The relation $\sim$ is defined on $\mathbb{Z}$ by $m\sim n$ if the HCF of $m$ and $n$ is $3$. a)Show that $\sim$ is neither reflexive or transitive.
b)Show that $\sim$ is symmetric.
If the HCF of $m$ and $n$ is 3, then the HCF of $n$ and $m$ is also $3$.
I think that my answer to b) is correct but I don't understand a). Can anyone point me in the right direction?
Thanks

Your answer for (b) does seem correct, as you said.
Reflexive means that every element is related to itself. In particular, we would have needed to show $\forall n \in \mathbb{Z}$, $n$~$n$. This is clearly not true. The HCF of any number with itself is the absolute value of itself: so the HCF of $n$ and $n$ is of course just $|n|$. There are many cases for which $|n| \neq 3$, hence this relation is not reflexive.
Transitive means that if I have $x, y, z \in \mathbb{Z}$, if $x$~$y$, and $y$~$z$, then $x$~$z$. Suppose that $x = z = 6$ and that $y = 3$. Clearly in this example, $x$~$y$ since the HCF of $x$ and $y$ is 3 and same for $y$~$z$. But the HCF of $x$ and $z$ is 6, so $x$ is not related to $z$ by ~.
