Define a relation on Z as xRy if |x−y|<1.

I have shown this relation is symetric and reflexive and i am pretty sure its transitive because this is the equality relation isnt it? thats my first question and my second is how to show it is transitive.

I attempted a direct proof but i dont know how to link the two inequalities together to get that |x−z|<1 (im trying to show if xRy and yRz then xRz).

Any help would be appreciated, thanks! I am looking for the proof of this last property (transitivity).

  • $\begingroup$ Thanks, but its a relation on z so only for integer values $\endgroup$ – Carlos Bacca Dec 3 '18 at 4:14
  • $\begingroup$ There are no integers satisfying this relation. $\endgroup$ – user58697 Dec 3 '18 at 4:41
  • $\begingroup$ what about x=4 y=4? $\endgroup$ – Carlos Bacca Dec 3 '18 at 4:43
  • $\begingroup$ Sorry I forgot to say distinct. Modulo is non-negative; the only non-negative less than 1 is 0. $\endgroup$ – user58697 Dec 3 '18 at 4:49
  • $\begingroup$ So is it transitive? $\endgroup$ – Carlos Bacca Dec 3 '18 at 4:53

If your relation is in fact on $\Bbb{Z}$, then $xRy$ is the same as writing $x=y$ because if $x\ne y$ then the distance between x and y is obviously larger than 1.
"=" is transistive thus R is also transistive because the two relations are equivalent

If $x\ne y$ then $y = x+k$, with $k\in \Bbb{Z}^*$
And $|x-y| = |k| \ge 1$, thus no two distinct integers verify this relation
If $xRy$ and $yRz$ that means that $x=y=z$ thus $xRz$ also. So yes the relation is transistive

  • $\begingroup$ Thanks, i understand that it is transitive but how woud you show this $\endgroup$ – Carlos Bacca Dec 3 '18 at 18:19

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