230exam2sol - Math 230 E Exam 2 Fri Nov 4 Fall 2011 Drew...

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Math 230 E Fall 2011 Exam 2 — Fri Nov 4 Drew Armstrong There are 4 problems, worth 5 points each. There is 1 bonus point for writing your name. This is a closed book test. Anyone caught cheating will receive a score of zero . 1. [5 points] Let a and b be integers, not both zero. (a) Accurately state the definition of the greatest common divisor gcd( a,b ). The greatest common divisor of a and b is the maximum integer d Z such that d | a and d | b . (b) Using your definition above, prove the following. If there exist integers x and y such that ax + by = 1 then gcd( a,b ) = 1. Proof. Let d = gcd( a,b ) and suppose that there exist integers x,y with ax + by = 1. Since d | a and d | b we have d | ax + by , hence d | 1, which implies d 1. But we know 1 d since 1 is a common divisor of a and b and d is the greatest common divisor. We conclude that d = 1. ± 2. [5 points] (a) Accurately state the Division Theorem. Given integers
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This note was uploaded on 01/08/2012 for the course MATH 461 taught by Professor Armstrong during the Fall '11 term at University of Miami.

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230exam2sol - Math 230 E Exam 2 Fri Nov 4 Fall 2011 Drew...

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