000 1 More generally, consider a unit fraction of the form where 푛 > 푚. Multiplying both the numerator 2푛.5푚 and denominator by 5푛−푚 gives: 1 1 5푛−푚 5푛−푚 5푛−푚 = × = = 2푛. 5푚 2푛. 5푚 5푛−푚 2푛. 5푛 10푛 Since the denominator is 10푛 it follows that the fraction contains 푛 terminating decimals. Similar arguments hold for 푚 > 푛 (where there will be 푚 terminating decimals) as well as for 푚 = 푛. 1 Let us now consider fractions of the form . In line with the above, fractions of this form can be 5푚 2푚 converted to the form from which it is then possible to write down the terminating decimal expansion 10푚 1 directly since the last of the 푚 decimals will simply be the 푚th power of 2. Thus will have 4 terminating 54 4 1 1 decimals, the final digits of which will be 16 (since 2 = 16), i.e. = 0,0016. Similarly, will have 8 54 58 8 1 terminating decimals, the final digits of which will be 256 (since 2 = 256), i.e. = 0,00000256. 58 6 http://www.amesa.org.za/LTM%2016_2.pdf Learning and Teaching Mathematics, No. 25, 2018, pp. 36-39 Page 37 1 Note that a similar argument holds for unit fractions of the form . In this case the terminating decimal 2푚 1 expansion will have 푚 digits, and the last of these 푚 decimals will be the 푚th power of 5. Thus will 24 have 4 terminating decimals, the final digits of which will be 625 (since 54 = 625). We can thus immediately 1 write = 0,0625. 24 NON-TERMINATING UNIT FRACTIONS If the prime factorisation of the denominator of a unit fraction contains factors other than 2 and/or 5, then the decimal expansion of the fraction will be non-terminating. More specifically, the decimal expansion will 1 be cyclic – i.e. it will have a repeating structure. Consider the unit fraction which can be expressed in 540 1 1 1 decimal form as 0,00185̅̅̅̅̅. Since 540 = 22. 5. 33 we can write in the form × , i.e. the product 540 22.5 27 of a terminating fraction and a non-terminating fraction. Note that the larger exponent in the numerator 1 1 of is 2, while = 0, 037̅̅̅̅̅. We thus notice that the larger exponent in the denominator of the 22.5 27 1 terminating fraction equates to the number of non-repeating digits in , i.e. two (00), while the number 540 1 1 of repeating digits in the non-terminating fraction equates to the number of repeating digits in , i.e. 27 540 three (185). 1 As a further example consider the unit fraction which can be expressed in decimal form 2625 1 1 1 as 0,000380952̅̅̅̅̅̅̅̅̅̅. Since 2625 = 53. 3.7 we can write in the form × , i.e. the product of a 2625 53 21 1 terminating fraction and a non-terminating fraction. Note that = 0,008 with three terminating digits as 53 1 expected, and = 0, 047619̅̅̅̅̅̅̅̅̅̅ with six repeating digits. As anticipated, the exponent in the denominator of 21 1 the terminating fraction equates to the number of non-repeating digits in , i.e. three (000), while the 2625 1 number of repeating digits in the non-terminating fraction equates to the number of repeating digits 21 1 in , i.e. six (380952). 2625 This suggests we can make the following general statement: Statement 1: 1 1 If is a terminating decimal fraction and is a fully/purely repeating decimal fraction, 푎 푏 1 1 then will have a non-periodic decimal length equivalent to that of and a repeating 푎푏 푎 1 decimal length equivalent to that of . 푏 Learning and Teaching Mathematics, No. 25, 2018, pp. 36-39 Page 38 1 It is important to note that this only holds true if the repeating decimal of commences directly after the decimal 푏 1 1 comma, i.e. must be a fully/purely repeating decimal fraction. Consider for example = 0,00260416̅ 푏 384 1 1 1 1 1 which can be written in the form × where = 0,0625 and = 0,0416̅. Clearly is not 24 24 24 24 24 purely repeating, but has a combination of three non-periodic digits (041) and one repeating digit (6). We 1 know that has four terminating decimals, and adding these four to the three non-periodic digits 24 1 1 in gives a total of seven, the number of non-periodic decimals in (i.e. 0026041). This suggests we 24 384 can make the following more general statement: Statement 2: 1 1 If is a terminating decimal fraction of length 푚, and is a non-terminating decimal fraction 푎 푏 1 comprising 푝 non-periodic digits and 푛 repeating digits, then will have 푚 + 푝 non- 푎푏 periodic digits and 푛 repeating digits where 푚 and 푛 are positive integers and 푝 is a non- negative integer (i.e. 푝 can equal zero). 1 Notice that in the above example we can force 푝 to be zero as follows. Instead of writing in the 384 1 1 1 1 form × we can write it in the form × . We can now apply Statement 1 to conclude 24 24 27 3 1 1 that has seven non-repeating digits (from the exponent in 27 ) and one repeating digit (since = 0, 3̅). 384 3 ODD NUMBERS 1 Let us now consider unit fractions where 푥 is an odd positive integer. If 푥 is a multiple of 5 then we 푥 1 can rewrite 푥 in the form 5푏. 푁 where 푁 is odd. We can thus write the unit fraction in the modified 푥 1 1 1 form × . From the previous discussion it should be clear that relates to the number of non- 5푏 푁 5푏 1 1 repeating decimal digits while relates to the number of repeating digits in the decimal expansion of . 푁 푥 1 1 Note that if we have 푏 = 0 then = and the repeating decimal digits commence directly after the decimal 푥 푁 comma. This leads to the following statement: Statement 3: Any odd number greater than 1, and not divisible by 5, when expressed as a unit fraction, will have repeating decimal digits that commence directly after the decimal comma. EVEN NUMBERS Statement 3 suggests that we can also make the following general claim: Statement 4: Any even number expressed as a unit fraction and having some repeating decimal(s) must have leading non-repeating decimal digits. Learning and Teaching Mathematics, No. 25, 2018, pp. 36-39 Page 39 1 We can prove Statement 4 as follows. Let have 푚 repeating decimal digits directly after the comma, in 푥 other words: 1 = 0, 푑 푑 … 푑 푑 푑 … 푥 1 2 푚 1 2 Multiplying through by 10푚 gives: 10푚 = 푑 푑 … 푑 , 푑 푑 … 푑 푑 푑 … 푥 1 2 푚 1 2 푚 1 2 We can thus see that: 10푚 1 10푚 − 1 − = = 푑 푑 … 푑 푥 푥 푥 1 2 푚 푚 If we let 푑1푑2 … 푑푚 = 푦 then 10 − 1 = 푥푦. Assuming 푥 to be even, then 푥 can be expressed as 2푡 where 푡 is a positive integer. It then follows that 10푚 − 1 = 2푡푦, which is impossible since the left- hand side, 10푚 − 1, is clearly odd. Our assumption is thus false, and any even number expressed as a unit fraction and having some repeating decimal(s) must have leading non-repeating decimal digits. A few examples relating to Statement 3 and Statement 4 are illustrated in the following table: Unit fraction Statement 3 Statement 4 Comments 1 27 is odd. The unit fraction has repeating 0, ̅037̅̅̅̅ decimal digits that commence directly after the 27 decimal comma. 1 189 is odd. The unit fraction has repeating 0, ̅005291̅̅̅̅̅̅̅̅̅ decimal digits that commence directly after the 189 decimal comma. 1 105 is odd but has a factor of 5. There is a 0,0̅095238̅̅̅̅̅̅̅̅̅ leading non-repeating decimal digit after the 105 decimal comma. 1 12 is even. There are leading non-repeating 0,083̅ 12 decimal digits after the decimal comma. 1 110 is even. There are leading non-repeating 0,0090̅̅̅̅ 110 decimal digits after the decimal comma.
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