A relation $ R $ is defined on $ \mathbb{Z} $ by $ xRy $ if $ x \cdot y \geq 0 $. Prove or disprove the following:
(a) $ R $ is reflexive
(b) $ R $ is symmetric
(c) $ R $ is transitive
(a) If $ xRx $ then $ x \cdot x \geq 0 $ for all $ x $ in $ \mathbb{Z} $. This is true because $ (-a)(-a) = a $, for all $ a $ in $ \mathbb{Z} $.
(b) If $ xRy $ then we want $ yRx $ for all $ x, y $ in $ \mathbb{Z} $. This is true because $ ab = ba $ for all $ a, b $ in $ \mathbb{Z} $.
(c) Now I am stuck because I know that "If $ xRy $ and $ yRz $ we want that $ xRz $. This is true because ..." but how do I say that multiplication is transitive.
Suppose $x*y \ge 0$ and $y*z\ge 0$.
Case 1: $x < 0$. $x*y \ge0 \implies $\frac 1x x*y \le \frac 1x *0$ so $y \le 0$.
Case 1a: $y < 0$ then $y*z \ge 0 \implies z \le 0$. So $x < 0$ and $z\le 0$ so $x*z \le 0$. So far so good.
Case 1b: $y = 0$ then $y*z \ge 0$ means $y*z=0$ and $z$... could be anything.
If $z\le 0$ we would have $x*z \le 0$.
But if $z > 0$ we would have $x< 0$ and $z > 0$ so $x*z < 0$ and that fails transitivity.
Counter example:
Let $- 1 R 0$ because $-1*0 \ge 0$. And $0 R 1$ because $0*1 \ge 0$. But $-1 \not R 1$ because $-1*1 < 0$.
So not transitive.