This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 )...
View
Full
Document
This note was uploaded on 12/24/2010 for the course CS CS 173 taught by Professor Fleck during the Spring '10 term at University of Illinois, Urbana Champaign.
 Spring '10
 fleck

Click to edit the document details