# If $a\mid c$ and $b\mid c$, must $ab$ divide $c$?

If $a\mid c$ and $b\mid c$, must $ab$ divide $c$? Justify your answer.

$a\mid c$, $c=ak$ for some integer $k$

$b\mid c$, $c=bu$ for some integer $u$

From here I wanted to try to check if there were counter examples I could use,

$c\ne(ab)w$ for some integer $w$

From here I got stuck because there is nothing I can plug into that equation so I know that I am probably missing something.

• $2,4,4 \qquad{}$ – user223391 Jan 29 '17 at 21:10
• $4\mid 12$ and $6\mid12$ but $4\times6\nmid12. \qquad$ – Michael Hardy Jan 29 '17 at 21:37
• Fun fact: If $a\mid c$ and $b\mid c$ for coprime $a$ and $b$, then $ab\mid c$. – ryanblack Jan 29 '17 at 21:40

This is not true.

Take for instance $a=b=2$ and $c=2$.

Then $a\mid c$ and $b\mid c$, but

$$ab=4\nmid 2=c.$$

$4\mid 12$ and $6\mid12$ but $4\times6\nmid12. \qquad$

The proposition is true when $\gcd(a,b)=1.$

• Womp. This was the example I was going to use. $+1$! – Cameron Williams Jan 29 '17 at 21:41

No it must not, you can give a simple counter-example like $$3\mid 9$$ and $$9\mid 9$$ but obviously $$ab=9\cdot3=27 \nmid9$$ Hope it helps you out!

• $a = 3, b = 5, c=5, gcd(a,b) = 1$ but the $ab \not \mid c$ – Rezwan Arefin Jan 29 '17 at 21:25
• True that, ill change this up in a second, I was thinking of an extension to Euclid's Lemma – Carlos Seda Jan 29 '17 at 21:29