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# am_lect_09 - r for some q ∈ Z Hence r ≥ 0 and r = a −...

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Announcements Reminder : Quiz 1 is on Wednesday, here, first 15 minutes of class http://www.cs.uiuc.edu/class/sp10/cs173

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The Division Theorem Recap : a b if n Z , b = a . n The Division Theorem : a Z , b Z + , ! q Z , ! r Z such that a = bq + r 0 r < b q is the quotient and r is the remainder We get “uniqueness” by the fact that the remainder r is in [0, b ) we write the remainder on dividing a by b as a mod b Examples : 12 mod 5 = 2 = 2 mod 5, 8 mod 3 = 1 Three important points about mod: 1. In this class, we will always use the above definition of mod 2. Use “mod” for this definition (not rem, %, etc. from programming) 3. When programming, check what your language does for negatives Wikipedia has a good page
Well Ordering Property Property : Every non-empty set of natural numbers has a smallest element Note that a similar property does not hold for real numbers! Proof of Division Theorem (existence) : Let S = set of non-negative numbers of the form a bq , where q Z S is non-empty, so it has a smallest value

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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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