Math 371 Lecture #1 x1.1,1.2: The Division Algorithm, Divisibility We begin with a principle you saw in Math 290. It is the principle that tames the infinite and underlies the theory we review today and next. Well-Ordering Axiom: any nonempty set of nonnegative integers has a smallest element. You remember that division is nothing more than repeated subtraction such as 22 divided by 4 is 22 − 4 = 18; 18 − 4 = 14; 14 − 4 = 10; 10 − 4 = 6; 6 − 4 = 2; so that 22 = 4 + 4 + 4 + 4 + 4 + 2 = 4 × 5 + 2: That this can always be done is a consequence of the Well-Ordering Axiom. Theorem 1.1 (The Division Algorithm). Let a and b be integers with b 6= 0 (the divisor). Then there exist unique integers q (the quotient) and r (the remainder) such that a = bq + r; 0 ≤ r < jbj: Proof. For integers a and b > 0 (other case b < 0 for you), consider the set of nonnegative integers S = fa − bx : x 2 Z and a − bx ≥ 0g: We are anticipating what the remainder r is here. Is the set S nonempty? You might guess that a 2 S because we can take x = 0 in a − bx to get a, but a might be negative. So we have to work a bit harder to show S is nonempty. We know that b ≥ 1 so that bjaj ≥ jaj ≥ −a, and so a + bjaj ≥ 0: Thus with x = −|aj, the integer a − bjaj = a − b(−|aj) belongs to S. Now by the Well-Ordering Axiom, the set S has a smallest nonnegative integer r = a−bx for some x that we call q. So we have a = bq + r; r ≥ 0: We argue that r < b by contradiction: suppose r ≥ b. Then 0 ≤ r − b = (a − bq) − b = a − b(q + 1) 2 S: Since b ≥ 1, we have r − b < r so that a − b(q + 1) = r − b < r: This implies that a − b(q + 1) is a smaller element of S, a contradiction to r being the smallest element of S. Thus r < b. Last, to show that q and r are unique, we suppose there are q1 and r1 such that a = bq1 + r1; 0 ≤ r1 < b: Combining this equation with a = bq + r gives bq + r = bq1 + r1 ) b(q − q1) = r1 − r: From 0 ≤ r < b we have −b < −r ≤ 0, and adding this to 0 ≤ r1 < b gives −b < r1 − r < b: Then since b(q − q1) = r1 − r, we have −b < b(q − q1) < b and since b > 0 we arrive at −1 < q − q1 < 1: Well there is only one integer strictly between −1 and 1, so that q − q1 = 0, or q = q1. Since b(q − q1) = r1 − r, we get r1 − r = 0, or r1 = r. An important case of the division algorithm is when the remainder r is 0, for then a = bq. Definition. Let a and b be integers with b 6= 0. We say that b divides a (or that b is a divisor of a, or that b is a factor of a) if a = bc for some c 2 Z. Symbolically we write b j a when b divides a, and b - a when b does not divide a. We know that 4 - 22 because 22 = 4 × 5 + 2, but 11 j 22 because 22 = 11 × 2. Some basic facts of b j a are: (1) a and −a have the same factors. Why? Because a = bc implies −a = b(−c). (2) every divisor of a 6= 0 is less than or equal to jaj, and a 6= 0 has only finitely many factors. Why? Because a = bc implies jaj = jbj jcj, so that jbj ≤ ja. Often it happens that an integer divides two integers, such as 3 divides both 12 and 30. Definition Let a and b be integers, not both 0. The greatest common divisor (gcd) of a and b is the largest integer d that divides both a and b. In other words the greatest common divisor of a and b is an integer d for which (a) d j a and d j b, and (b) if an integer c is any common divisor of a and b, then c ≤ d. Symbolically we denote the greatest common divisor of a and b as (a; b). Some facts about the greatest common divisor are: (1) (a; b) exists and is unique. Why? There are only finitely many common divisors, and one of them is the largest. (2) (a; b) ≥ 1. Why? Because 1 a common divisor of a and b. Definition. Two integers a and b, both not zero, are relatively prime if (a; b) = 1. The integers 6 and 35 are relatively prime while 8 and 10 are not. The greatest common divisor has a remarkable property. Theorem 1.2. Let a and b be integers, both not zero. If d = (a; b), then there exist integers u and v (not necessarily unique) such that d = au + bv: Warning. This Theorem does NOT say that if d = au + bv for some integers u and v, that then d = (a; b). That is to say, this Theorem is not an if and only if kind of theorem..