Question

if R is relation on z define by xRz=|x-y|<or=1 then R is​

Answer 1

Let x be a element in Z,

then ∣x−x∣=0≤1.

So every element of Z is related to itself, Thus R is a reflexive relation .

Let x,y be two element in Z such that ∣x−y∣≤1,

then ∣y−x∣≤1.

So, xRy⇔yRx and thus R is a symmetric relation .

Now let's prove that R is not transitive by an example to contradict,

(2,1)⇒∣2−1∣≤1 is in R and (1,0)⇒∣1−0∣≤1 is also in R but (2,0)⇒∣2−0∣≥1 is not in R.

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Answer 2

thanks for free points