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.

  • 1
    $\begingroup$ In a you have the implication backwards. You should say that $x \cdot x \ge 0 $, so $xRx$ $\endgroup$ May 18, 2020 at 21:40

2 Answers 2


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.