f A={1,4,5} and the relation R defined on the set A as aRb if a+b < 6 checkwhether the relation R is an equivalence relation

To determine whether the relation R is an equivalence relation, we need to check three properties: reflexivity, symmetry, and transitivity.

1. Reflexivity: For a relation to be reflexive, every element in the set A must be related to itself. In this case, we need to check if aRa holds for all a in A.

Let's check:

• For a = 1: 1 + 1 = 2 < 6, so 1R1 holds.
• For a = 4: 4 + 4 = 8 > 6, so 4R4 does not hold.
• For a = 5: 5 + 5 = 10 > 6, so 5R5 does not hold.

Since not all elements in A satisfy the reflexivity property, the relation R is not reflexive.

2. Symmetry: For a relation to be symmetric, if a is related to b, then b must also be related to a. In this case, we need to check if aRb implies bRa for all a, b in A.

Let's check:

• For a = 1 and b = 4: 1 + 4 = 5 < 6, so 1R4 holds. However, 4 + 1 = 5 < 6, so 4R1 also holds.

Since aRb implies bRa for all a, b in A, the relation R is symmetric.

3. Transitivity: For a relation to be transitive, if a is related to b and b is related to c, then a must be related to c. In this case, we need to check if aRb and bRc imply aRc for all a, b, c in A.

Let's check:

• For a = 1, b = 4, and c = 5: 1 + 4 = 5 < 6, so 1R4 holds. Also, 4 + 5 = 9 > 6, so 4R5 does not hold.

Since aRb and bRc do not imply aRc for all a, b, c in A, the relation R is not transitive.

In conclusion, the relation R is not an equivalence relation because it fails to satisfy the reflexivity and transitivity properties. However, it does satisfy the symmetry property.