###### Question 7978 – GATE 2017 [Set-2]

November 16, 2023###### Question 10288 – Set-Theory

November 16, 2023# Question 14237 – Set-Theory

A relation R is said to be circular of aRb and bRc together imply cRa. Which of the following options is/are correct?

Correct Answer: A

Question 3 Explanation:

Theorem: A relation R on a set A is an equivalence relation if and only if it is reflexive and circular.

For symmetry, assume that x, y ∈ A so that xRy, lets check for yRx.

Since R is reflexive and y ∈ A, we know that yRy. Since R is circular and xRy and yRy, we know that yRx. Thus R is symmetric.

For transitivity, assume that x, y, z ∈ A so that xRy and yRz. Check for xRz. Since R is circular and xRy and yRz, we know that zRx. Since we already proved that R is symmetric, zRx implies that xRz. Thus R is transitive.

If a relation S is reflexive and circular, then S is an equivalence relation.

If a relation S is transitive and circular, then S is an equivalence relation.

If a relation S is circular and symmetric, then S is an equivalence relation.

If a relation S is reflexive and symmetric, then S is an equivalence relation.

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