Relations and Functions
Relations and Functions carries the exam's definitions: relations as subsets of A × B, the three relation properties, and functions classified as one-one (injective), onto (surjective) or bijective. Karnataka papers frequently pair it with counting questions and with inverse and composite function algebra.
Core rules
- A relation R from A to B is any subset of A × B. Domain is the set of first elements, range the set of second elements, codomain is all of B.
- Reflexive: (a, a) ∈ R for every a ∈ A. Symmetric: (a, b) ∈ R implies (b, a) ∈ R. Transitive: (a, b) and (b, c) in R imply (a, c) ∈ R. Reflexive + symmetric + transitive gives an equivalence relation.
- With |A| = m and |B| = n: number of relations = 2^(mn); number of functions = n^m; one-one functions need m ≤ n; bijections exist only when m = n.
- (f ∘ g)(x) = f(g(x)). f is invertible exactly when it is bijective, and then f(f⁻¹(x)) = x.
Worked example
f: R − {2} → R − {1} is defined by f(x) = (x − 1)/(x − 2). Show f is invertible and find f⁻¹. Put y = (x − 1)/(x − 2). Then y(x − 2) = x − 1, so xy − 2y = x − 1 and x(y − 1) = 2y − 1, giving x = (2y − 1)/(y − 1). Hence f⁻¹(x) = (2x − 1)/(x − 1) on R − {1}. The excluded values explain the codomain: 1 is never attained, and 2 is not in the domain.
Traps
- Writing the range when the codomain is asked, or the reverse.
- Checking only two properties and calling the relation an equivalence relation.
- Claiming an inverse for a function that is one-one but not onto.
- Reversing composition order: f ∘ g is usually not g ∘ f.