Reflexive: for any integer a, a - a = 0, which is divisible by 3, so (a, a) in R. Symmetric: if (a - b) is divisible by 3, then (b - a) = -(a - b) is also divisible by 3, so aRb implies bRa. Transitive: if (a - b) and (b - c) are both divisible by 3, then their sum (a - b) + (b - c) = (a - c) is also divisible by 3, so aRb and bRc imply aRc. Since R is reflexive, symmetric and transitive, it is an equivalence relation.

One-one: suppose f(x1) = f(x2). Then 2x1 + 3 = 2x2 + 3, so 2x1 = 2x2 and x1 = x2; hence f is one-one. Onto: let y be any real number in the codomain. Solving y = 2x + 3 gives x = (y - 3)/2, which is a real number, and f((y - 3)/2) = y. So every y has a pre-image, and f is onto. Being both one-one and onto, f is a bijection.

Since f is a bijection, its inverse exists. Let y = f(x) = 2x + 3. Solve for x in terms of y: y - 3 = 2x, so x = (y - 3)/2. Therefore the inverse function is f inverse (y) = (y - 3)/2, or writing in terms of x, f inverse (x) = (x - 3)/2. We can check that f(f inverse (x)) = x.

f inverse (x) = (x - 3)/2