am_lect_09 - r for some q Z Hence r 0 and r = a bq i.e., a...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Announcements Reminder : Quiz 1 is on Wednesday, here, first 15 minutes of class http://www.cs.uiuc.edu/class/sp10/cs173
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
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
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the 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) &lt; 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.

Page1 / 4

am_lect_09 - r for some q Z Hence r 0 and r = a bq i.e., a...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online