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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).

Carlos Bacca
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1 Answers1

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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

Proof:
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

TheD0ubleT
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