Proof of an equivalence relation

Question

Let S be the relation on R defined by $xSy \Leftrightarrow x=|y|$, $\forall x,y\in\Re$

Is the relation reflexive, symmetric and/or transitive?

By my proof that

1) $x=|y| \Rightarrow |y|=x$ (reflexive) 2) $x=|y| \wedge|x|=y$ (symmetric) 3) But not transitive because not exist $z:xSy \wedge ySz s.t. xSz$

Right?

Answer 1

I hope no one minds if I assemble the various comments by Cloudscape and coffeemath (and myself) into an answer.

• For reflexivity, you want to show that $xSx$, or $x=|x|$ for all $x\in\mathbb{R}$. Clearly this is not true for any negative values of $x$, so $S$ is not reflexive.
• For symmetry, you want to show that $xSy$ implies $ySx$, or that whenever $x=|y|$, we also have $y=|x|$. This also fails whenever $y$ is negative, so $S$ is not symmetric.
• For transitivity, you want to show that $xSy$ and $ySz$ imply $xSz$, or in other words that if $x=|y|$ and $y=|z|$, then $x=|z|$. This one is true, since if $y=|z|$ then $y$ must be non-negative, so it follows that $|y|=y$. Then you have $x=|y|=y=|z|$, so transitivity holds.