Relations and functions · Answer guide

Lesson 1 · Test the properties of a relation

Use counterexamples to test reflexive, symmetric and transitive relations.

1. Which pairs are required for reflexivity on {a,b}?

   Answer: (a,a),(b,b). Every element must be related to itself.

2. If (a,b)∈R, symmetry requires…

   Answer: (b,a)∈R. Symmetry requires the reversed pair.

3. An equivalence relation must be…

   Answer: Reflexive, symmetric and transitive. All three relation properties are required.

Discuss: Test the relation “less than or equal to” on {1,2,3}. Which properties hold, and what counterexample refutes symmetry?

Teaching note: Checking a few successful triples is not a proof for an infinite relation. A single failing triple can disprove transitivity. Open responses need discussion; they are not automatically marked.

Lesson 2 · Separate domain, codomain and range

Recognise a function and test whether it is one-one and onto.

1. For 1↦a,2↦a,3↦b, the range is…

   Answer: {a,b}. Only a and b are actually reached.

2. Why is this assignment not one-one?

   Answer: Two distinct inputs map to a. 1 and 2 share the output a.

3. For codomain {a,b,c}, why is it not onto?

   Answer: c is not reached. Onto requires every member of the codomain to be reached.

Discuss: For x↦x², explain why restricting the domain to {0,1,2} changes one-one behaviour.

Teaching note: Several inputs may share an output in a function. What is forbidden is one input receiving two distinct outputs. Open responses need discussion; they are not automatically marked.

Lesson 3 · Justify a bijection

Check one-one and onto separately and explain reversible matching.

1. Which map from {1,2} to {a,b} is bijective?

   Answer: 1↦a,2↦b. It has no shared output and reaches both targets.

2. For f(x)=2x−5 on ℝ, which input gives output 9?

   Answer: 7. Solve 2x−5=9: 2x=14, x=7.

3. Reversing a bijection produces…

   Answer: A function. Each former output has exactly one former input.

Discuss: Show why x↦x+1 from the positive integers to the positive integers is one-one but not onto.

Teaching note: Equal domain and codomain sizes do not by themselves make an arbitrary assignment bijective. The function still needs the mapping property. Open responses need discussion; they are not automatically marked.

Final check · Set 1

Try one set after the lessons. Use the other set after reviewing any errors.

1. If (1,2),(2,3)∈R but (1,3)∉R, which property fails?

   Transitivity. The two connected pairs require (1,3) for transitivity.

2. Equality on a nonempty set is…

   An equivalence relation. Equality is reflexive, symmetric and transitive.

3. Can a many-to-one assignment still be a function?

   Yes, if each input has one output. Several inputs may share an output without violating the function rule.

4. Which information is necessary when checking onto?

   The codomain. Surjectivity means the range equals the specified codomain.

5. Why is x↦x+1 on positive integers not onto positive integers?

   It misses 1. No positive integer input maps to 1.

6. If f(a)=f(b) implies a=b, this establishes…

   One-one. Distinct inputs cannot share an output.

Final check · Set 2

Try one set after the lessons. Use the other set after reviewing any errors.

1. “Less than” on {1,2} fails reflexivity because…

   1 is not less than itself. Reflexivity would require 1<1 and 2<2, both false.

2. One missing reversed pair can…

   Disprove symmetry. Symmetry requires the reversal for every pair in the relation.

3. x↦x² from {−1,0,1} to {0,1} is…

   Onto but not one-one. Both codomain values are reached, but −1 and 1 share output 1.

4. An input with two distinct assigned outputs…

   Violates the function rule. A function must have exactly one output for each input.

5. For f(x)=3x+1 on ℝ, solve y=f(x) for x.

   x=(y−1)/3. Subtract 1, then divide by 3.

6. Equal finite set sizes alone guarantee a given function is bijective.

   False. A constant function between two two-element sets is a counterexample.