This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Lecture Notes, Concepts of Mathematics (21-127) Lecture 1, Recitation AD, Spring 2008 3 Congruences 3.1 Congruence READ: Textbook pages 5763. EXERCISES: Pages 8288, #126, 5668. Definition 3.1 Let m be a fixed positive integer. If a, b Z , we say that a is congruent to b modulo m and write a b ( mod m ) whenever m | ( a- b ) . If m- ( a- b ) , we write a 6 b ( mod m ) . The condition for a to be congruent to b modulo m is equivalent to: there exists an integer k such that a = b + km . Example 3.1 We have 7 3 ( mod 4) ,- 6 14 ( mod 10) 4 ( mod 10) , 121 273 ( mod 2) 1 ( mod 2) - 1 ( mod 2) . Also 5 6 4 ( mod 3) and 21 6 10 ( mod 2) . Congruence arithmetic is perhaps most familiar as a generalization of the arithmetic of the clock. Since there are 60 minutes in an hour, minute arithmetic uses a modulus of m = 60. If one starts at 40 minutes past the hour and then waits another 35 minutes, 40 + 35 = 15 (mod 60), so the current time would be 15 minutes past the (next) hour. Similarly, hour arithmetic on a 12-hour clock uses a modulus of m = 12, so 10 oclock (a.m.) plus five hours gives 10 + 5 = 3 (mod 12), or 3 oclock (p.m.). We will find many other uses for congruences in the rest of this chapter. Proposition 3.1 (3.11) Let a, b, c be integers. Then (i) a a ( mod m ) . (ii) If a b ( mod m ) , then b a ( mod m ) . (iii) If a b ( mod m ) and b c ( mod m ) , then a c ( mod m ) . Proof: (i) Since a- a = 0 and m | 0, it follows that a a (mod m ). (ii) If a b (mod m ), then m | ( a- b ). This implies m | (- 1)( a- b ) = ( b- a ), or b a (mod m ). (iii) If a b (mod m ) and b c (mod m ), then m | ( a- b ) and m | ( b- c ), so m | (( a- b ) + ( b- c )) = ( a- c ), therefore a c (mod m ). 1 Proposition 3.2 (3.12) If a a ( mod m ) and b b ( mod m ) , then (i) a + b a + b ( mod m ) (ii) a- b a- b ( mod m ) (iii) a b a b ( mod m ) . Proof: Assume a a (mod m ) and b b (mod m ). Then there exist integers k, ` such that a = a + km and b = b + `m . Then we have a + b = a + b + ( k + ` ) m a- b = a- b + ( k- ` ) m a b = a b + ( kb + `a + k`m ) m. Since k + ` , k- ` , and kb + `a + k`m are all integers, the results (i)(iii) follow. Proposition 3.3 (3.13) If ac bc ( mod m ) and gcd( c, m ) = 1 , then a b ( mod m ) . Proof: If ac bc (mod m ), then m | ( ac- bc ) = c ( a- b ). If gcd( c, m ) = 1, then from Proposition 2.28 we have m | ( a- b ) so a b (mod m )....
View Full Document