week03 - Congruences (Part 1) Original Notes adopted from...

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Original Notes adopted from September 25, 2001 (Week 3) © P. Rosenthal , MAT246Y1, University of Toronto, Department of Mathematics typed by A. Ku Ong Modular Arithmetic a, b Z , m >1 "a is congruent to b modulo m" means m|(a-b). 0) 1) If a b (mod m) then (a+c) (b+d) (mod m) & c (mod m) Proof: a b (mod m) means a – b = mq, some q Z Also c – d = mr some r Z. We Want: (a-b) – (b+d) is a multiple of m a = b + mq c = d + mr (a+c) – (b+d) = ( b + nq + (d + mr)) – b = b + d + mq + mr - b – d = mq + mr = m( q + r) , a multiple (a + c) (b+d) (mod m) Eg. 7 3 (mod 4) -2 6 (mod 4) 5 9 (mod 4) 2) If a b (mod m) & c d (mod n) then ac bd (mod m) Proof: a = b + qm c = d + rm, some r,q ac – bd = (b + qm) ( d + rm) – bd = bd + brm + qmd +qrm 2 – bd = m (br + qd + qrm) Divisible by m, so ac bd (mod m) Corollary: If a b (mod m), then a 2 b 2 (mod m) Proof: Case where a = c & b = d 3) If a b (mod m) and n N then an = bn (mod m) Proof: Use Mathematical Induction. Case n = 1 True
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week03 - Congruences (Part 1) Original Notes adopted from...

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