Relations and functions · Learner pages

Lesson 1 · Test the properties of a relation

Use counterexamples to test reflexive, symmetric and transitive relations.

Name or initials: ____________   Date: ____________

A relation on A is a subset of A×A: its members are ordered pairs.

1. 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}.
2. Test reflexivity. Every number has the same parity as itself, so (1,1),(2,2),(3,3) are included.
3. Test symmetry. If a and b have the same parity, b and a also do. Thus (1,3) comes with (3,1).
4. 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.

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

   1. (a,a),(b,b) · 2. (a,b) only · 3. (b,a) only

2. If (a,b)∈R, symmetry requires…

   1. (b,a)∈R · 2. (a,a)∉R · 3. a=b

3. An equivalence relation must be…

   1. Only symmetric · 2. Reflexive, symmetric and transitive · 3. One-one and onto

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.

Name or initials: ____________   Date: ____________

Each input of a function has exactly one output in its codomain.

1. 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.
2. Read the range. The actual outputs are a and b. Thus the range is {a,b}; the codomain remains {a,b,c}.
3. Test one-one. 1 and 2 are distinct inputs with the same output a. Therefore f is not one-one (injective).
4. 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.

1. For 1↦a,2↦a,3↦b, the range is…

   1. {1,2,3} · 2. {a,b} · 3. {a,b,c} always

2. Why is this assignment not one-one?

   1. There are three inputs · 2. Two distinct inputs map to a · 3. The range is nonempty

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

   1. c is not reached · 2. a is reached twice · 3. The domain is finite

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.

Name or initials: ____________   Date: ____________

A bijection is a function that is both one-one and onto.

1. 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.
2. Check for collisions. No two distinct inputs share an output, so f is one-one.
3. Check all targets. All three codomain members are reached, so f is onto. Both tests pass: f is bijective.
4. 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.

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

   1. 1↦a,2↦b · 2. 1↦a,2↦a · 3. 1↦b,2↦b

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

   1. 2 · 2. 7 · 3. 14

3. Reversing a bijection produces…

   1. A function · 2. Two outputs per input · 3. An empty relation

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

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?

   Reflexivity necessarily / Symmetry necessarily / Transitivity

2. Equality on a nonempty set is…

   An equivalence relation / Never reflexive / Never transitive

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

   Yes, if each input has one output / No / Only with an empty domain

4. Which information is necessary when checking onto?

   The font used / The codomain / Only the formula name

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

   It misses 1 / It misses every even number / It has repeated outputs

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

   Onto / One-one / Neither

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<2 / 1 is not less than itself / 2>1

2. One missing reversed pair can…

   Disprove symmetry / Prove symmetry / Prove reflexivity

3. x↦x² from {−1,0,1} to {0,1} is…

   One-one and onto / Onto but not one-one / Not a function

4. An input with two distinct assigned outputs…

   Violates the function rule / Proves onto / Proves one-one

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

   x=(y−1)/3 / x=3y+1 / x=y−3

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

   True / False / Only if the size is even