Prove or show a counter example for: $\forall a, b, c \in \mathbb Z$, if $ab\mid c$ then $a\mid c$ and $b\mid c$ I'm working on my Discrete Mathematics homework and they are asking me this:
Prove or show a counterexample for: $\forall a, b, c \in \mathbb Z$, if $ab\mid c$ then $a\mid c$ and $b\mid c$
I'm not completely sure how to prove it. I was thinking to find a counterexample like zero but I'm not sure if that it's going to work. Or maybe try to approach it backwards? 
I'm a little bit lost
 A: Hint: If $ab\mid c$, then, since $a\mid ab$, …
A: think about what they are saying and maybe do a simple example just to get the concept down.
Take $a = 5; b=7;$ and $c = 9$ then if $ab = 35$ and if $35|9$ .... oh, wait, that won't work because $35$ doesn't divide $9$.  So what does?  Well, $2*35 = 70$ so $35|70$.  
So Take $a = 5; b=7$ and $c = 70$ then if $35|70$ (it does $70 = 2*35$) then does $5|70$ (it does $70= 5*14$) and does $7|70$ (it does $70= 7*10$).
Did that have to happen?  Well, .... what do you think? 
What if  we wrote them out as visible products rather than number.
We have $5*7|2*5*7$ and we have $5|2*5*7$ and we have $7|2*5*7$.  Did that have to happen?
I'm hoping it's pretty clear.  How'd you put it in words?  Well, I'd say some thing like "Well, $2*5*7$ is made out of $5*7$ and $5*7$ is made out of $5$ and $7$ so is something is made out of something that is made of smaller things, it must be made out of the smaller things themselves, right?"
And that's the basic idea.
But of course, that's not math.  That's just thinking aloud.  BUT, and it's an  important but, that is the reason.  If something is made out of something that is made out of smaller things, than that something is made out of the smaller things.  That IS why the statement must be true.
So let's make it math:
What does $m|n$ mean.  It means that there is an integer $k$ so that $n = m*k$.  So we are asked to prove if $ab|c$ then and so there exists and integer, $k$ so that $c = (ab)k$ then we must prove $a|c$ and $b|c$ or that there are integers $j,m$ so that $c = aj$ and $c = bm$.
So we KNOW $c = (ab)k$ and we want to show that $j,m$ exist so that $c = (ab)k = aj = bm$.
Well, if $(ab)k = aj$ then the integer that we need is $j = (bk)$.  Let's take it.  If $j = bk$ then $aj = a(bk) = (ab)k = c$ and $a|c$ QED.
And if $(ab)k = bm$ then the integer that we need is $m= ak$.  Let's take it.  If $m = ak$ then $bm = b(ak) = (ab)k = c$ and $b|c$ QED.
That's that.
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I'm not completely sure how to prove it. I was thinking to find a counterexample like zero

It's not either/or.  You can't just decide to find a counter-example because you don't know how to prove it.
Either it is true, or it isnt.
If it is true then there are not counter examples.
If it isn't ture then you will not be able to prove it.
Counter-examples can only exist if it is false.  A proof can only exist if it is true.  And whether it is true or false isn't a choice left to us.
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And as a tangent.  This is a bit off topic but $0$... What about $0$?
Every number divides $0$.  If $a$ is an number then $a*k = 0$ if $k = 0$.  So $a|0$.
But zero does not divide any number other than $0$.  If $p\in \mathbb N$ then there is no integer $k$ so that $0*k = p$.  SO $0\not \mid p$.
But $0|0$.
I think.....
Some books define $m|n$ if $\frac nm\in \mathbb Z$.  If your book does it that way then all numbers, $a$, except $0$ divide $0$ so $a|0$ if $a\ne 0$.  And $0$ does not divide any numbers.
A: I know how it feels when someone is so unsure about what to do or so lost deep in theorems and lemmas and whatnot that no first step seems reasonable.
However, one can always try some small examples provided it doesn't take much effort to work them through. For example, try to assign values to $a,b$ and $c$.
Say you would take $a = b = 1$. That doesn't seem fruitful, because then $a$ and $b$ will divide any number, right? Then try changing $b$ to something else. Then try a few different values of $c$. That should give you a hint as to whether you should try to prove or disprove the claim.
If you stumble upon a counter-example, great. If you try a lot of small examples and they all comply with the claim, then maybe it is true.
