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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.

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    $\begingroup$ $2,4,4 \qquad{}$ $\endgroup$
    – user223391
    Jan 29, 2017 at 21:10
  • $\begingroup$ $4\mid 12$ and $6\mid12$ but $4\times6\nmid12. \qquad$ $\endgroup$
    – Michael Hardy
    Jan 29, 2017 at 21:37
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    $\begingroup$ Fun fact: If $a\mid c$ and $b\mid c$ for coprime $a$ and $b$, then $ab\mid c$. $\endgroup$
    – ryanblack
    Jan 29, 2017 at 21:40

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This is not true.

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

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

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

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

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

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    $\begingroup$ Womp. This was the example I was going to use. $+1$! $\endgroup$
    – Cameron Williams
    Jan 29, 2017 at 21:41
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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!

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  • $\begingroup$ $a = 3, b = 5, c=5, gcd(a,b) = 1$ but the $ab \not \mid c$ $\endgroup$
    – Rezwan Arefin
    Jan 29, 2017 at 21:25
  • $\begingroup$ True that, ill change this up in a second, I was thinking of an extension to Euclid's Lemma $\endgroup$
    – Carlos Seda
    Jan 29, 2017 at 21:29

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