Division Algorithm B1 Introduction to the Division Algorithm (Procedure) quotient remainder

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1 A Survey of Divisibility Page 1 SECTION B Division Algorithm By the end of this section you will be able to apply the division algorithm or procedure Our aim in this section is to show that for any given integer a there is a unique way of expressing this integer in terms of a quotient and remainder. (You will see what these terms mean below.) B1 Introduction to the Division Algorithm (Procedure) To introduce the concept of the quotient and remainder we go back to long division, let us look at the simple case of 5 divided 4, we have: We can write this as 5 = 6( 4) Rem Here the number 6 is called the quotient and 1 the remainder. Let s look at a geometric interpretation of this example: Figure 1 Showing 5 = 6( 4) + 1 Of course the above can also be written as: 5 = 7( 4) - 5 =- 5( 4) = 1( 4) + 1 In other words, there are an infinite number of ways that we can write 5 as a multiple of 4 plus a remainder. The question we wish to address is: is there a unique way of writing our division? The answer is yes, and we achieve this by placing a restriction on the remainder, we require that the remainder be greater than or equal to zero but less than what we are dividing by, which in this case is 4. With this restriction we have a unique way of dividing 5 by 4 which is our first answer,

2 A Survey of Divisibility Page 1 5 = 6( 4) + 1. Before we look at proving this for the general case, we will consider some other examples. Example 9 Express the following numbers in terms of quotient and remainder where the remainder is the smallest positive number or zero. (a) 7 divided by 5 (b) 65 divided by 7 (c) 159 divided by (a) We have 7 = 5( 5) +. (b) Similarly 65 = 5( 7) + 1. (c) Finally = - 5( ) + 0. Where in each of the above we have place the quotient outside of the brackets. We now generalize this result. Let a and b be any given integers with b ³ 1. Then there exists a quotient q and a remainder r such that a = bq + r The quotient q and remainder r for those cases given in Example 9 are: (a) q 5, r (b) q 5, r 1 (c) q 5, r 0 If we restrict the reminder r such that 0 r < b then the expression a = bq + r is unique, which we prove below. We can look at this geometrically: Figure 1 - Geometric interpretation of the quotient and remainder expression of the integer a. We choose our remainder r to be the smallest non-negative integer. Note that in the above example the remainders are the smallest non-negative integers. The following is a demonstration of why we are interested in the smallest nonnegative remainder. Example 10 We have 51 litres of drink and we can get 4 portions out of each litre. How many portions should we serve each person to be fair to everyone?

3 A Survey of Divisibility Page 14 We can use the division algorithm to solve this problem. The total number of portions we have are 4 51= 04. Writing this in the form a = bq + r where 0 r < b, with a = 04 and b = 7 : 04 = 5( 7) This means that in order to have a fair distribution of drinks we should serve each person 5 portions. We are now in a position to write down the general form of the Division Algorithm and prove it. The proof is quite lengthy and makes use of a result and notation from Set Theory (see introductory chapter). Recall that a set is denoted by braces { } and is used to illustrate a grouping. We can define a set using a rule, for example { x : x is an even number} such that which would be read as x such that x is an even number. This set would be equivalent to {,4,6,8, } where three dots show that the pattern repeats and the above colon represents such that. Theorem (1.7). Division Algorithm. Given any integers a and b where b 1 there exists unique integers q called the quotient and r called the remainder such that We need to prove this result. How? The clue is in the statement, we need to show two things. 1) the existence of q and r and ) the uniqueness of these integers. Proof. 1) Existence Let S be the set S { a, a b, a, b a, b } = Remember, b is our divisor so we can take away multiples of it until we get to a remainder between 0 and b. We can write the set S in compact notation as { : is an integer and 0} S = a -mb m a - mb ³ Note that the set S will only contain positive integers between 0 and a, we can plot these as points in the real number line:

4 A Survey of Divisibility Page 15 Figure - The Set S is contains the integers between 0 and a. In order to show that q and r exist, we need to show that the set S is non-empty, (in other words, that the set has some elements.) How do we show S is non-empty? Select an integer m so that a - mb ³ 0. Let m =- a. Why? Because substituting m =- a into a - mb gives a - mb = a + a b ³ 0 ébecause b ³ 1ù êë úû Which shows that S is non empty. We can now apply the Well Ordering Principle (WOP) (see introductory chapter) to show that S has a least element. WOP: Let S be a non-empty set of positive integers and zero. Then the set S has a least element. This means that there exists a least element of the set S, say r, which is of course an integer. We have r is a member of the set S, and furthermore it is the smallest number in the set S, which means there exists an integer m = q (where q is the largest number we can multiply by) such that: r = a - mb = a - qb ³ Showing that r is in the set is only part of the task, we also need to show that r How? By contradiction, suppose r ³ b. However, we are given that b ³ 1 so r - b < r But r is the smallest element of S and we have found a smaller element r - b which is also in S meaning we have a contradiction. Our supposition r therefore, r ) Uniqueness < b. How do we prove that the integers q and r are unique? Suppose there also exists integers q ' and 0 r ' such that a = bq' + r' 0 r' < b (*) < b. ³ b must be wrong We need to show that We already have q = q' and r = r'. (**)

5 A Survey of Divisibility Page 16 Subtracting (* *) and (*), gives 0 = bq ( - q') + ( r-r') r' - r = b( q -q') r' - r = q - q' (1) b Next we add the inequalities, 0 and, 0 to give 0 +- ( b) < r - r' < b + 0 implies - b < r - r' < b Dividing by then gives r - r' - 1< < 1 b However, from equation (1) we see that 1 1 And the only integer in this range is 0 which implies that is the integer. Thus we have q = q'. This result when substituted back into equation (1) gives, thus we see that both and are unique. We have therefore proved not only that both q and r exist, but that they are also unique. This completes the proof of the Division Algorithm. The division algorithm can also be applied to algebraic expressions in order to prove certain statements about integers, as the following examples will demonstrate. Example 11 Show that the square of any integer is of the form m or m 1. Let n be any integer. How do we prove this result? By applying the above Division Algorithm (1.7): Given any integers a and b with b 1 there exists unique integers q and r such that With b we can write any integer a n as: Squaring both sides of this gives n q r 9q 6qr r What values can r take? n q r 0 r where Factorizing q qr r k r k q qr

6 A Survey of Divisibility Page 17 Squaring r gives r r 0, 1 and because 0 r 0, 1 and simplifying r 0, 1 and 4 Substituting these r 0,1 and 4 values into the above Note that n k k k In the last n k k k, 1and 4respectively n k r gives, 1, 4says that any square number is of the form: n integer, integer 1, integer 4 n integer 4we can take a factor of out and this can be written as n k k integer 1. That is, it has the same form as k+1. Therefore, the square of any integer has the form integer ( ) or integer ( ) 1 + or m or m + 1, where m is an integer. Example 1 Show that the cube of any integer has one of the following forms: 9,9 k k + 1 or 9k + 8 How do we show this result? In the same manner as Example 11. Let n be any integer and we use the Division Algorithm (1.7): Given any integers a and b 1 there exists unique integers q and r such that With b 9 and a n where n is any integer: n = 9q + r 0 r < 9 Taking the cube of this by using the binomial theorem or expanding by multiplying out: ( ) ( ) ( ) ( ) n = 9q + r = 9q + 9q r + 9q r + r = 9k + r By binomial Since 0 r 9 and r is an integer, it can take the following values: r = 0, 1,,,, 8 = 9k Finding the remainder r ' for each of these cubes r : gives r ' gives r ' gives r ' 8

7 A Survey of Divisibility Page gives r ' gives r ' gives r ' gives r ' gives r ' gives r ' 8 We see that r can only have remainder values 0, 1 and 8 after dividing by 9. Hence the cube of any integer has the form Where k is an integer. n k r k k k = 9 + = 9, or Example 1 Show that a 1 is never a perfect square. [Hint: The square of any integer is either of the form k or k 1. See Example 11.] By the hint every square is of the form k or k 1. First consider the case where the square is of the form k : Suppose a 1 is a perfect square. Then a 1 k a k 1 This is impossible because integer 1. Similarly consider the other case where the square is of the form k 1: a 1 k 1 a k Re-arranging We cannot have integer. Hence a 1 is never a perfect square. To close this statement we extend the Division Algorithm (1.7) to a more general case where we no longer require that b ³ 1, but rather that b ¹ 0. This is given in Corollary (1.8) below. Corollary (1.8). Given any integers a and b with b 0 there exists unique integers q and r such that

8 A Survey of Divisibility Page 19 Proof. See Exercise 1(b). SUMMARY Suppose we have any integers a and b ³ 1 then there are unique integers q and r such that

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