Finding a counterexample

Question

Why is this wrong? Doesn't $a | b$ mean $a$ divides $b$ so $b = a(x)$?

So then if $a=3, b=4, d=7$ then

$d | ab$ will mean $12 = 7(12/7)$, $d | a$ will mean $3 = 7(3/7)$, $d | b$ will mean $4 = 7(4/7)$

so $d | ab$ is not equal to $d | a$ or $d | b$

doesn't that mean this is a counterexample?

I must be not understanding it correctly.

Can someone please explain?

Answer 1

$a|b$ means that there exists a number $z\in\mathbb{Z}$ such that $az=b$. For example, $3|18$ is true because there is an interger $z$ ($6$ in this case) such that $3z=18$. $4|5$ is false because there is no interger $z$ such that $4z=5$. So we say that $4\nmid 5$. $7|-28$ is true because $z=-4$ (which is clearly an interger) satisfies $7$x$z=-28$

As a counter example you can consider $a=2, b=6,$ and $d=4$. So we have that $ab=12$ and $4|12$ (since $4$x$3=12$). But $4\nmid 2$ and $4\nmid 6$

Answer 2

A valid counterexample is a pair $(a,b,d)$ for which $d \mid ab$ but $d$ divides neither $a$ nor $b$. You took $a = 3$, $b = 4$, and $d = 7$. Observe that $d$ does not divide $ab = 12$, so the implication in question is vacuously true.