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Let $a$ and $b$ be integers. Then $(ca,cb)=c(a,b)$ for any positive integer $c > 0$.

I have proved this using Bezout's lemma. However, I am trying to prove this again by following the sketch of the proof from The Theory of Numbers: A Text and Source Book of Problems which states:

Note that the smallest positive integer which is a linear combination of $ca$ and $cb$ is simply $c$ times the least positive integer which is a linear combination of $a$ and $b$. Now apply (1.5)

Theorem 1.5 in this case is:

Suppose $a$ and $b$ are not both $0$, and let $d=(a,b)$. Then $d$ is the smallest positive integer that can be expressed as a linear combination of $a$ and $b$.

This is the proof I came up with, but the argument feels like it is missing something.

Proof. Let $a,b,c \in \mathbb{Z}$ such that $c > 0$ and $a$ and $b$ are not both zero. Let $S = \{cak + cbm : k,m \in \mathbb{Z}\}$. Now, consider the subset $S' = \{x \in S: x > 0\}$. $S'$ has a smallest element by the Well Ordering Principle, and by definition this element is $d=(ca,cb)=c(ak+bm)$. Since $d>0$ and $c>0$, this means that $ak + bm > 0$. Note that multiplication over the integers preserves ordering, and since $d$ is the smallest element of $S'$ this means that $ak + bm$ is the smallest positive linear combination of $a$ and $b$. By definition, this is $(a,b)$. Hence, $(ca,cb)=c(a,b)$. $\blacksquare$

wildcat
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  • Related. – PinkyWay Oct 16 '21 at 21:10
  • Does this answer your question? Prove that $(ma, mb) = |m|(a, b)\ $ [GCD & LCM Distributive Law] – PinkyWay Oct 16 '21 at 21:12
  • Depends on how you are defining the GCD. If it's defined as the smallest positive integer that is a linear combination of the inputs, this works. If the GCD is defined by the classical definition, you need Bezout for this to work. – Rushabh Mehta Oct 16 '21 at 21:12
  • See the answer by Bill Dubuque in the thread. – PinkyWay Oct 16 '21 at 21:12
  • @DonThousand Yes, the book defines $(a,b)$ to be the smallest positive integer that is a linear combination of $a$ and $b$. – wildcat Oct 16 '21 at 21:22
  • @wildcat In that case, this is valid. – Rushabh Mehta Oct 16 '21 at 21:22
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    I have looked at Bill Dubuque's answer and I understand the inductive and Bezout proof. I haven't taken abstract algebra yet so I'm not sure I get the other proofs completely, but ultimately I want to prove this using the approach specified in the book I am working through. – wildcat Oct 16 '21 at 21:23
  • @wildcat If your books defines $(a,b)$ as the smallest linear combination, it's strange that this fact is also a Theorem (theorem 1.5). – subrosar Oct 16 '21 at 21:25
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    @subrosar Clarification: The book gives multiple definitions/theorems to characterize $(a,b)$. The first definition is simply that $(a,b)$ is the largest common divisor of $a$ and $b$. They later prove Theorem 1.5 which states $(a,b)$ is the smallest positive linear combination of $a$ and $b$. The third definition they give is that $(a,b)$ is a unique positive integer that divides both $a$ and $b$, and if any integer $e$ divides $a$ and $b$, then it also divides $(a,b)$ (although I would say that this is a more rigorous version of the first definition). – wildcat Oct 16 '21 at 21:44
  • The key idea is simple: $, c(a\Bbb Z + b\Bbb Z) = ca\Bbb Z + cb\Bbb Z,$ are equal sets (with positive elements) so they have the same least positive element - which yields the sought gcd equality by applying the characterization of the gcd given by Theorem $1.5$. This structural viewpoint will become clearer when one studies ideals in ring theory. – Bill Dubuque Oct 18 '21 at 21:48
  • One way to prove the claimed set equality is as below $$\begin{align},n \in ca\Bbb Z + cb\Bbb Z \iff &\exists, i,j!:\ \ \ \ \ n = ca i + cb j\ \iff &\exists, i,j!:\ n/c =\ \ a i, +, b j\ \ \ &\ \ \ c\mid n\ \iff &\ \ \ \ \ \ \ \ \ \ \ , n/c\ \in\ a\Bbb Z + b\Bbb Z\ \iff &\ \ \ \ \ \ \ \ \ \ \ \ n\ \in\ c(a\Bbb Z + b\Bbb Z) \end{align}$$ – Bill Dubuque Oct 18 '21 at 22:07

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