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Let S(a,b) = {na + mb : n, m € Z}. Problem 0.1. If e is a common divisor of a and b then cs for all s e S(a, b) Problem 0.2. If s € S(a,b) then ged(a, b)|s. Problem 0.3. If s € S(a,b) then sI € S(a, b) for all I €Z

Let S(a,b) = {na + mb : n, m € Z}. Problem 0.1. If e is a common divisor of a and b then cs for all s e S(a, b) Problem 0.2. If s € S(a,b) then ged(a, b)|s. Problem 0.3. If s € S(a,b) then sI € S(a, b) for all I €Z

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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 imagine 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 two 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 \( S(a, b) = \{na + mb : n, m \in \mathbb{Z}\} \). **Problem 0.1.** If \( c \) is a common divisor of \( a \) and \( b \) then \( c|s \) for all \( s \in S(a, b) \). **Problem 0.2.** If \( s \in S(a, b) \) then \( \text{gcd}(a, b)|s \). **Problem 0.3.** If \( s \in S(a, b) \) then \( xs \in S(a, b) \) for all \( x \in \mathbb{Z} \). **Problem 0.4.** If \( S(a, b) = \mathbb{Z} \) if and only if 1 \(\in S\). **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 \(\mathbb{Z}\
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 imagine 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 two 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 \( S(a, b) = \{na + mb : n, m \in \mathbb{Z}\} \). **Problem 0.1.** If \( c \) is a common divisor of \( a \) and \( b \) then \( c|s \) for all \( s \in S(a, b) \). **Problem 0.2.** If \( s \in S(a, b) \) then \( \text{gcd}(a, b)|s \). **Problem 0.3.** If \( s \in S(a, b) \) then \( xs \in S(a, b) \) for all \( x \in \mathbb{Z} \). **Problem 0.4.** If \( S(a, b) = \mathbb{Z} \) if and only if 1 \(\in S\). **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 \(\mathbb{Z}\
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