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Unformatted text preview: r for some q ∈ Z Hence r ≥ 0 and r = a − bq i.e., a = bq + r Suppose r ≥ b . Then r − b ∈ S because r − b ≥ 0 and r − b = a − bq − b = a − b ( q +1) < r This contradicts the fact that r is the smallest value in S Proof of Division Theorem (uniqueness) : Try a proof by contradiction! Congruence mod k Definition : a ∈ Z is congruent to b ∈ Z mod k if k ( a − b ) — Notation: a ≡ b (mod k ) Examples : 17 ≡ 5 (mod 12) − 3 ≡ 4 (mod 7) 3 ≡ − 3 (mod 7) Theorem : For all a , b , k, a ≡ b (mod k ) ↔ ( a mod k ) = ( b mod k ) Claim : For all a , b , c , d , k , if a ≡ b (mod k ) and c ≡ d (mod k ) then a + c ≡ b + d (mod k ) Proof : By definition, k ( a − b ) and hence ∃ n , ( a − b ) = kn and k ( c − d ) and hence ∃ m , ( c − d ) = km . So ( a + c − ( b + d )) = ( a − b ) + ( c − d ) = kn + km = k ( n + m )...
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 Spring '10
 fleck
 Remainder, Wikipedia, 15 minutes

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