Determine whether the relation R defined on the set R of all real numbers as R = {(a, b) : a, b ∈ R and a – b + 3–√ ∈ S, where S is the set of all irrational numbers}, is reflexive, symmetric and transitive.

To determine whether the relation R is reflexive, symmetric, and transitive, we need to test it against the definitions of these properties.

1. Reflexive: A relation R is reflexive if for every element a in the set, (a, a) is in R. In this case, for (a, a) to be in R, a - a + 3^(-1/2) must be irrational. However, a - a + 3^(-1/2) simplifies to 3^(-1/2), which is irrational. Therefore, the relation R is reflexive.

2. Symmetric: A relation R is symmetric if for every pair (a, b) in R, (b, a) is also in R. In this case, for (a, b) to be in R, a - b + 3^(-1/2) must be irrational. However, if we switch a and b to get (b, a), we get b - a + 3^(-1/2). This is not necessarily equal to a - b + 3^(-1/2) because subtraction is not commutative. Therefore, the relation R is not symmetric.

3. Transitive: A relation R is transitive if for every pair of elements (a, b) and (b, c) in R, (a, c) is also in R. In this case, for (a, b) and (b, c) to be in R, a - b + 3^(-1/2) and b - c + 3^(-1/2) must both be irrational. However, this does not guarantee that a - c + 3^(-1/2) is irrational, so the relation R is not transitive.

In conclusion, the relation R is reflexive, but not symmetric or transitive.