# Deteremining whether the relation R on the set of all Real Numbers is Reflexive, Symmetric, Antisymmetric, Transitive, and/or Irreflexive

I am attempting to work out a problem from my Discreet Mathematics Textbook and am a little stuck on part of this one question. I was wondering if someone could walk me through (b) and (c) on the below?

The question: Determine whether the relation R on the set of all real numbers is R-reflexive, S-symmetric, AS-antisymmetric, T-Transitive, and/or I-irreflexive, where $(x,y) \in R$ if and only if:

• (a). $y = \pm x$
• (b). $x-y$ is a rational number.
• (c). $xy \ge 0$
• (d). $x = 1$

For (a) and (d) I grew out the graph and referenced my definitions of the above properties to try and determine which the set qualified under, but for (b) and (c) I'm having a difficult time conceptualizing what is being presented so as to determine which properties match up.

Can someone help me with (b) and (c)?

• b) Is an equivalence relationship. – Henricus V. Apr 7 '16 at 14:00
• ...Which means? – Analytic Lunatic Apr 7 '16 at 14:15

$b)$

Reflexive: Let $x \in \mathbb{R}$. Then $x-x=0 \in \mathbb{Q}$

Transitive: Let $x,y,z \in \mathbb{R}$ with $x-y \in \mathbb{Q}$ and $y-z \in \mathbb{Q}$. Then $x-z = (x-y) + (y-z) \in \mathbb{Q}$.

Symmetric: Let $x,y \in \mathbb{R}$ with $x-y \in \mathbb{Q}$. Then $y-x = -(x-y) \in \mathbb{Q}$ (if $z \in \mathbb{Q}$ then $-z \in \mathbb{Q}$)

Not antisymmetric: Let $x=1$ and $y=2$. Then $x-y \in \mathbb{Q}$ and $y-x \in \mathbb{Q}$ but $x \neq y$.

Not irreflexive, since it's reflexive.

$c)$

Symmetric: Let $x,y\in \mathbb{R}$ with $xy \geq 0$. Then $yx = xy \geq 0$.

Reflexive: Let $x \in \mathbb{R}$. Then $x^2 \geq 0$.

Not transitive: Let $x=1, y=0, z=-1$. Then $xy=0 \geq 0, yz=0 \geq 0$, but $xz = -1 < 0$

Not antisymmetric: Let $x=2,y=1$. Then $xy=yx=2 \geq 0$, but $x \neq y$

Not irreflexive, since it's reflexive.

• This definitely made the picture more clear. Thank you so much for the in-depth response! – Analytic Lunatic Apr 7 '16 at 21:07