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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.

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    $\begingroup$ 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

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

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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.

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