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Let S(a, b) = {na + mb : n, m e Z}. Problem 0.1. If c is a common divisor of a and b then c|s for all s E S(a,b)

Let S(a, b) = {na + mb : n, m e Z}. Problem 0.1. If c is a common divisor of a and b then c|s for all s E S(a,b)

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.1: Real Numbers
Problem 38E
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There is another surprising way of characterizing the gcd. For two numbers a and b, we think about all the numbers you can get by adding multiples of a and b together. We can imaging this by thinking of a and b as dollar values of bills and then asking what prices can paid with them. For example, if your country only issues a 6 dollar bill and a 14 dollar bill, can you buy something that costs 10 dollars? Yes - you pay with two 14 dollar bills and get three 6 dollar bills back in change. Can you buy something that costs 15 dollars? No - all the bills are worth an even number of dollars so there is no way to get an odd net transaction. Formulated more abstractly: Let S(a, b) = {na + mb : n, m E Z}. Problem 0.1. If c is a common divisor of a and b then cs for all s E S(a, b) Problem 0.2. If s E S(a,b) then gcd(a,b)|s. Problem 0.3. If s E S(a,b) then sx E S(a, b) for all x E Z Problem 0.4. If S(a,b) = Z if and only if 1 E S Problem 0.5. The set S(0,0) is {0}. For any other a and b the set S(a,b) is infinite. Problem 0.6. If a|b then S(a, b) is precisely the set of multiples of a. The main fact we are aiming to prove in the next few steps is a more general version of the last statement: Theorem 0.7. For any a and b in Z, not both zero, the set S(a, b) is precisely the set of multiples of gcd(a, b). Problem 0.8. Show that proposition 0.7 is equivalent to the statement that gcd(a,b) e S(a, b). The key ingredient in the proof of proposition 0.7 is this: Problem 0.9. Let s e S(a, b) and write s = aq + r using the division algorithm, then r E S(a, b) (note that the same works if we divide by b instead of a). Using that you can prove: Problem 0.10. If c is the smallest positive element of S(a,b) then c is a common divisor of a and b. This is almost enough to finish the proof of proposition 0.7. The only missing piece is to show that this c must be the greatest common divisor, not a smaller one. But using the earlier propositions we know that c must be divisible by gcd(a, b). You will be writing up the details carefully for homework. 1
Transcribed Image Text:There is another surprising way of characterizing the gcd. For two numbers a and b, we think about all the numbers you can get by adding multiples of a and b together. We can imaging this by thinking of a and b as dollar values of bills and then asking what prices can paid with them. For example, if your country only issues a 6 dollar bill and a 14 dollar bill, can you buy something that costs 10 dollars? Yes - you pay with two 14 dollar bills and get three 6 dollar bills back in change. Can you buy something that costs 15 dollars? No - all the bills are worth an even number of dollars so there is no way to get an odd net transaction. Formulated more abstractly: Let S(a, b) = {na + mb : n, m E Z}. Problem 0.1. If c is a common divisor of a and b then cs for all s E S(a, b) Problem 0.2. If s E S(a,b) then gcd(a,b)|s. Problem 0.3. If s E S(a,b) then sx E S(a, b) for all x E Z Problem 0.4. If S(a,b) = Z if and only if 1 E S Problem 0.5. The set S(0,0) is {0}. For any other a and b the set S(a,b) is infinite. Problem 0.6. If a|b then S(a, b) is precisely the set of multiples of a. The main fact we are aiming to prove in the next few steps is a more general version of the last statement: Theorem 0.7. For any a and b in Z, not both zero, the set S(a, b) is precisely the set of multiples of gcd(a, b). Problem 0.8. Show that proposition 0.7 is equivalent to the statement that gcd(a,b) e S(a, b). The key ingredient in the proof of proposition 0.7 is this: Problem 0.9. Let s e S(a, b) and write s = aq + r using the division algorithm, then r E S(a, b) (note that the same works if we divide by b instead of a). Using that you can prove: Problem 0.10. If c is the smallest positive element of S(a,b) then c is a common divisor of a and b. This is almost enough to finish the proof of proposition 0.7. The only missing piece is to show that this c must be the greatest common divisor, not a smaller one. But using the earlier propositions we know that c must be divisible by gcd(a, b). You will be writing up the details carefully for homework. 1
This characterization of the god turns out to be very useful. In particular we frequently Lse both of the following: Corollary 0.11. If the integers a and b have no common factors then there are integers n and m so that an + bm = 1 Corollary 0.12. If d is a common divisor of a and b then digcd(a, b) Put all of the above steps together to write down a complete and careful proof of proposition 0.7. You will be graded as much on clarity as well as Correctness so make sure all the steps are justified, all the notation is defined and clear, and the the argument flows naturally without irrelevant, repetitive, or explained steps.
Transcribed Image Text:This characterization of the god turns out to be very useful. In particular we frequently Lse both of the following: Corollary 0.11. If the integers a and b have no common factors then there are integers n and m so that an + bm = 1 Corollary 0.12. If d is a common divisor of a and b then digcd(a, b) Put all of the above steps together to write down a complete and careful proof of proposition 0.7. You will be graded as much on clarity as well as Correctness so make sure all the steps are justified, all the notation is defined and clear, and the the argument flows naturally without irrelevant, repetitive, or explained steps.
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