Notes on Chapter 1, Section 2 Arithmetic and Divisibility

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1 Notes on Chapter 1, Section 2 Arithmetic and Divisibility August 16, Arithmetic Properties of the Integers Recall that the set of integers is the set Z = f0; 1; 1; 2; 2; 3; 3; : : :g. The integers have the following basic arithmetic properties: 1. The set of integers is closed under addition and closed under multiplication. This means that if a and b are any integers, then a + b and ab are also integers. 2. Addition and multiplication are associative. This means that for any integers a; b; and c, it is true that a + (b + c) = (a + b) + c and a (bc) = (ab) c. 3. Addition and multiplication are commutative. This means that for any integers a and b, it is true that a + b = b + a and ab = ba. 4. The set of integers contains an additive identity element. This additive identity element is the number 0. It is called an additive identity element because if a is any integer, then a + 0 = a. 5. The set of integers contains a multiplicative identity element. This multiplicative identity element is the number 1. It is called a multiplicative identity element because if a is any integer, then 1 a = a. 1

2 6. Given any integer a, there is another integer which is the additive inverse of a. This additive inverse is the number a. It is called an additive inverse because a + ( a) = The integers obey the distributive law of multiplation over addition. This means that if a; b; and c are any integers, then a (b + c) = ab + ac. All of the other familiar properties of integer arithmetic can be deduced from these six properties. For example, these properties can be used to prove that if a is any integer, then 0 a = 0. Although you may not feel the need to prove that 0a = 0 because you are already very comfortable with this fact and have been using it since elementary school, you should consider this to be an introduction to the art of theorem proving. A theorem is a statement of the form If A is true, then B must be true. In the Theorem that we are about to prove, the A is the set of properties 1 7 listed above which we are assuming to be true, and the B is the statement that 0 a = a. In general, the given information, A, is called the hypothesis or set of hypotheses, and the to be proved informaion, B, is called the conclusion. Note that, in proving that 0 a = 0, we actually use the word Proposition rather than Theorem. A Proposition is a mini theorem or a simple fact one that requires proof, but can be deduced fairly quickly and easily from the given set of hypotheses. The word Theorem is usually reserved for deeper results whose proofs require a more complex and clever assembly of the hypotheses. Make sure, in reading the proof of the following proposition, that you understand every step. You should observe that only the given information (properties 1 7) is being used to arrive at the conclusion. In the homework exercises, you will have the opportunity to develop your theorem proving skills by proving similar propositions. Proposition 1 If a is any integer, then 0 a = 0. Proof. Let a be an integer. Since 0 is the additive identity element of the integers, we know that = 0. By multiplying both sides of the above equation by the integer a, we obtain a (0 + 0) = a 0. 2

3 Due to the distributive property of multiplication over addition, we now see that a 0 + a 0 = a 0. Since the set of integers is closed under multiplication and a and 0 are both integers, we know that a 0 must also be an integer. Since a 0 is an integer and every integer has an additive inverse, we know that the integer (a 0) exists. Adding this integer to both sides of the above equation, we obtain (a 0 + a 0) + ( (a 0)) = a 0 + ( (a 0)). as By the associative property of addition, we can write the above equation a 0 + (a 0 + ( (a 0))) = a 0 + ( (a 0)) Since a 0 + ( (a 0)) = 0 (by Property 6), we now have a = 0. Since 0 is the additive identity element, we know that a = a 0. Upon comparing the previous two equations, we deduce that a 0 = 0. Finally, since multiplication is commutative, we arrive at the conclusion that 0 a = 0. 2 Divisibility and the Division Algorithm De nition 2 Let a and b be integers with b 6= 0. We say that a is divisible by b (or that a is a multiple of b or that b is a factor of a) if there exists an integer q such that a = qb. If the integer a is divisible by the integer b, then we say that b divides a and we write b j a. 3

4 Example 3 84 is divisible by 3 because 84 = Thus we can say that 3 divides 84 and we can write 3 j ; 921 is divisible by 357 because 18; 921 = is divisible by 6 because 30 = is not divisible by 3 because there is no integer q such that 50 = q 3. Whenever an integer a is not divisible by an integer b, we can still write a as a multiple of b plus a remainder. Example 4 17 is not divisible by 7 but we can write 17 = : 24 is not divisible by 5 but we can write 24 = Some basic results involving divisibility are given in the following Proposition. Lemma 5 Suppose that a; b; and c are integers. 1. If a divides b and b divides c, then a divides c. 2. If a divides both b and c, and if m and n are any integers, then a divides mb + nc. Proof. First we prove assertion 1: Suppose that a divides b and b divides c. This means that there are integers s and t such that b = sa and c = tb. It follows from this that c = t (sa) = (ts) a. Since ts is an integer, we see that a divides c. Now we prove the second assertion of the lemma: Suppose that a divides both b and c. This means that there are integera s and t such that b = sa and c = ta. Now let m and n be any given integers. Then mb + nc = m (sa)+n (ta) = (ms) a+(nt) a = (ms + nt) a. Since mb+nc = (ms + nt) a and since ms + nt is an integer, we see that a divides mb + nc. A general fact about divisibility is that if a and b are any integers with b > 0, then we can always nd integers q and r such that a = qb + r. However, given any particular a and b, there is always more than one to choose q and r to make the above equation be true. For example, for a = 17 and b = 7, we know that 17 = 27+3, but it is also true that 17 = 37+( 4). Thus we could use q = 2 and r = 3 or we could just as well use q = 3 4

5 and r = 4. The following very important theorem, called The Division Algorithm, states that we can always choose the numbers q and r in such a way that 0 r < b and that there is only one way in which to do this. The tool needed to prove the Division Algorithm is the General Well Ordering Principle (Chapter 1, Section 1). Theorem 6 (The Division Algorithm) Let a and b be integers with b > 0. Then there are integers q and r such that a = qb + r and 0 r < b. Furthermore, there is only one choice of q and r that will make a = qb + r and 0 r < b both be true. Proof. Let the integers a and b with b > 0 be given and de ne the set S as S = fa qb j q 2 Z and a qb 0g. We want to apply the General Well Ordering Principle to the set S to conclude that S has a smallest member. In order to do this, we must show that S 6= ; and that there is some integer n 0 such that every member of S is greater than or equal to n 0. It is easy to see that we can take n 0 = 0, because part of the de nition of the set S is that every one of its members must be greater than or equal to 0. But we still need to make sure that S 6= ;. To see why this is so, we consider two possibilities: One possibility is that a 0. In this case, the number a 0 b is in S because 0 2 Z and a 0 b = a 0. The other possibility is that a < 0. In this case, the number a a b is in S because a 2 Z and a a b = a (1 b), and since a < 0 and b 1, it must be the case that a (1 b) 0. This reasoning shows that, in any case, S 6= ;. By the General Well Ordering Principle, S must have a smallest member. We will call this smallest member r. Since r 2 S, there must be some integer q such that a qb = r and it must be true that r 0. We have almost nished proving that rst assertion of the Division Algorithm Theorem, because we proved that there exist integers q and r such that a = qb + r and r 0. We still have to prove, though, that r < b. To prove this, we observe that r b = a qb b = a (q + 1) b. From the above equation, we can deduce that the number r b must be negative. Why? If it were the case that r b 0, then since q + 1 is an integer and r b = a (q + 1) b, it would be true that r b 2 S. However, 5

6 clearly r b < r (because b > 0), and so it is not possible that r b 2 S because r is the smallest member of S. We conclude that r b must be a negative number and hence that r < b. This completes the proof of the rst assertion of The Division Algorithm. We have proved that if a and b are any integers with b > 0, then there must exist integers q and r such that a = qb + r and 0 r < b. We are still left to prove the second assertion of the theorem, which is that the number q and r can be chosen in only one way in order to satisfy both of these criteria. Suppose that there are two di erent choices of q and r that both satisfy the conditions a = qb + r and 0 r < b. Call these di erent choices (q 1 ; r 1 ) and (q 2 ; r 2 ). Thus a = q 1 b + r 1, a = q 2 b + r 2, 0 r 1 < b, and 0 r 2 < b. For de niteness, let us suppose that r 1 r 2. Then 0 r 1 r 2 < b. Also which implies that q 1 b + r 1 = q 2 b + r 2 (q 1 q 2 ) b = r 2 r 1. Now observe that r 2 r 1 0 and also, since r 2 < b and r 1 0, then r 2 r 1 < b. Therefore 0 r 2 r 1 < b. From the above equation, we conclude that 0 (q 1 q 2 ) b < b. Since b > 0, we may divide all parts of the above inequality by b (without reversing the order of the inequality) to obtain 0 q 1 q 2 < 1. Since q 1 q 2 is an integer, it must be true that q 1 q 2 = 0. Thus q 1 = q 2. It then follows from the equation (q 1 q 2 ) b = r 2 r 1 that r 1 = r 2. This proves that there is only one choice of q and r for which both conditions a = qb + r and 0 r < b are satis ed. Our proof of the Division Algorithm Theorem is now complete. We will refer to the numbers q and r obtained in the division algorithm as the quotient and the remainder obtained in the division of a by b. Example 7 Apply the division algorithm to a = 356 and b = 56. (By apply the division algorithm, we mean to nd q and r such that 356 = q 56 + r where 0 r < 56.) 6

7 Solution 8 We might guess that 56 goes into 356 about 6 times. Let s check = 20. Our guess was right because we obtained a remainder of 20 which satis es 0 20 < 56. Example 9 Apply the division algorithm to a = 7 and b = 3. Solution 10 Since these are small numbers, no experimentation is required. We see that 7 = We know that this is the correct division algorithm represenation of 7 3 because the remainder is 1 and 0 1 < 3. Example 11 Apply the division algorithm to a = 3 and b = 7. Solution 12 7 goes into 3 zero times, so it looks like the quotient should be 0 and the remainder should be 3. In fact, we see that 3 = Example 13 Apply the division algorithm to a = 356 and b = 56. Solution 14 In a previous example, we found that It follows from this that 356 = = ( 20). However, remember that we want to nd an expression a = qb + r in which 0 r < b. Thus we don t want to have a negative remainder. What if we change the 6 to a 7? = 36 gives us 356 = and this is what we were looking for because 0 36 < Homework In Chapter 1, Section 2, pages 13 14, do all of the exercises (numbers 1 11). 7

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