Faculty of Mathematics Waterloo, Ontario N2L 3G Centre for Education in Mathematics and Computing Grade 6 Math Circles November st /2 nd Egyptian Mathematics Ancient Egypt One of the greatest achievements of the ancient Egyptian civilization was their sophisticated development of mathematics. As early on as 2700 (BC), perhaps earlier, a base 0 number system had already emerged for agricultural and religious reasons. In addition, they pioneered some of the most robust techniques of their time to solve very practical problems. Though their use of mathematics was extensive, they had little regard for formulas or complex mathematical relations. They often employed trial and error and used close approximations as oppose to exact answers for their calculations. Luckily, the ancient Egyptians left behind the Rhind Papyrus which explains how they did arithmetic and geometry. It gives explicit demonstrations of how multiplication and division was performed at that time and provides insight on their knowledge of unit fractions, composite and prime numbers, arithmetic, geometric, and harmonic means, and quadratic equations. In this week s circles, we will be exploring some the techniques the Egyptian implored in their daily lives. Egyptian Multiplication The ancient Egyptians had an interesting way of multiplying of two numbers. Unlike us, the ancient Egyptians did not possess nor memorize their multiplication table. They were, however, capable of multiplying and adding by 2. In other words, they only knew their 2 times tables. However, with that ability alone, they were able to multiply any two numbers! They key was to reduce all multiplication problems into a problem that involved multiplying by 2.
Example. Let s multiply 2 by 8 (using the notation we see today). 2 8 8 2 36 4 72 8 44 6 288. Create a chart with two columns, with the the first and second number on top of the first and second column respectively 2. In the first column, write down all the powers of 2 that are smaller or equal below 2 3. In the second column, starting from 8 write down multiples of 8 until you reach the last row 4. Starting at the bottom number of the first column. determine which powers of 2 add up to the 2. To figure out which powers of 2 that are add to 2. We include the biggest power of 2 that is less than 2, so 6. We then subtract 2 6 = 5 and find the next largest power of 2 that is equal to or less than 5. In this case it is 4. Repeating this process, with 5 4 =, the next power of 2 equal to or less than is itself. So we cross out all powers of 2 that we did not use and the numbers in the column next to it. The remaining numbers should be except 6, 4, and and the numbers in the column beside it 2 8 8 2 36 4 72 8 44 6 288 We add the remaining numbers on the right column 8 + 72 + 288 = 378. So we have that 2 8 = 378 2
Exercise. Evaluate the following using Egyptian Multiplication.. 22 2. Solution. We first construct our chart with 22 and 2 on top of the first and second column respectively. We write down all powers of 2 less than 22 on the first column and keep doubling 2 on the second column. 22 2 2 2 42 4 84 8 68 6 336 After crossing off the unused terms, we add the remaining terms on the right 42 + 84 + 336 = 462. Therefore 22 2 = 462 as needed. 2. 33 24. Solution. Similar to the first example, we construct a chart with 33 and 24 on the first and second column respectively. We write down all powers of 2 less than 33 on the left column and keep doubling 24 on the second column. 33 24 24 2 48 4 96 8 92 6 394 32 768 Adding the remaining terms on the right we have 34 + 768 = 792. Therefore 33 24 = 792 Egyptian Division The Egyptians did division in a very similar manner as multiplication. Example. Let s divide 25 by 4. 3
. Create a chart with two columns, with the the dividend i.e. 25 on top of the first column and the divisor i.e. 4 on top of the second column 2. Below the divisor (in this case 4), keep doubling 4 on each row until you reach the largest possible number that is less than 25. 3. Below 25, starting with, keep doubling until you reach the same row as the largest multiple of 4. 4. Subtract the largest possible number from the right column from 25. If you have something remaining, repeat this procedure until all the numbers on the right column is larger than what you have left over. 25 4 4 2 8 4 6 Now we subtract 25 with the largest possible number on the right column. Afterwards, we repeat the process with what is left over until we can no longer subtract. First, 25 6 = 9. Since we are left we 9, the biggest number we can subtract from the right column is 8, so 9 8 =. We are left with, with no possible others numbers to subtract from since all numbers on the right column, 4, 8, and 6 are greater than 4. Then we cross off the remaining terms that aren t used. 25 4 4 2 8 4 6 To obtain our final answer we add the remaining numbers on the left column 2 + 4 = 6. So we have a remainder of with a quotient of 6. Exercise. Using the Egyptian Method evaluate 32 2 32 2 2 2 24 4 48 8 96 4
Adding the numbers remaining on the left column + 2 + 8 =. The final answer is. Egyptian Fractions The Egyptians of 3000 BC incorporated the use of fractions in a very interesting way. They exclusively used unit fractions (a fraction where the numerator is always ) i.e 2, 4, 7. They did NOT write fractions like 2 5 or 3 4 like we do today. Egyptian Notation The Egyptians used the hieroglyph above a number to represent a fraction with a numerator of. For example, we have that How do they represent other fractions? It turns that they represent all other fractions as a sum of different unit fractions. Example. 3 4 = 2 + 4 However, 2 3 = 3 + 3 was not allowed since it was the sum of the same unit fraction. Why use Egyptian Fractions at all? Motivating Problem. Suppose you are in ancient Egypt and you have 5 loaves of bread to share among 8 workers. How would you ensure that you distribute 5 pieces of bread into 8 equal portion? Writing it as 5 does not help, that just simply restates the problem. If we 8 convert 5 into a decimal, we get 0.625, but how much is 0.625 of 5 loafs of bread anyway? 8 Think about this for a second. How would you divide 5 pieces of bread amongst 8 people? Draw out your solution below. Suppose instead we had 4 loafs of bread instead of 5. How would we distribute this? If we cut up our 4 breads in half, everyone can get half a piece. We can start with distributing a 5
half to everyone from our 4 loaves of bread. We now have loaf of bread left over which we can cut up in 8 pieces, one for each of the 8 workers. We can express this mathematically 5 8 = 2 + 8 Notice that the numerator is, because each person gets one portion. It is speculated the Egyptians almost exclusively to facilitate in distribution of food. Exercise. How about 3 loaves share among 2 people? Can you use what you have seen in the previous answer to help you out? The next question naturally should be, how did the ancient Egyptians break a fraction into a sum of different unit fractions? Secondly, is there one way to break into different unit fractions? Let s make some interesting observations first. Pattern. Every Unit Fraction is a Sum of Two Distinct Unit Fractions Evaluate the following. Did you notice a pattern?. 2. 3. 4. 5. 4 + 2 = 3 5 + 20 = 4 3 + 6 = 2 7 + 42 = 6 6 + 30 = 5 Can you use the pattern you observed to evaluate 20 + 380 In general for for any positive integer n: n = n + + n(n + ) Problem. We already know that 3 4 = 2 + 4. Is it possible to express 3 4 distinct fractions? 4 distinct units fractions? Express your answer below. as a sum of 3 6
In general, is it possible to write 3 as a sum of n distinct unit fractions, where n is the 4 number of terms or fractions? Justify your answer. Pattern 2. Egyptian Fractions with Even Denominators Evaluate the following expressions below. Can you spot another pattern?. 2. 3. 8 + 24 = 6 6 + 2 = 4 0 + 40 = 8 Every even number n is equal to a product of 2 and another integer m i.e. n = 2m. For example 6 = 2 3, and 0 = 2 5. For any even denominator 2m, we get the following relation 2m = 2m + 2 + 2m(m + ) Fibonacci s Greedy Algorithm Suppose we have a fraction where a b, where a is not a unit fraction i.e the numerator is b not. Fibonacci developed a method to express a as a sum of distinct unit fractions. This b particular method involves finding the biggest fraction we can take from a. Afterwards, we b repeat the process with what is left over. Example. Using the Greedy Algorithm, express 7 5 as a sum of unit fractions. We need to find the biggest unit fraction that is equal to or less than 7. We can see that 5 2 is bigger than 7 5. The next biggest unit fraction is 3 which is indeed less than 7 5. So we can represent 7 5 as 7 5 = 3 + R where R is the remainder of 7 5 3 = 2 5, so R is 2 5 7
7 5 = 3 + 2 5 Since 2 isn t a unit fraction, we repeat this process and find the largest unit fraction that 5 is equal to or less than 2. After some analysis, we can see that is the largest unit fraction 5 7 that is less than 2. We can simplify the above expression as: 5 7 5 = 3 + 8 + R We find the remainder once again 2 5 8 = 20 7 We can finally express our final as a sum of distinct unit fractions. 5 7 5 = 3 + 8 + 20 Exercise. Using the Greedy Algorithm, express 4 5 as a sum of unit fractions 4 5 = 2 + 3 0 = 2 + 4 + 20 The greedy algorithm always works, but sometimes, we may end up with horrendously large denominators. If we use the Greedy Algorithm to reduce 5 2 into a sum of unit fraction we will get: 5 2 = 25 + 757 + 763309 + 8739608093 + 5276279564209348846225 8
Problem Set. Use the Egyptian Method of Multiplication to evaluate the following a. 28 24 b. 5 8 c. 2 3 a. We construct our chart with 28 and 24 on top of the first and second column respectively. We write down all powers of 2 less than 28 on the first column and keep doubling 24 on the second column. 28 24 24 2 48 4 96 8 92 6 384 After crossing off the unused terms, we add the remaining terms on the right 96 + 92 + 384 = 672. Therefore 28 24 = 672 as needed. b. Constructing a chart again, we have 5 8 8 2 6 4 32 8 64 Interestingly, none of the terms are crossed off so we add all the terms on the right column 8 + 6 + 32 + 64 = 20. Therefore 5 8 = 20 c. After constructing our chart, we have 2 3 3 2 26 4 52 8 04 Adding the remaining terms on the right column 52 + 04 = 56 so we have 2 3 = 56 9
2. Use the Egyptian method of Division to evaluate the following a. 25 6 b. 23 8 c. 77 25 After constructing our charts in the same manner as the examples a. 4 Remainder b. 5 Remainder 3 c. 7 Remainder 2 3. Can you come up with a sum of distinct Egyptian Fractions that add up to. Hint: It may help to know that 6 = 3 + 2 + 6 = 3 + 2 + = 3 + 2 + 6 = 3 6 + 2 6 + 6 = 2 + 3 + 6 4. What is the largest unit fraction that is less than the following Egyptian fractions? (a) 2 5 (b) 7 9 (c) (d) 26 00 3 0 a. 3 b. 2 c. 4 d. 4 5. Express the following unit fractions as a sum of two distinct unit fractions. a. b. 8 0 0
a. 8 = 9 + 72 b. 0 = + 0 c. = 2 + 2 6. Use Fibonacci s Greedy Algorithm to express the following fractions as a sum of distinct unit fractions. a. 2 a. 2 = 2 + 3 + 2 b. 2 5 = 3 + 5 b. 2 5 7. Rhind Papyrus Problem 28 when translated into English reads as follows: Think of a number, and add 2 3 of this number to itself. From this sum subtract 3 its value and say what your number is. Suppose the answer is 0. Then take away 0 of this 0 giving 9. Then 9 was the first number thought of. The scribe, Ahmes, then gives the following argument: If the original number was 9, then 2 is 6, which added makes 5. Then 3 3 which on subtraction leaves 0. This is how you do it. of 5 is 5 If you read this carefully, you will find that Ahmes was trying to prove the equation below: ( n + 2n 3 ) 3 ( n + 2n 3 ) 0 [( n + 2n 3 ) 3 ( n + 2n 3 )] = n (a) Use n = 9 and verify that the left and right side of the equation do equal to each other (b) Algebraically simplify the left side of the equation and show that it is indeed equal to the right side for all values of n. Hint: Notice that a certain expression in brackets appear multiple times. If you simplify it once you can use it throughout the whole expression. We will first simplify the expression n + 2n 3 = 3n 3 + 2n 3 = 5n 3 We will substitute this into the left side of the equation.
( n + 2n ) ( n + 2n ) [( n + 2n 3 3 3 0 3 = 5n 3 ( ) 5n [ 5n 3 3 0 3 ( )] 5n 3 3 ( ) 5n ) 3 ( n + 2n 3 )] = 5n 3 5n 9 0 3 5n 9 ( 5n ) = 5n 3 5n 9 0 = 5n 3 5n 9 ( 0 = 5n 9 5n 9 n 9 = 0n 9 n 9 = 9n 9 = n 9 5n 9 ) 0n 9 8. The Inheritance Puzzle A man who had 2 horses and 3 children wrote on his will to leave of his horses to Pat, to Chris, and to Same. However, just after he died 2 3 2 one of his horses died too. How would they divide the remaining horses so as to fulfill term of the will? The answer is that a friend calls by and offers to add his own horse to the others. Now they can split the horses with 2 3 = 6 horses going to Pat = 4 horses going to Chris 2 = horse going to Sam What is bizarre is that there is total of horses exactly as specified in the will but leaving one horse over - the horse the friend brought. The friend could leave taking his horse with him and the children too go home happy. What happened here? The answer lies in the peculiar conditions of the will since 2 + 3 + 4 = 2 and not 2 2 It works because the denominators 2, 3 and 2 are all factors of 2 2
9. Suppose we have a shelf of identical books. If we lay them down one on top of another, what is the maximum overhang we can achieve? Can the top book ever completely overhang the bottom book? For two books we can get an overhang of a book length 2 The two books now have a centre of gravity in the centre of their overlap, so the bottom one can overhang and extra of a book length. The centre of gravity of the three 4 books means they can have an overhang of, and so on for more books. The extra 6 overhang for four 8 and so forth for more books. + 2 + 3 + 4 2 = 25 2 2 = 25 24 How many books will it take before the overhang exceeds the length of one book? Since 25 is bigger than, we can conclude that 5 stacked books are sufficient for the top one 24 to overhang the bottom one. 3