MACM 101 Lecture 26: Lecture 26 Part 1_ Common Divisors

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browncricket987

21 Dec 2018

School

Simon Fraser University

Department

Math & Comp Sci

Course

MACM 101

Professor

Andrei Bulatov

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For integers a and b, a positive integer c is said to be a common divisor of a and b if c | a and c | b. Let a, b be integers such that a 0 or b 0. The greatest common divisor of a and b is denoted by gcd(a,b) For any positive integers a and b, there is a unique positive integer c such that c is the greatest common divisor of a and b. First try: take the largest common divisor, in the sense of usual order. Given a, b, let s = { as + bt | s,t z, as + bt > 0 }. Since s , it has a least element c. we show that c = gcd(a,b) We have c = ax + by for some integers x and y. If d | a and d | b, then d | ax + by = c.

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ABSTRACT ALGEBRA PROOFPlease provide a super detail proof for part (a). Please be detail and clear each step of the proof. Only solve if you fully understand. Do not provide an incomplete proof. Thanks.

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MACM 101 Lecture 26: Lecture 26 Part 1_ Common Divisors

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