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if $m|a$ and $m|b$ then $(a/m,b/m)=(a,b)/m$

proof

show $(a/m,b/m)\leq (a,b)/m$ and $(a/m,b/m) \geq (a,b)/m$

Let $(a,b)=d$, so by bezout's identity there exists intergers x,y such that ax+by=d

$$ax+by=d$$ $$ax/m+by/m=d/m$$

This implies that $gcd(a/m,b/m)|d/m$, so $gcd(a/m,b/m)\leq d/m$

I just could show that direction. Need help on the other one.

TheMathNoob
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  • $gcd(a,b) = \max {k \ \mid \ k | a, k | b}$ so that $gcd(am,bm) = \max {k \ \mid \ k | ma, k | m b} =\max { m k \ \mid \ k | a, k | b} = m \ gcd(a,b)$ – reuns Jun 22 '16 at 06:20
  • so this doesn't have to do anything with my approach right?. It's a new proof. I have to prove gcd(a/m,b/m)=gcd(a,b)/m – TheMathNoob Jun 22 '16 at 06:22
  • the point is that I started from a definition of $gcd(a,b)$ (you didn't) and I showed it, assuming the middle step $ \max {k \ \mid \ k | ma, k | m b} =\max { m k \ \mid \ k | a, k | b} $ is obvious (why ?) – reuns Jun 22 '16 at 06:27
  • for what I understand, you are showing $gcd(ma,mb)=mgcd(a,b)$ right? – TheMathNoob Jun 22 '16 at 06:29

3 Answers3

1

Hint $\ $ You can unify both directions and simplify and generalize the proof as follows

$$ d\mid (a,b)/m \!\iff\! dm\mid (a,b) \!\iff\! dm\mid a,b \!\iff\! d\mid a/m,b/m \!\iff\! d\mid(a/m,b/m)$$

So $\ (a,b)/m\, =\, (a/m,b/m)\ $ since they have same divisors $\,d,\,$ so they divide each other.

Remark $\ $ See here for a few more proofs of this gcd Distributive Law.

Community
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Bill Dubuque
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First, We need to prove that $\gcd(ka,kb)=k\gcd(a,b)$. Set $d=\gcd(a,b)$. since $d\mid a$ and $d\mid b$, then $kd\mid ka$ and $kd\mid kb$, so $kd\mid \gcd(ka,kb)$.

conversely, $\gcd(ka,kb)\mid ka$ and $\gcd(ka,kb)\mid kb$, so $\gcd(ka,kb)\mid kax+kby$ for every $x,y\in Z$. Let $x_0,y_0\in Z$ such that $d=ax_0+by_0$, so in particular, $\gcd(ka,kb)\mid kax_0+kby_0$, that is $\gcd(ka,kb)\mid kd$.

Now, your claim follows almost immediately, since
$$ m\gcd(a/m,b/m)=\gcd(a,b) $$

boaz
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  • I already proved that and took the same approach. How is that important to the proof we are trying to do? – TheMathNoob Jun 22 '16 at 06:36
  • sorry, I don't see how it follows so easy. – TheMathNoob Jun 22 '16 at 06:40
  • ok I think now I see something. We proved that $ gcd(ka,kb)=kgcd(a,b)$, so if we pull out an imaginary m from a,b we get $mgcd(a/m,b/m)$ which equals $gcd(a,b)$. – TheMathNoob Jun 22 '16 at 06:43
  • Generally, when working with $\gcd$ i don't recommend using inequalities. – boaz Jun 22 '16 at 06:45
  • I saw a similar proof online and I thought I could apply it in this case. I really liked this proof because even though it was long, it was very clear to me.https://math.dartmouth.edu/~m31x13/Proof%20Examples.pdf – TheMathNoob Jun 22 '16 at 06:47
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The other direction is almost the same:
By Bézout identity, there are $x, y$ such that $(a/m)x+(b/m)y=\gcd(a/m,b/m).$ Then multiply the equation by $m$ and obtain $$ax+by=m\cdot\gcd(a/m,b/m).$$ Hence $\gcd(a,b)\mid m\cdot\gcd(a/m,b/m).$
Combined with what you proved, this proves the statement.

Hope this helps.

awllower
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