Answer to Question #222564 in Discrete Mathematics for mostafa

Question #222564

let z be the set of integers and R be the relation on Z defined as: aRb if and only if 1+ab>0 then


1
Expert's answer
2021-08-03T13:16:50-0400

Given relation is aRbaRb is 1+ab>0.1+ab>0.

Considering both aa and bb are real numbers, we know that ab=baab=ba


aRb=1+ab>0=>bRa=1+ba=1+ab>0aRb=1+ab>0=>bRa=1+ba=1+ab>0

Then RR is a symmetric relation.



aRa=1+a2>0aRa=1+a^2>0

Then RR is a reflexive relation.


Let a=0.5,b=0.5,a=0.5, b=-0.5, and c=4.c=-4. Then


aRb=1+ab=1+0.5(0.5)=0.75>0aRb=1+ab=1+0.5(-0.5)=0.75>0

bRc=1+bc=1+(0.5)(4)=3>0bRc=1+bc=1+(-0.5)(-4)=3>0

But


aRc=1+ac=1+0.5(4)=1<0aRc=1+ac=1+0.5(-4)=-1<0

aRcaRc is not a relation.


Hence RR is not a equivalence relation, but is a reflexive and symmetric relation.


reflexive and symmetric relation.



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