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  1. Show that if $d$ is the $\gcd$ of $a_1,a_2,\dots,a_n$. then $d=\sum t_ia_i$, where $t_i$ is in $\mathbb{Z}$, for all $i$ with $1\leq i\leq n$.

I thought in using induction, instead of generalizing the typical proof of using the Least Integer Axiom, but I can't see how to ge it.

  1. Show that $\gcd(a,b,c)$ is equal to $\gcd(a,\gcd(b,c))$.

Here I would like to ask if the proof is correct.

Set $f=\gcd(b,c)$, $d=\gcd(a,f)$. Therefore:

$$ d=s_1a+s_2f = s_1a+s_2(r_1b+r_2c)=s_1a+(s_2r_1)b+(s_2r_2)c, $$

On the other hand, if $d$ is $\gcd(a,b,c)$, then

$$ d=t_1a+t_2b+t_3b. $$

If we identify $t_2=s_2r_1$ and $t_3=s_2r_2$, then $d=gcd(a,b,c)=gcd(a,gcd(b,c))$.

EDIT: Another approach.

Suppose $d=gcd(a,b,c)$, then $d|a$, $d|b$, $d|c$. Now, the $gcd$ of $b$ and $c$ is, say $f$. Since $d=gcd(a,b,c)$, then $f|d$, and thus $d|gcd(b,c)$.

Conversely, if $d=gcd(a,gcd(b,c))$, then $d|a$ and $d|gcd(b,c)$, then $d|b$ and $d|c$. Therefore $d=gcd(a,b,c)$.

user2820579
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    What is being asked in the first problem? – SvanN Jul 31 '17 at 17:04
  • I think you omitted the question (or part of it). – lulu Jul 31 '17 at 17:04
  • Not at all, it is to show that that the $gcd$ of $n$ integer numbers exist, then there is a linear combination $d=\sum t_ia_i$. – user2820579 Jul 31 '17 at 17:07
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    Your proof in part $2$ looks incomplete. All you show is that $d=\gcd(a,\gcd (b,c))$ can be written as a linear combination of $a,b,c$. That shows that $\gcd(a,b,c)$ divides $d$, not that they are equal. – lulu Jul 31 '17 at 17:12
  • For the first question, if the $t_i$ are allowed to be $0$, then it suffices to show the result for $n=2$. – Aurel Jul 31 '17 at 17:13
  • How do you define the g.c.d.? – Bernard Jul 31 '17 at 17:19
  • For those who ask if the questions are incomplete, I got them from Rotman's A first course in abstract algebra, 3rd edition, question 1.65 (i). The second one is question 1.66 (i). – user2820579 Jul 31 '17 at 17:21
  • @lulu and now is it complete? – user2820579 Jul 31 '17 at 17:27
  • I don't understand the new part of the argument, and using "$d$" to mean two a priori different things is very confusing. Side note: the reason I and (I presume) the other commenter said your question was incomplete was that your first version omitted the "$d=$ ". You corrected that in a subsequent edit. – lulu Jul 31 '17 at 17:35
  • I would structure your proof for part $2$ in two parts. Let $d=\gcd (a,\gcd (b,c))$ and let $\delta = \gcd (a,b,c)$. You have shown that $\delta ,|,d$. Now show that $d,|,\delta$. – lulu Jul 31 '17 at 17:36
  • that's true about the $d$, I kept at the first writting the question as it is in the book. For the proof, let me correct it with another approach (without using linear combinations), and that uses what you are suggesting. – user2820579 Jul 31 '17 at 17:37
  • You should also respond to the question @Bernard asked. You appear to believe that you can define $D=\gcd (A,B)$ by saying that $D=\lambda A+\mu B$ but this is not true. You need to add that $D$ is the minimal natural number that can be so written. Or, alternatively, just use the old definition as the greatest common divisor of $A,B$. – lulu Jul 31 '17 at 17:40
  • Your EDIT is not correct. We have $d|f$, not the other way around. – Ross Millikan Jul 31 '17 at 19:24
  • What do you want the other way around. It is true that $d\leq f$, but the fact that $f$ is bigger is immaterial, because $d|a$. Besides, the Euclidean algorithm gives $f$, and using it again it will give $d=gcd(a,f)$. – user2820579 Jul 31 '17 at 22:24

2 Answers2

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The set $ \,S\,$ of integers of form $ \,a_1\,x_1 + \cdots + a_n x_n,\ x_i\in \mathbb Z,\,$ is closed under subtraction so, by the Lemma, every positive $ \,k\in S\,$ is divisible by $ \,d = $ least positive $ \in S.\,$ Therefore $ \,a_i\in S$ $\,\Rightarrow\,$ $ d\mid a_i,\,$ i.e. $ \,d\,$ is a common divisor of all $ \,a_i,\,$ and greatest: $ \ c\mid a_i$ $\Rightarrow$ $ \,c\mid d = a_!\,x_1\!+\!\cdots\!+\!a_nx_n$ $\Rightarrow$ $ \,c\le d$

Lemma $\ \ $ Let $\, S\ne\emptyset \,$ be a set of integers $>0\,$ closed under subtraction $> 0,\,$ i.e. for all $ \,n,m\in S, \,$ $ \ n > m\ \Rightarrow\ n-m\, \in\, S.\,$ Then the least $ \:\ell\in S\,$ divides every element of $\, S.$

Proof ${\bf\ 1}\,\ $ If not there is a least nonmultiple $ \,n\in S,\,$ contra $ \,n-\ell \in S\,$ is a nonmultiple of $ \,\ell.$

Proof ${\bf\ 2}\, \,\ \ S\,$ closed under subtraction $ \,\Rightarrow\,S\,$ closed under remainder (mod), when it is $\ne 0,$ since mod is simply repeated subtraction, i.e. $ \ a\ mod\ b\, =\, a - k b\, =\, a\!-\!b\!-\!b\!-\cdots\! -\!b.\,$ Thus $ \,n\in S\,$ $\Rightarrow$ $ \, (n\ mod\ \ell) = 0,\,$ else it is in $\, S\,$ and smaller than $ \,\ell,\,$ contra minimality of $ \,\ell.$

Remark $\ $ In a nutshell, two applications of induction yield the following inferences

$ \rm\begin{eqnarray} S\ closed\ under\ {\bf subtraction} &\:\Rightarrow\:&\rm S\ closed\ under\ {\bf mod} = remainder = repeated\ subtraction \\ &\:\Rightarrow\:&\rm S\ closed\ under\ {\bf gcd} = repeated\ mod\ (Euclid's\ algorithm) \end{eqnarray}$

Interpreted constructively, this yields the extended Euclidean algorithm for the gcd.

For the GCD associative law see this answer. Hint: first prove the universal property of the gcd: $\ d\mid a_1,a_2,a_3,\ldots\! \iff d\mid (a_1,a_2,a_3,\ldots)\,$ using the above general Bezout Lemma.

Bill Dubuque
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You can assume your numbers are nonnegative. In this case, the relation $$ a\mid b \qquad\textit{for}\qquad \text{there exists $c$ such that $b=ac$} $$ is a partial order relation that defines a semilattice structure on $\mathbb{N}$. Indeed, $d=\gcd(a,b)$ has the property that

  1. $d\mid a$ and $d\mid b$
  2. for all $c$, if $c\mid a$ and $c\mid b$, then $c\mid d$

which means that $d$ is the greatest lower bound of $\{a,b\}$. Existence is guaranteed by Euclid’s algorithm.

Actually, this defines a lattice structure, where the least upper bound is the least common multiple.

Now, that $\gcd(a,\gcd(b,c))=\gcd(a,b,c)$ is easily shown with lattice techniques; more generally, you can prove by induction that $$ \gcd(a_1,\dots,a_n,a_{n+1})=\gcd(\gcd(a_1,\dots,a_n),a_{n+1}) \tag{*} $$ in the sense that the right-hand side is the greatest lower bound of $\{a_1,\dots,a_n,a_{n+1})$.

Over the natural numbers, there is the Bézout’s identity. The first statement can then be indeed proved by induction, using (*). The statement is obvious for $n=1$. Suppose it holds for $n$ numbers. Then \begin{align} \gcd(a_1,\dots,a_n,a_{n+1}) &=\gcd(\gcd(a_1,\dots,a_n),a_{n+1})\\ &=\gcd(r_1a_1+\dots+r_na_n,a_{n+1})\\ &=r(r_1a_1+\dots+r_na_n)+sa_{n+1} \end{align} which also shows how to find suitable integers: $$ t_i=rr_i \quad (i=1,\dots,n),\qquad t_{n+1}=s $$

egreg
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