Define the relation $\mathscr{R}$ on $\Bbb Z$ by $x \mathscr{R} y$ iff $xy >0$. I wish to examine several properties of this relation. Considering on $\mathbb{Z}$ the following relation:

$$\mathscr{R} = \left \{ (x,z) \in \mathbb{Z} \times \mathbb{Z} \quad \mbox{ such that } \quad xz &gt; 0  \right \}$$

i.e. $\forall x,z \in \mathbb{Z}, x \mathscr{R} z \iff \mbox{ the product } \, \, x \cdot z &gt; 0 $

Check if the relation $\mathscr{R}$ is:  

i) reflexive;
ii) symmetrical;
iii) transitive;
iv) anti-symmetrical;
v) of order;
vi) of equivalence.

What I have done is the following:
i) is it a reflexive relation?
$\forall x,x \in \mathbb{Z}, \quad x \mathscr{R} x \iff x\cdot x = x^2 &gt; 0 $
since a square is always $&gt; 0$ the relation is reflexive.

ii) is it a symmetrical relation?
$\forall x,z \in \mathbb{Z}, \quad x \mathscr{R} z \iff x \cdot z &gt; 0$ by hypothesis,
since the operation of multiplication is commutative on $\mathbb{Z}$ the following is true:
$\forall x,z \in \mathbb{Z}, \quad z \mathscr{R} x \iff z \cdot x &gt; 0$
hence, the relation is symmetrical.

iii) is it a transitive relation?
$\forall x,z \in \mathbb{Z}, \quad x \mathscr{R}z \iff x \cdot z &gt; 0$ by hypothesys,
if we take another ordered pair
$\forall z,m \in \mathbb{Z}, \quad z \mathscr{R} m \iff z \cdot m &gt; 0$
hence,
if $x \mathscr{R}z, \quad z \mathscr{R} m \Rightarrow x \cdot m &gt; 0$
i.e. if $x \cdot z &gt; 0, \quad z \cdot m &gt; 0 \Rightarrow x \cdot m &gt; 0$
so the relation is transitive.

iv) is it a anti-symmetrical relation?
No. Because it is not true that $z \mathscr{R}x \iff x=z$, as we have seen in the symmetrical relation.

v) is it a order relation?
No. Because it is not anti-symmetrical.

vi) is it a equivalence relation?
Yes, because it is reflexive, symmetrical and transitive.

What do you think about it?
Please, can you give me any suggestions? Many thanks!
$
\newcommand{\RR}{\mathscr{R}}
\newcommand{\Z}{\mathbb{Z}}
$
Note, intuitively, what we have. This relation $\RR$ on $\Z$ essentially means $x \RR y$ if and only if $x,y$ have the same sign and are nonzero. (Same sign ensures nonnegative, nonzero ensures positive.) Hopefully this makes the intuition clearer.
Onto the errors with your approach:

You were wrong on reflexivity. Yes, $x^2 &gt; 0$ for any $x \in \Z$, but only if $x \ne 0$.
Your justification for transitivity is unclear, in my opinion.
Your justification for antisymmetry seems to amount to &quot;it's symmetrical and thus not antisymmetrical.&quot; The properties are not necessarily opposed; a relation can be both, or neither. You're right in that this relation is not antisymmetrical, though, but even so. (In particular, a relation on $A$ of the form $\{(x,x) \mid x \in X\}$ for some $X \subseteq A$ is both antisymmetric and symmetric.)

Onto the overall right ideas:

Reflexivity: This does not hold, because $(0,0) \not \in \RR$. This is because $0 \cdot 0 = 0 \not &gt; 0$.

Symmetry: This does hold under commutativity of integer multiplication. Assume $(x,z) \in \RR$. Then $xz &gt; 0$. But then $xz = zx &gt; 0$. Thus, $(z,x) \in \RR$ on the premise $(x,z) \in \RR$.

Transitivity: This does hold. It's easiest to refer back to the intuitive idea: $(a,b) \in \RR$ if and only if $a,b$ are nonzero and share the same sign. Suppose $(a,b),(b,c) \in \RR$. Then $a,b$ share the same sign, and are nonzero; likewise, the same is true of $b,c$. Thus, $a,b,c$ are nonzero and have the same sign. Thus, $a,c$ have the same sign and are nonzero, and thus $(a,c) \in \RR$.
Of course, this can be verified algebraically too. Since $(a,b),(b,c) \in \RR$, then $ab&gt;0$ and $bc&gt;0$. Multiply the inequalities and you have $ab^2c &gt; 0$. Since $b \ne 0$ (otherwise neither inequality would hold) and $b^2 &gt;0$ as a result, we may divide both sides by $b^2$ and see $ac &gt; 0$. Thus, $(a,c) \in \RR$.

Antisymmetry: Anti-symmetry requires that, when $(a,b) \in \RR$ and $(b,a) \in \RR$, that $a=b$. However, commutativity of multiplication allows them to be distinct. More specifically: $(1,2),(2,1) \in \RR$ since $1 \cdot 2 = 2 = 2 \cdot 1 &gt; 0$, but obviously $1 \ne 2$.

Order: An order relation requires antisymmetry, so we don't have this.

Equivalence: Since we don't have reflexivity, we don't have equivalence.
$x\cdot x\geq0$  (is always greater than zero), it is reflexive.

$x\cdot z\geq0,\,z\cdot x&gt;0$, it is symmetry

$x\cdot z&gt;0,\,z\cdot y&gt;0,\,x\cdot y&gt;0,\,x\cdot z\geq0$ therfore fore $x\cdot z=x\cdot y$ it is transitivity