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.

  • $\begingroup$ To show symmetric you must show if xRy then yRx $\endgroup$ – J. W. Tanner Apr 30 '19 at 13:54
  • $\begingroup$ All good except for the proof of symmetry. Check the definition again. $\endgroup$ – A. Pongrácz Apr 30 '19 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 $|y-x|=|x-y|\le 2$.


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