Is my understanding of Binary relations correct? On my assignment it asks 

Determine with reason if the binary relation is reflexive, symmetric,
  antisymmetric or transitive.
$$R = \{(a, b) \in \mathbb{Z} \times \mathbb{Z} \mid a \text{ is an
 integer multiple of }b\}$$

I believe it is reflexive because a number is a multiple of itself
But I think it is not symmetric because take $(12,3)$, $12$ is a multiple of $3$, so $(12,3)$ exists in $R$, but $3$ is not a multiple of $12$ therefore $(3,12)$ is not in $R$. Making it not symmetric
The part that I can't make up my mind is about is whether or not it's antisymmetric, our textbook says a binary relation $R$ on a set $A$ is antisymmetric if and only if $a,b \in R$ and both $(a,b)$ and $(b,a)$ are in $R$, then $a=b$.
And in that case would it be antisymmetric if I use $(1,1)$ as $(a,b)$? since $1$ is a multiple of $1$, then $(a,b)$ and $(b,a)$ are both in $R$, and $a=b$, making it antisymmetric?
I would really appreciate it if someone can help me out on this, thanks.
 A: You are correct about $R$ being reflexive and not symmetric. 
However, you have not proved that it is antisymmetric. You have only given a case that is antisymmetric, namely $(1,1)$, but you would need to show that it is true for all cases. You would need to prove the following statement:

For all $(a,b)\in \mathbb{Z}\times \mathbb{Z}$ the following holds: If $(a,b)\in R$ and $(b,a) \in R$, then $a=b$. 

However, you will not be able to prove the statement because it is false. To demonstrate that it is false you need only one counter example. Can you see why any case of the form $(-a,a)$, where $a\neq 0$ makes it false? 
A: Let $(a,b) \in \mathbb{Z}^2$. 
If $(a,b) \in R$, then $a = m b$ for some $m \in \mathbb{Z}$.
If $(b,a) \in R$, then $b = n a$ for some $m \in \mathbb{Z}$.
So if $(a,b), (b, a) \in R$ then
$$
a = m b = m (n a) = (m n) a \iff m n = 1 \iff m = n = 1 \vee m = n = -1
$$ 
The first case means $a = b$ and the second means $a = -b$.
The second case gives us counter examples.
e.g. $(1, -1) \in R$ and $(-1, 1) \in R$, but $-1 \ne 1$. 
