# Finding and proving if a relation is reflexive/transitive/symmetric/anti-symmetric

Let R be binary relation on N (natural numbers) defined by xRy if and only if x − 2 ≤ y ≤ x + 2. Is R reflexive? Is R symmetric? Is R antisymmetric? Is R transitive?

I'm not sure if I'm understanding this right or how I can prove it. This is what I currently have:

Reflexive: Yes, because $$x-2≤x≤x+2$$ so $$xRx$$ (How can I prove this for all $$x$$ though?)

Symmetric: Yes, because $$xR(x+2)$$ therefore $$(x+2)R((x+2)-2)$$ (I can tell this is wrong, but basically I'm trying to say since x always relates to x+2, x+2 will always relate to x).

Anti-Symmetry and Transitive just need examples of them not working, so I think these work:

Anti-Symmetric: No, e.g. $$2R4$$ and $$4R2$$, but $$4 ≠ 2$$

Transitive: No, e.g. $$2R4$$ and $$4R6$$, but $$2! R 6$$

Thanks for any help.

• To show symmetric you must show if xRy then yRx – J. W. Tanner Apr 30 at 13:54
• All good except for the proof of symmetry. Check the definition again. – A. Pongrácz Apr 30 at 14:03

Note that $$R$$ is just $$|x-y| \le 2$$, that is the distance between two points are less than or equal to $$2$$.
It is reflexive because, we have $$-2 \le 0 \le 2$$, add $$x$$ to the inequalities and we have the result.
It is symmetric, if $$|x-y| \le 2$$, then $$|y-x|=|x-y|\le 2$$.