1. ## [SOLVED] Simple proof

Show that if $a\mid{c}$ and $b\mid{c}$ then $ab\mid{c}$

I got started and have Let c=ae and c=bd where d and e are integers.
c=ae=bd and a and b are not equal then e=b and d=a thus $ab\mid{c}$ where the result would be 1.
I am pretty sure my logic is really messed up though.

2. Originally Posted by diddledabble
Show that if $a\mid{c}$ and $b\mid{c}$ then $ab\mid{c}$
$3\mid12~\&~6\mid 12\text{ but is it true that }18\mid 12?$

3. I forgot to add that gcd (a,b)=1 Sorry.

4. In your equations c=ae and c=bd try solving for a and b then look and the product of the two.

That should get ya there

5. So then $a=c/e$ and $b=c/d$ then $ab=e/d$ Where $e/d=c$ Is that correct?

6. ## Proof

$(a,b)=1\Rightarrow \exists s,t\in \mathbb{Z} s.t. as+bt=1$

Multiply through by c to arrive at
$asc+btc=c$

ab|asc because a|a and b|c

ab|btc because a|c and b|b

so it divides the sum which is c as desired.

7. Originally Posted by diddledabble
So then $a=c/e$ and $b=c/d$ then $ab=e/d$ Where $e/d=c$ Is that correct?

I take back my original post. Let's try another approach.

Since gcd(a,b) = 1 then there exists integers $x,y$ such that $1=ax+by$

Multiply this equation by c to yield
1) $c*1 = c = c(ax+by)=axc+byc$

Now $c=ak=bl$ where $k,l$ are integers

Substitute these values in equation 1)
$c=ax(bl)+by(ak)=abxl+abyk=ab(xl+yk)=abm$ where $m$ is an integer

Thus $c=abm$ or equivalently $ab|c$