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# lecnotes3 - Lecture Notes Concepts of Mathematics(21-127...

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Lecture Notes, Concepts of Mathematics (21-127) Lecture 1, Recitation A–D, Spring 2008 3 Congruences 3.1 Congruence READ: Textbook pages 57–63. EXERCISES: Pages 82–88, #1–26, 56–68. 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 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 4 ( mod 3) and 21 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 o’clock (a.m.) plus five hours gives 10 + 5 = 3 (mod 12), or 3 o’clock (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

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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 ). Proposition 3.4 (3.14) a b ( mod m ) if and only if a and b have the same remainders when divided by m .
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lecnotes3 - Lecture Notes Concepts of Mathematics(21-127...

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