let $J_m$ be the set of all integral multiples of the integer $m$. 1) Prove that $J_{g.c.d(m,n)}$ contains all the elements of $J_m$ and $J_n$.

2) Prove that if $J_r$ contains all the elements of $J_m$ and $J_n$, then $J_r$ contains all the elements of $J_{g.c.d(m,n)}$.

for part 1) the elements of $J_{g.c.d(m,n)}$ are of the form $c*g.c.d(m,n)$, where $c\in \mathbb{Z}$.

Now if $x$ is in $J_m$, then ∃ $a\in \mathbb{Z}$ | $am=x$. So we want $x=am=b*g.c.d(m,n)$ for some $b\in \mathbb Z$. If we pick $b=am/g.c.d(m,n)$, then our task is done.

Now for part 2) ∀ $a,b\in \mathbb Z$, $am,bn\in J_r$

To show that this implies that ∀ $a\in \mathbb Z$ $a*g.c.d(m,n)$ is in $J_r$ I am assuming that something like $am*g.c.d.(m,n)$ which is in $J_r$ can be shown to go through all the integers; meaning, if $am$ can be shown to go through all integers I would be done. But right now I am not seeing the solution.

Any hint or help would be appreciated. Thank you in advance.


Hint $ $ Note $\, J_n = n\Bbb Z,\ $ and $\ m\Bbb Z \supseteq n\Bbb Z \iff m\mid n\ \ $ [contains = divides]

Translating $\ \gcd(m,n)\Bbb Z\supset m\Bbb Z,n\Bbb Z\iff \gcd(m,n)\mid m,n$

and $\ r\Bbb Z\supseteq m\Bbb Z,n\Bbb Z\,\Rightarrow\,\gcd(m,n)\Bbb Z\ \iff\ r\mid m,n\,\Rightarrow\,r\mid \gcd(m,n)$

The first follows from the fact that $\,\gcd(m,n)\,$ is, by definition, a common divisor of $\,m,n.\,$

The second follows by $\,r\mid m,n\,\Rightarrow\, r\mid jm\!+\!kn = \gcd(m,n)\,$ for some $\,j,k,\in\Bbb Z,\,$ by Bezout.

Remark $\ $ Both can be combined into the following universal property of the gcd

$$ r\mid m,n\iff r\mid \gcd(m,n)$$

  • $\begingroup$ ok I am still confused as I do not understand parts of the hint. $\endgroup$ – Vladimir Louis Feb 23 '15 at 0:41
  • $\begingroup$ @Vladimir What parts? $\endgroup$ – Bill Dubuque Feb 23 '15 at 0:42
  • $\begingroup$ First line, the condition for the superset is that m|n? I do not see that right away. $\endgroup$ – Vladimir Louis Feb 23 '15 at 0:43
  • $\begingroup$ @Vladimir $\ n\in n\Bbb Z \subseteq m\Bbb Z\,$ so $\ \ldots$ $\endgroup$ – Bill Dubuque Feb 23 '15 at 0:46
  • $\begingroup$ ok I will study the hint. If I have any question I will let you know, thanks again! $\endgroup$ – Vladimir Louis Feb 23 '15 at 0:49

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