Attempting to write proof for the given question. Are the proofs sufficient? Would appreciate any feedback and criticisms.
The relation $\star$ is defined on $\mathbb{R}-\{0\}$ by $x \star y$ iff $x y \geq 0 $.
Proof of reflexivity:
For any non-zero real number $x \in \mathbb{R} - \{0\}$, we have $x^2 \geq 0$. Thus, $\star$ is reflexive.
Proof of symmetry:
Since real number multiplication is commutative, if $x, y \in \mathbb{R} - \{0\}$ and $x \star y$ satisfies $xy \geq 0$, then the relation $y \star x$, which is equivalent to $yx \geq 0$, is symmetric.
Proof of anti-symmetry:
Anti-symmetry states that when $(x, y) \in \mathbb{R} - \{0\}$ and $(y, x) \in \mathbb{R} - \{0\}$, then $x = y$. However, due to the commutativity of multiplication, this property does not hold. For example, when we take $x = 1$ and $y = 2$, the relation $x \star y$ implies $x \times y = 1 \times 2 = 2 = 2 \times 1 \geq 0$. Nevertheless, it is evident that $2 \neq 1$. Thus, the relation is not anti-symmetric.
Proof of transitivity:
When considering $(x, y) \in \mathbb{R} - {0}$ such that $xy \geq 0$, it is necessary for both $x$ and $y$ to share the same sign. Similarly, for $(y, z) \in \mathbb{R} - {0}$, $y$ and $z$ must have the same sign to satisfy $yz \geq 0$. Given that $x$, $y$, and $z$ share the same sign and are non-zero, it becomes evident that $x \star z$, where $xz \geq 0$, is achievable. Hence, the relation is transitive.