# Prove or disprove that R is an equivalence relation

A relation R is defined on Z by xRy if x · y ≥ 0. Prove or disprove the following:
(a) R is reflexive
(b) R is symmetric
(c) R is transitive

(a) If xRx then x*x >= 0 for all x in Z. This is true because (-a)(-a) = a, for all a in Z.
(b) If xRy then we want yRx for all x,y in Z. This is true because ab = ba for all a, b in 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.

• In a you have the implication backwards. You should say that $x \cdot x \ge 0$, so $xRx$ May 18, 2020 at 21:40

HINT: Is $$-1\,R\,0$$? Is $$0\,R\,1$$? Is ... ?

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.