Relations and Functions
Relations and Functions
This domain tests whether you can move between a relation as a set of ordered pairs and a function as a rule, and whether you can justify injectivity, surjectivity and invertibility. ASSEB questions often combine domain restrictions with composition or inverse.
Core rules
- A relation R from A to B is a subset of A×B; domain = first coordinates, range = second coordinates, codomain is the declared target set.
- R is reflexive if (a,a)∈R for all a∈A; symmetric if (a,b)∈R⇒(b,a)∈R; transitive if (a,b),(b,c)∈R⇒(a,c)∈R. A relation with all three is an equivalence relation.
- f:A→B is one-one if f(x1)=f(x2)⇒x1=x2; onto if every b∈B has some a∈A with f(a)=b; bijective if both.
- For a bijection, f^{-1} exists and f(f^{-1}(x))=x, f^{-1}(f(x))=x on the appropriate domains.
- (g∘f)(x)=g(f(x)); composition is not generally commutative.
Worked example
Let f(x)=(3x+2)/(x−1), x≠1. Find f^{-1}. Put y=(3x+2)/(x−1). Then y(x−1)=3x+2, so xy−y=3x+2, x(y−3)=y+2, hence x=(y+2)/(y−3). Therefore f^{-1}(x)=(x+2)/(x−3), x≠3. Check: f(f^{-1}(x)) = [3(x+2)/(x−3)+2]/[(x+2)/(x−3)−1] = [(3x+6+2x−6)/(x−3)]/[(x+2−x+3)/(x−3)] = (5x)/(5)=x. Thus the inverse is correct on the stated domain.
Traps
Do not call a relation a function when one input has two outputs. Do not confuse range with codomain. An inverse exists only after proving bijectivity, and its domain is the original codomain. In composition, order matters: f∘g and g∘f can differ.