A chapter companion introducing relation properties and testing one-one, onto and bijective functions with finite examples. General proofs and the full exercise map remain additional work.
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LESSON 1
Test the properties of a relation
Use counterexamples to test reflexive, symmetric and transitive relations.
A relation on A is a subset of A×A: its members are ordered pairs.
Choose a finite set
Let A={1,2,3}. Relate two elements when they have the same parity. The odd block is {1,3}; the even block is {2}.
Test reflexivity
Every number has the same parity as itself, so (1,1),(2,2),(3,3) are included.
Test symmetry
If a and b have the same parity, b and a also do. Thus (1,3) comes with (3,1).
Test transitivity
If a and b have the same parity and b and c do too, a and c have the same parity. The relation has all three properties, so it is an equivalence relation.
Worked example: On {1,2,3}, R={(1,1),(2,2),(3,3),(1,2)} is reflexive. It is not symmetric because (1,2) is present but (2,1) is absent, so it is not an equivalence relation.
Watch for this: Checking a few successful triples is not a proof for an infinite relation. A single failing triple can disprove transitivity.
Discuss: Test the relation “less than or equal to” on {1,2,3}. Which properties hold, and what counterexample refutes symmetry?
LESSON 2
Separate domain, codomain and range
Recognise a function and test whether it is one-one and onto.
Each input of a function has exactly one output in its codomain.
Assign one output per input
Let A={1,2,3}, B={a,b,c}, with f(1)=a, f(2)=a and f(3)=b. Every input has exactly one output, so f is a function.
Read the range
The actual outputs are a and b. Thus the range is {a,b}; the codomain remains {a,b,c}.
Test one-one
1 and 2 are distinct inputs with the same output a. Therefore f is not one-one (injective).
Test onto
No input reaches c. Therefore f is not onto (surjective) B. If the codomain were changed to {a,b}, this same assignment would be onto.
Worked example: For g:{−2,−1,0,1,2}→{0,1,4}, g(x)=x² is onto because 0,1,4 are all reached. It is not one-one because g(−1)=g(1)=1. Domain and codomain are essential to this conclusion.
Watch for this: Several inputs may share an output in a function. What is forbidden is one input receiving two distinct outputs.
Discuss: For x↦x², explain why restricting the domain to {0,1,2} changes one-one behaviour.
LESSON 3
Justify a bijection
Check one-one and onto separately and explain reversible matching.
A bijection is a function that is both one-one and onto.
Make a matching
Let f:{1,2,3}→{a,b,c} send 1 to a, 2 to b and 3 to c. Each input has one output.
Check for collisions
No two distinct inputs share an output, so f is one-one.
Check all targets
All three codomain members are reached, so f is onto. Both tests pass: f is bijective.
Explain reversal
The inverse sends a to 1, b to 2 and c to 3. It is a function because every target had exactly one original input. For finite sets of equal size, injectivity and surjectivity imply one another; this counting shortcut needs those conditions.
Worked example: For f:ℝ→ℝ defined by f(x)=2x−5, suppose f(a)=f(b). Then 2a−5=2b−5, hence a=b, proving one-one. For any real y, x=(y+5)/2 gives f(x)=y, proving onto. Its inverse is y↦(y+5)/2.
Watch for this: Equal domain and codomain sizes do not by themselves make an arbitrary assignment bijective. The function still needs the mapping property.
Discuss: Show why x↦x+1 from the positive integers to the positive integers is one-one but not onto.
