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Considering on $\mathbb{Z}$ the following relation:

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

i.e. $\forall x,z \in \mathbb{Z}, x \mathscr{R} z \iff \mbox{ the product } \, \, x \cdot z > 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 > 0 $ since a square is always $> 0$ the relation is reflexive.

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

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

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    $\begingroup$ (i) is wrong because $z=0$ does not satisfy $z\cdot z>0$. $\endgroup$ – vadim123 Feb 19 '17 at 20:35
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    $\begingroup$ ii) is phrased strangely. The third line as it is written does not follow from the fact that multiplication is commutative but rather follows directly from the definition of the relation. What you mean to write is $x\mathcal{R}z\iff x\cdot z>0\iff z\cdot x>0\iff z\mathcal{R}x$ $\endgroup$ – JMoravitz Feb 19 '17 at 21:15
  • $\begingroup$ @amwhy yes... my argument does. His however as it is originally written seems incoherent and doesn't appear to link $x\mathcal{R} z$ to $z\mathcal{R} x$ but instead goes off on a tangent and just restates the original definition of the relation with different labels. He never stated that $x\mathcal{R} z$ implies or doesn't imply $z\mathcal{R} x$. What he wrote is equivalent to $x\mathcal{R} z \implies (z\mathcal{R} x\iff z\cdot x>0)$ which is not the same thing as $x\mathcal{R} z \implies z\mathcal{R} x$. $\endgroup$ – JMoravitz Feb 19 '17 at 21:31
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$x\cdot x\geq0$ (is always greater than zero), it is reflexive.

$x\cdot z\geq0,\,z\cdot x>0$, it is symmetry

$x\cdot z>0,\,z\cdot y>0,\,x\cdot y>0,\,x\cdot z\geq0$ therfore fore $x\cdot z=x\cdot y$ it is transitivity

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    $\begingroup$ well $0\cdot 0\not>0$. $\endgroup$ – thesmallprint Jun 7 '18 at 14:49

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