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for example, I have two numbers: 36 and 48. their gcd is 12. 12 can be divided by any common divisors of these two numbers: 1, 2, 3, 4, 6, 12. Is there a general proof for this? Thanks!

Bernard
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giorgi shengelaia
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    It can be proved building the gcd with prime factorization. – Crostul Sep 19 '17 at 15:37
  • You can find the standard proof in the linked dupe. It works by showing that the gcd has linear form $,\gcd(a,b) = ja+kb,$ which makes it clear that $,c\mid a,b,\Rightarrow,c\mid \gcd(a,b).\ $ – Bill Dubuque Sep 19 '17 at 15:46

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Let $\gcd(m,n) = d$. Then $m= d*m'$ and $n=d*n'$ for some integers $m',n'$.

Now if $m', n'$ had any factor in common (other than $1$), and we called that factor $f$, then $f*d$ would be a common divisor that is larger then $d=\gcd(m,n)$. So we can safely conclude $m',n'$ (the remains as it were) have no factors in common[$*$].

If $a$ is a common factor of $m$ and $n$ it can not share any factors (other than $1$) with either $m'$ nor $n'$ as $m',n'$ have no factors (other than $1$) in common.

Now $a|m$ so $\frac ma = \frac {dm'}a = k$ is an integer. $\frac na = \frac {dn'}a = j$ is an integer. If $a\not \mid d$ then $a $ will have factors that do not divide $d$ and must therefore divide $m'$ and $n'$. But $m'$ and $n'$ have no factors in common.

That's it.

[$*$] Note, it would be a mistake to conclude that $m'$ and/or $n'$ will have no factors in common with $d = \gcd(m,n)$ or with $a$. It is possible that one of $m'$ or $n'$ will have some factors in common with $d$ or with $a$. but it is impossible that both $m'$ and $n'$ will have factors in common.

fleablood
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  • Thanks, I get it now :) but I think it should state: "m|a" and "d∤a" instead of "a|m" and "a∤d". isn't "|" same as "/" (division slash)? – giorgi shengelaia Sep 20 '17 at 11:44
  • No. In fact they are almost opposites. $a|d $ means "a divides into d". Such as $5|15$ means 5 divides into 15. $\frac ad $ means the number you get when you divide a by d. So $\frac 5 {15}=\frac 13$. We know: $a|d \iff \frac da$ is an integer. – fleablood Sep 21 '17 at 06:21
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Claim: If $a\vert b$ and $b\vert c$, then $a\vert c$.

Proof: $c=kb=k(ma)=(km)a$.

Laars Helenius
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The definition of $d=\gcd(a,b)$ is that it is the positive generator of the ideal $(a,b)\subset\mathbf Z$.

So there exist integers $u,v$ such that $ua+vb=d$. Now if $c$ divides both $a$ and $b$, we can write $a=ca'$, $b=cb'$ -$a',b'\in\mathbf Z$. We deduce at once $$d=uca'+vcb'=c(ua'+vb').$$

Bernard
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