lecture08&iuml;&frac14;ˆcomplete&iuml;&frac14;‰

lecture08&iuml;&frac14;ˆcomplete&iuml;&frac...

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1 MA1100 Lecture 8 Mathematical Proofs Congruences Proof by Cases Absolute Values Chartrand: section 3.4, 4.2, 4.3

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Lecture 8 2 Announcement ± No Homework due next week ± Tutorial 2 to 4 ² changes in problem sets T2 2.31, 2.32, 2.37, 2.40, 2.48, 2.49, 3.10, 3.16, 3.18, 3.29 T3 3.20, 3.22, 3.27, 3.28, 4.4, 4.12, 4,15, 4.23, 4.33, 4.42 T4 5.4, 5.10, 5.13, 5.17, 5.22, 5.23, 5.26, 5.32, 5.33 3.10, 3.18, 3.29 2.62, 2.67, 2.68 4.23, 4.33, 4.42 X XX
Lecture 8 3 Calendar September 2009 Sun Mon Tue Wed Thu Fri Sat 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 1 2 3 Homework due dates Mid-term test How are numbers within the same columns related? Oct Term break

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Lecture 8 4 Congruence Definition Let a, b and n be integers with n > 1. If n divides a – b , we say that a is congruent to b modulo n Notation a ª b mod n Example 24 ª 10 mod 7 -2 ª 8 mod 5 84 ª 0 mod 12 Negation a ª b mod n 5 ª 2 mod 6 1 ª 0 mod 4 Two numbers along the same columns of a calendar month are congruent to each other modulo 7 Do not confuse congruence with logical equivalence
Direct proof Lecture 8 5 Congruence Proposition Let a, b and n be integers with n > 1. Proof Hence a - b = nk for some integer k. Given a ª b mod n . This means n | a – b . So a = b + nk for some integer k. hypothesis definition of congruence definition of divide algebraic manipulation conclusion The converse of this proposition is also true there exists an integer k such that a = b + nk. If a ª b mod n, then a = b + nk for some integer k. Convert ª to = for algebraic manipulation

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Lecture 8 6 Congruence a ª b mod n Example ( \$ k œ Z ) ( a = b + nk ) To list integers congruent to b modulo n , We only need to add/subtract multiples of n to/from b. multiple of n Let b = 3 and n = 4 All integers congruent to 3 modulo 4 : 3 7 11 15 -1 -5 All integers congruent to 0 modulo 4 : … -8 -4 4 8 12 … multiples of 4 0
Lecture 8 7 Congruence Proposition Let a and n be integers with n > 1. a ª 0 mod n if and only if n | a Prove this biconditional statement Exercise This gives a criteria for divisibility a ª 0 mod n n | a n | a a ª 0 mod n Both parts can be proven using “ direct proof

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-13 = 5(-2) - 3 Lecture 8 8 Congruence and Remainder Fact Let a, n be integers with n > 1. Every integer is congruent to exactly one of 0, 1, 2, … , n-1 modulo n. If r is the remainder of a when it is divided by n, then a ª r mod n . Example Division by 5 13 = 5(2) + 3 13 ª 3 mod 5 -13 = 5( -3 ) + 2 -13 ª 2 mod 5 For division by n, the remainder is between 0 and n-1 Converse is not true 3 is the remainder 2 is the remainder Note: -13 ª -3 mod 5 too!
Lecture 8 9 Properties of Congruence (1) For every integer a, a ª a mod n (2) For all integers a and b, if a ª b mod n, then b ª a mod n (3) For all integers a, b and c, if a ª b mod n and b ª c mod n, then a ª c mod n reflexive property symmetric property transitive property Proposition Let n be an integer with n > 1. ª ” behaves like “=”

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Lecture 8 10 a divides b (1) For every integer a, a | a (2) For all integers a and b, if a | b, then b | a (3) For all integers a, b and c, if a | b and b | c, then a |c reflexive property symmetric property transitive property Does the “divide” relation a | b satisfy reflexive , symmetric and transitive properties?
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This note was uploaded on 04/15/2010 for the course MATHS MA1101R taught by Professor Vt during the Spring '10 term at National University of Singapore.

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lecture08&iuml;&frac14;ˆcomplete&iuml;&frac...

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