Reflexive, Symmetric, And Transitive Define a relation R on $\Bbb{Z}$ as follows:
$$
(x, y) \in R \qquad \text{if and only if} \qquad x \cdot y > 0
$$


*

*Is R reflexive?

*Is R symmetric?

*Is R transitive?
I think it's not reflexive because $x \cdot y \ngtr x \cdot y$ and not symmetric because $0 \ngtr x \cdot y$.
I am not sure about transitivity.
Please provide explanations.
 A: I’m afraid that you’ve badly misunderstood reflexivity and symmetry.
Reflexivity of $R$ means that $\langle x,x\rangle\in R$ for each $x\in\Bbb Z$, i.e., that $x^2>0$ for every $x\in\Bbb Z$. This is almost true: $x^2>0$ for every $x\in\Bbb Z$ except $0$. $0^2=0\not>0$, so $\langle 0,0\rangle\notin R$, and $R$ is not reflexive.
Symmetry of $R$ means that if $\langle x,y\rangle\in R$, then $\langle y,x\rangle\in R$ as well. Suppose that $\langle x,y\rangle\in R$; that means that $xy>0$. Of course $yx=xy$, so $yx>0$, too, and therefore $\langle y,x\rangle\in R$. Thus, $R$ is symmetric.
Transitivity of $R$ means that if $\langle x,y\rangle\in R$ and $\langle y,z\rangle\in R$, then $\langle x,z\rangle\in R$. Suppose that $\langle x,y\rangle\in R$ and $\langle y,z\rangle\in R$; that means that $xy>0$ and $yz>0$. Does this guarantee that $xz>0$? Yes, so $R$ is transitive. There are at least two ways to see this. 


*

*One is to observe that if $xy>0$ and $yz>0$, then $(xy)(yz)>0$, i.e., $xzy^2>0$. We know that $y\ne 0$, because if $y$ were $0$, $xy$ would also be $0$, and it isn’t. Therefore $y^2>0$, and we can divide the inequality $xzy^2>0$ by $y^2$ to find that $xz>0$.

*The other is to realize that since $xy>0$, $x$ and $y$ must have the same algebraic sign: either both are positive, or both are negative. Similarly, $yz>0$ implies that $y$ and $z$ have the same algebraic sign. But then $x$ and $z$ both have the same sign as $y$, so they must have the same sign as each other; this ensures that their product is positive, i.e., that $xz>0$, and hence that $\langle x,z\rangle\in R$.
