Prove by contradiction if $a$, $b$ and $c$ are integers and $bc$ is not divisible by $a$, then $b$ is not divisible by $a$ So I'm new to proofs as a whole topic and have been set a series of proofs to complete. All in all I have felt confident in my responses, however there is one that has been tripping me up.

"Prove by contradiction if $a$, $b$ and $c$ are integers and $bc$ is not divisible by $a$, then $b$ is not divisible by $a$."
I've come to assume... if $bc$ is divisible by $a$ then $b$ is not divisible by $a$.
If $a$, $b$ and $c$ are even then $a = 2k$, $b = 4k$, $c = 6k$
$bc = 4k * 6k = 24k$
$bc/a = 24k/2k = 12$
Therefore, if $bc$ is divisible by $a$, $b$ is divisible by $a$ so if $bc$ is not divisible by $a$ then $bc$ is not divisible by $a$.

It's all a little bit messy and I'm not sure if I'm correct to assume that $a$, $b$ and $c$ are even since the same case may not work with odd numbers... but I don't know what else to do.
Thanks!
 A: You're right in your suspicion that assuming $a,b,$ and $c$ to be even is too strong for the proof. You want to show it for all integers, not just the even ones.
First we assume that $a,b,$ and $c$ are integers and that $bc$ is not divisble by $a$ but $b$ is divisble by $a$. We will show that our assumption that $b$ is divisble by $a$ contradicts our assumption that $bc$ is not divisble by $a$.
So, what does it mean that $b$ is divisble by $a$? Intuitively, we know that if $b$ is divisble by $a$ then when I divide I should get an integer result. Another way to write this is that $b = d\cdot a$ for some integer $d$. This is just a way of writing that $\frac{b}{a} = d$ where $d$ is an integer. With this representation of $b$, we can revisit $bc$ to get $bc = (d\cdot a)c$. Since multiplication is associative, we have $(d\cdot a)c=dac$. Now, using commutativity of multiplication we can further write $dac = adc$ and since $d$ and $c$ are both integers we have $adc = a\cdot z$. In the end, this means that $bc = a\cdot z$ where $z$ is some integer. But this is precisely the condition for $bc$ to be divisble by $a$! We can rewrite $bc = a\cdot z$ as $\frac{bc}{a} = z$ and it's clear we've contradicted our initial assumptions that $\frac{bc}{a}$ would not be an integer.
A: Suppose that $a$ does not divide $bc$. Suppose $a\mid b$; that is, that $a$ divides $b$. Then $b=ak$ for some natural $k$. Now $bc=(ak)c=a(kc)$, which by definition means that $a$ divides $bc$ since $kc$ is an integer. This is a contradiction. Thus $a$ does not divide $b$.
