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)?

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

Maybe a straightforward proof may help you with this.


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.


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.

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

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.