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 sa e S(a, b) for all æ € Z
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 sa e S(a, b) for all æ € Z
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Transcribed Image Text:There is another surprising way of characterizing the ged. 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 c|s 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 a € Z
Problem 0.4. If S(a, b) = Z if and only if 1 E S
%3D
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 alb then S(a, b) is precisely the set of multiples of a.
Expert Solution
Step 1
The GCD of two integers is the largest positive integer that divides both the integers.
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