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We defined a new class of relations “GO” on a Set.

A relation $\text{R}$ on a set $\text{A}$ is said to be GO iff $\forall a,b [ (a\text{R}b \wedge b\text{R}a) \leftrightarrow (a=b) ],$ where $a,b \in \text{A}.$

Which of the following options is/are correct about relation GO?

  1. Every GO relation is reflexive.
  2. Every GO relation is symmetric.
  3. Every GO relation is anti-symmetric.
  4. Every GO relation is transitive.
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2 Answers

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A relation R on a set A is said to be GO iff ∀a,b [ (aRb ∧ bRa) ↔ (a = b) ], where a,b∈A.

  1. a, b are related to each other iff a = b; AND
  2. a not related to b OR b not related to a iff a ≠ b.

Option A : From point 1, we can say that aRa, where a∈A. Thus, reflexive.

Option B : From point 2, we can say that at most one of aRb and bRa is possible, where a,b∈A and a≠b. Thus, not symmetric.

Option C : From point 1, we can say that aRb and bRa iff a=b, where a,b∈A . Thus, anti-symmetric.

Option D : Let aRb and bRc, then also it is possible that a is not related to c, where a,b,c∈A and a≠b≠c. (See point 2). Thus, not transitive.

Answer :- A, C

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0 votes

A relation R on a set A is said to be GO iff ∀a,b [ (aRb ∧ bRa) ↔ (a = b) ], where a,b∈A.

This is definition of Antisymmetric Relation. Hence GO is antisymmetric.

As antisymmetric relation may or may not be Reflexive,Symmetric,Transitive. Relation GO is only Antisymmetric.

Let A = {x,y,z} and R = {(y,z),(x.x)} also the relation R satisfies iff ∀a,b [ (aRb ∧ bRa) ↔ (a = b) ] but it is not Reflexive.

Hence Correct Answer is Only C.

Answer:

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