A relation is transitive if (x, y) epsilon R and (y, z) epsilon R implies (y. z)

Question

A relation is transitive if (x, y) epsilon R and (y, z) epsilon R implies (y. z) E R for all x, y. y epsilon A A relation is transitive if x R y and y R z implies x R z for all x, y, z epsilon A The three properties are independent, in the sense that a relation may possess any of the 8 possible combinations of the three. We consider an equivalence relation an axiom system. Prove that each axiom is independent in an equivalence relation. Why is the following argument fallacious The reflexive property of an equivalence relation is redundant, because the symmetric property tells us that if a R b then R a. But if a R b and R a we know by transitivity that a R a. Determine whether the given relation is an equivalence relation on the set. x R y in Z if 2 divides a + b a R b in Z if there exists a k in Z such that a = bk..

Explanation / Answer

(a)

For a relation to be equivalence relation it must be reflexive,symmetric and transitive

Check if reflexive

For x an integer

2|(x+x) as x+x=2x

Hence, R is reflexive

Check if symmetric

xRy implies 2|x+y ,hence, 2|y+x

Hence, yRx

So , R is symmetric

Check if R is transitive

xRy, yRz

2|x+y,2|y+z

x+z=x+y-y+z=(x+y)-2y+(y+z)

2|x+y,2|2y,2|y+z

hence, 2|x+z

Hence, R is transitive

Hence, R is an equivalence relation.

(b)

It is not an equivalence relation

Consider: 2 and 6

6R2 because: 6=2*3

But 2R6 is not true because

2=6k for k an integer is not possible.

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