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 ged(a, b)|s. Problem 0.3. If s e S(a, b) then sr E S(a, b) for allr € Z Problem 0.4. If S(a, b) = Z if and only if 1 €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 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 ged(a, b).

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Put all of the above steps together to write down a complete
and careful proof of proposition 0.7

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 e is a common divisor of a and b then cs for all se 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 € Z
Problem 0.4. If S(a,b) = Z if and only if 1 € 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 alb 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 ged(a, b).
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 e is a common divisor of a and b then cs for all se 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 € Z Problem 0.4. If S(a,b) = Z if and only if 1 € 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 alb 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 ged(a, b).
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