lecture21(complete)

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Chartrand: 11.4, 11.5, 11.6 MA1100 Lecture 21 Number Theory Linear Combination Relatively Prime Prime Factorization Fundamental Theorem of Arithmetic
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Lecture 21 2 Lesson Plan (countdown: 25 days) Tue Lecture Fri Lecture Tutorial Week 11 20. Number Th. HW4 21. Number Th. Tut 8 Functions Week 12 22. Number Th. 23. Cardinality Tut 9 Number Th. Week 13 24. Revision lecture HW5 25. Revision lecture (tentative) Tut 10 Number Th./ Cardinality Week 14 READING WEEK Week 15 FINAL EXAM (NOV 24)
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Linear combination vs gcd Proposition B Let a and b be integers, not both 0. Corollary Every multiple of gcd(a, b) is a linear combination of a and b. 3 Lecture 21 Chartrand Theorem 11.7 Then gcd(a, b) is the smallest positive linear combination of a and b. ax + by for some integers x and y n is a multiple of gcd(a, b) n is a linear combination of a and b
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Linear combination vs gcd Proof (Non-constructive) 4 Lecture 21 Let S be the set of all positive linear combinations of a and b . i.e. S = {ax + by | ax + by > 0 and x, y œ Z } S is a non-empty subset of N . By Well ordering principle , S has a smallest element. Let this smallest element be d = ax 0 + by 0 > 0 We shall show d = gcd(a, b) ---(*) . We already have gcd(a,b) | d . Try to show d | gcd(a,b) , which will imply (*) . If a | b and b | a, then a = b gcd(a, b) is the smallest positive linear combination of a and b gcd(a, b) = ax + by for some integers x and y
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Linear combination vs gcd(a, b) Proof (Non-constructive) 5 Lecture 21 To show d | gcd(a,b) : Apply Division algorithm to a and d: a = dq + r where 0 § r < d . Then r = a – dq If r 0 , This contradicts d is the smallest in S. So r = 0 and thus d | a . Similarly we have d | b . So d is a common divisor of a, b. Hence d | gcd(a, b). d = ax 0 + by 0 > 0 : smallest element in S By proposition A S = {ax + by | ax + by > 0 and x, y œ Z } we show d | a and d | b = a – (ax 0 + by 0 )q = a(1-qx 0 )+ b(-qy 0 ) . then r œ S and r < d .
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Reversing Euclidean Algorithm Example a = 42 and b = 234 gcd(234, 42) = 6 234 = 42 ä 5 + 24 (1) 42 = 24 ä 1 + 18 (2) 24 = 18 ä 1 + 6 (3) 18 = 6 ä 3 + 0 Euclidean Algorithm From (3): 6 = 24 - 18 ä 1 Find integers x, y such that 6 = 234x + 42y From (2): 18 = 42 - 24 ä 1 6 = 24 ( 42 - 24 ä 1 ) ä 1 6 = 42 ä ( ) + 24 ä ( ) From (1): 24 = 234 - 42 ä 5 6 = 42 ä (-1) + ( 234 - 42 ä 5 ) ä 2 6 = 234 ä ( ) + 42 ä ( ) ì -1 2 2 -11 6 Lecture 20 Chartrand p.254 The answer is not unique
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Lecture 21 7 Relatively Prime Definition Let a and b be two nonzero integers. If
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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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lecture21&amp;iuml;&amp;frac14;ˆcomplete&amp;iuml;&amp;frac...

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