Relations and Functions · Lesson 4 of 4
Chapter Summary and Practice
“Connect the complete chapter through shared proof methods, visual checks and mixed worked problems.”
• Connect relations, equivalence classes, functions, bijections and inverses. • Recall all main definitions while respecting their underlying sets. • Choose between a general proof, a counterexample and a finite count. • Solve mixed questions about classes, output behaviour and composition. • Recognise misconceptions involving notation, finite sets and function operations. • Use a complete chapter comparison table to organise independent revision.
The chapter as one connected idea
The chapter begins with the broad freedom to select pairs from a Cartesian product. It then asks which selections have useful structure: an equivalence relation groups objects consistently, while a function assigns a unique output to every input. A bijection adds enough structure to reverse that assignment, and composition lets assignments work in successive stages.
These ideas share a common habit of reasoning: state the set first, interpret the rule precisely, and check a requirement for every allowed element. A definition is not merely a label. It tells you what a proof must establish and what kind of counterexample would defeat the claim.
A relation on X can contain no pairs, all pairs, or any intermediate selection. A function f : X → Y is a relation with exactly one pair (x, y) for each x ∈ X. An equivalence relation is usually discussed on one set X, because its self-links, reversed links and two-link chains all compare elements of that same set. The properties belong to different questions, so being a function does not automatically make a relation an equivalence relation.
A complete comparison of the main ideas
The table brings the chapter together without replacing its reasoning. Read each condition as an instruction for a proof: identify the arbitrary elements, state the required pair or input, and show that it is allowed. The final column gives a useful check when you are unsure how to begin.
| Chapter idea | Defining condition or relationship | Useful proof or counterexample check |
|---|---|---|
| Relation from A to B | R ⊆ A × B | Check both entries and membership in the chosen subset. |
| Empty and universal relations | R = ∅ or R = A × A | Ask whether no allowed pairs, or all allowed pairs, satisfy the rule. |
| Reflexive | (a, a) ∈ R for every a | A missing diagonal pair disproves it. |
| Symmetric | (a, b) ∈ R implies (b, a) ∈ R | A selected pair without its reverse disproves it. |
| Transitive | (a, b), (b, c) ∈ R imply (a, c) ∈ R | Two existing links without their shortcut disprove it. |
| Equivalence relation | Reflexive, symmetric and transitive | Prove all three separately. |
| Equivalence class and partition | [a] consists of elements related to a; distinct classes partition X | Classes are nonempty, disjoint when distinct, and cover X. |
| Function | Each declared input has exactly one output | No input may be omitted or have two different outputs. |
| One-one / injective | f(x₁) = f(x₂) implies x₁ = x₂ | Find a collision to disprove it. |
| Onto / surjective | Every y in the codomain has a preimage | Solve f(x) = y and check x lies in the domain. |
| Bijective | One-one and onto | Every required output has exactly one preimage. |
| Finite equal-size sets | For a map between equal-size finite sets, one-one ⇔ onto | Count distinct outputs; do not apply this automatically to infinite sets. |
| Composition | (g ∘ f)(x) = g(f(x)) | Apply the right-hand function first and check compatibility. |
| Inverse | f⁻¹ ∘ f = I_X and f ∘ f⁻¹ = I_Y | Verify both directions on their stated sets; f is invertible ⇔ bijective. |
| Connected constructions | Intersections and equal-value relations preserve equivalence | Track both memberships or equality of function values. |
| Finite bijection counts | n! self-bijections on an n-element set | Assign different outputs without replacement. |
| Function addition and equality | Sums may lose injectivity or surjectivity; equal functions agree on every allowed input | Test the resulting rule and retain the declared domain and codomain. |
Two logical distinctions deserve special attention. Symmetry and transitivity are implications: they start from pairs that already belong to the relation. Reflexivity demands a pair for every element, whether or not any other pair is present. For functions, one-one limits the number of preimages to at most one, while onto demands at least one. A bijection satisfies both requirements.
From equivalence classes to function behaviour
The same-value relation a R b defined by f(a) = f(b) provides a bridge between the chapter’s two main halves. It partitions the domain into groups of inputs that share outputs. The size of each group reveals collisions, while the range tells us which codomain values have any group at all.
Problem
Let A = {−2, −1, 0, 1, 2}, B = {0, 1, 4}, and f : A → B be f(x) = x². Classify f and find the classes for a R b when f(a) = f(b).
- 1.The outputs at −2, −1, 0, 1, 2 are 4, 1, 0, 1, 4. Every element of B is produced, so f is onto.
- 2.The different inputs −2 and 2 share 4, and −1 and 1 share 1. Therefore f is not one-one and has no inverse function defined from all of B to A.
- 3.The equal-value relation is an equivalence relation because equality is reflexive, symmetric and transitive.
- 4.Its distinct classes are {−2, 2}, {−1, 1} and {0}. They are disjoint and cover A.
- 5.If the domain is restricted to {0, 1, 2} with the same B, the collisions disappear while all outputs remain. The restricted function is bijective, and its equal-value classes are singletons.
Problem
On the plane, define P R Q when P and Q have the same distance from the origin. Describe the class containing P = (3, 4).
- 1.Every point has the same distance as itself, the statement reverses, and equality of distances passes through an intermediate point. The relation is therefore an equivalence relation.
- 2.The distance of (3, 4) from the origin is √(3² + 4²) = √25 = 5.
- 3.The class consists of all points (x, y) with √(x² + y²) = 5, equivalently x² + y² = 25.
- 4.This is the circle centred at the origin with radius 5. A different representative on that circle names the same class.
- 5.The origin is different: its class is {(0, 0)}, since only the origin is at distance zero from itself.
Numerical remainder classes, geometric circles and equal-output groups all express the same pattern. The elements may look different, but the relation identifies a feature whose value is shared. The feature determines the groups, and the three property tests justify treating those groups as equivalence classes.
Using definitions as a proof plan
The structure of a claim usually tells us how to work. A property asserted for every element needs general reasoning; a failed property needs one explicit witness. A question about how many objects exist needs a method that exhausts the possibilities without counting any twice.
Problem
On ℝ define a R b by a ≤ b². Show that this relation is neither reflexive nor symmetric nor transitive.
- 1.For reflexivity, choose a = 1/2. The diagonal pair would require 1/2 ≤ (1/2)² = 1/4, which is false.
- 2.For symmetry, 0 R 1 is true because 0 ≤ 1, but 1 R 0 is false because 1 ≤ 0 does not hold.
- 3.For transitivity, choose a = 1/2, b = −1 and c = 0. We have a ≤ b² since 1/2 ≤ 1, and b ≤ c² since −1 ≤ 0.
- 4.But a ≤ c² would require 1/2 ≤ 0, which is false. Thus the two premises hold while the shortcut fails.
- 5.Each property has been disproved by its own explicit counterexample; one failure alone would not settle the other two.
Problem
Let f : ℝ → ℝ be f(x) = 3 − 4x. Show that it is bijective and determine its inverse.
- 1.If 3 − 4x₁ = 3 − 4x₂, subtract 3 and divide by −4 to get x₁ = x₂. Thus f is one-one.
- 2.Given any real y, solve y = 3 − 4x: 4x = 3 − y, so x = (3 − y)/4.
- 3.This x is real and f((3 − y)/4) = 3 − (3 − y) = y. Thus f is onto.
- 4.The reverse rule is f⁻¹(y) = (3 − y)/4. Check f⁻¹(f(x)) = (3 − (3 − 4x))/4 = x.
- 5.Also f(f⁻¹(y)) = y for every y, as already shown. Both identity checks hold.
Notice the difference between solving an equation and proving onto. The solved expression is only a candidate input until we check its domain membership for every allowed output. This matters especially for functions on integers, excluded real inputs, or piecewise intervals. An algebraic answer outside the domain does not supply a preimage.
Problem
On A = {1, 2, 3}, how many equivalence relations contain (1, 2)?
- 1.Symmetry forces (2, 1), so 1 and 2 must lie in the same class.
- 2.Element 3 either forms its own class or joins their class. These give the partitions {{1, 2}, {3}} and {{1, 2, 3}}.
- 3.The first relation contains the three diagonal pairs together with (1, 2) and (2, 1). The second is the universal relation.
- 4.There are exactly two possibilities. This partitions-based count automatically accounts for all the pairs forced by equivalence.
Problem
On {1, 2, 3}, count reflexive and symmetric but not transitive relations containing (1, 2) and (1, 3).
- 1.Reflexivity forces the three diagonal pairs. Symmetry forces (2, 1) and (3, 1).
- 2.With these seven pairs, (2, 1) and (1, 3) are present but (2, 3) is absent. Similarly (3, 1) and (1, 2) lack (3, 2). The relation is not transitive.
- 3.The only undecided pairs are (2, 3) and (3, 2). Symmetry requires us to include both or neither.
- 4.Including both gives the universal relation, which is transitive and therefore forbidden. Including neither gives the seven-pair relation above.
- 5.Exactly one relation meets the stated conditions.
Composition and inverses belong in the revision too
Function types do not complete the chapter by themselves. Composition asks how two assignments connect, and an inverse asks whether an assignment can be reversed everywhere. These ideas use the same domain, codomain and preimage checks, so they should be revised together rather than treated as isolated formulas.
Problem
Let f : ℝ → ℝ be f(x) = 2x + 1 and g : ℝ → ℝ be g(x) = x − 3. Find both compositions and verify the inverse of f.
- 1.For g ∘ f, apply f first: g(f(x)) = (2x + 1) − 3 = 2x − 2.
- 2.For f ∘ g, apply g first: f(g(x)) = 2(x − 3) + 1 = 2x − 5.
- 3.The outputs differ, for example at x = 0, so the two compositions are unequal.
- 4.Solve y = 2x + 1 to obtain x = (y − 1)/2. Every real y gives a real input, and the linear rule is one-one. Thus f⁻¹(y) = (y − 1)/2.
- 5.Now f⁻¹(f(x)) = (2x + 1 − 1)/2 = x, and f(f⁻¹(y)) = 2((y − 1)/2) + 1 = y. Both directions are verified.
Do not replace this two-direction check with a reciprocal or with a computation at only one input. A reverse rule that works for a few numbers may fail elsewhere or on part of the codomain. The equivalence “bijective if and only if invertible” gives a useful alternative proof when an explicit reverse formula is difficult to write.
Misconceptions worth resolving together
Most errors in this chapter come from changing the question without noticing it. Confusing the range with the codomain changes an onto test; confusing a reversed pair with a shortcut changes a relation test. The following reminders keep the question attached to its exact definition.
One-one does not mean onto, except under a valid additional condition such as equal-size finite sets. A function can have distinct outputs while leaving codomain elements unused, as doubling does on ℕ.
The notation [a] depends on context: for an equivalence relation it denotes a class, while [x] for the greatest integer function denotes a number. State the meaning before using the brackets. Likewise, f⁻¹ denotes a reverse function, not 1/f.
Adding two onto functions can produce a sum that misses outputs, and adding two injective functions can create a collision. These claims must be checked for the new rule itself. Also, two different expressions can define equal functions on a finite domain; equality outside that domain is irrelevant to that particular function comparison.
How the function idea became more general
The historical development reflects a mathematical change in viewpoint. Early discussions tied functions closely to geometrical curves or expressions built from operations. Later viewpoints focused on the dependence of one quantity on another, and finally on a precise assignment between sets.
The chapter connects this development with Descartes, Gregory, Leibniz, Bernoulli, Euler, Lagrange, Dirichlet and Cantor. The useful lesson is the increasing generality of the idea: a function need not be a simple polynomial or a graph that can be drawn in the plane. Our finite arrow assignments and piecewise rules are functions because of the input–output requirement, not because they have a particular kind of formula.
| Historical thread | Connection with the present chapter |
|---|---|
| Descartes and Gregory: curves, powers and quantities formed by operations | Formulas provide familiar examples, but do not exhaust the meaning of a function. |
| Leibniz: varying geometric quantities and dependence on a variable | The language of input and output expresses dependence precisely. |
| Bernoulli, Euler and Lagrange: development of function notation | Symbols such as φ, f and f(x) let us name an assignment independently of a particular numerical input. |
| Dirichlet: a more general assignment viewpoint; Cantor: set theory | The set-based definition records domain, codomain and the unique output required for each input. |
Check your complete chapter understanding
These questions mix the chapter ideas deliberately. Decide what is being tested before calculating: relation properties, classes, function types, a composition, an inverse or a finite count. The same short formula can support several different questions.
Quiz
On a nonempty set X, which relation is always an equivalence relation?
For the even-difference relation on ℤ, what is [4]?
If f : X → Y is bijective, how many preimages does each y ∈ Y have?
If f(x) = x² from ℝ to [0, ∞), which statement is correct?
How many bijections map a set of four elements to itself?
For f(x) = x − 2 and g(x) = 3x on ℝ, what is (f ∘ g)(2)?
Which is the inverse of f : ℝ → ℝ, f(x) = 5x + 2?
What is the clearest way to disprove transitivity?
For f : X → Y, the relation f(a) = f(b) on X is:
Two functions with the same domain and codomain are equal when:
Practice Problems
- On A = {1, 2, 3, 4}, compare the empty, universal and equality relations. Test all three relation properties and describe any equivalence classes.
- On ℤ, prove that 5 divides a − b defines an equivalence relation. List its distinct classes and identify [−2].
- Construct one symmetric but nontransitive relation and one reflexive, transitive but nonsymmetric relation on {1, 2, 3}. Supply counterexamples for the failed properties.
- On the plane, describe the classes for “same distance from the origin”. Explain why the zero-distance class differs from the positive-distance classes.
- Classify x² and x³ with domain and codomain ℕ, then with domain and codomain ℝ. Give a reason for every successful or failed test.
- Classify the reciprocal function on ℝ* → ℝ*, and write its inverse with both identity checks.
- For f : ℝ ∖ {3} → ℝ ∖ {1}, f(x) = (x − 2)/(x − 3), derive the reverse rule and verify that every allowed output produces an allowed input.
- With f(x) = x + 2 and g(x) = x² on ℝ, find both compositions and explain why only one of the stages is invertible on ℝ → ℝ.
- For a function on {−3, −2, −1, 0, 1, 2, 3} given by x², choose a codomain that makes it onto and list all equal-output equivalence classes.
- Count equivalence relations on {1, 2, 3} containing (1, 2), then count reflexive, symmetric but nontransitive relations containing (1, 2) and (1, 3). Explain the difference in methods.
- Count bijections from {1, 2, 3, 4, 5} to itself. Explain why the same count applies to onto self-maps of this finite set.
- Give an infinite-set example where one-one does not imply onto, and another where onto does not imply one-one.
- For f(x) = sin x and g(x) = cos x on [0, π/2], explain how their sum loses injectivity. Compare this with the sum of two identity functions on ℕ losing surjectivity.
- Explain why two formulas may agree as functions on a chosen finite domain but disagree as functions on ℝ. Provide your own checked example.
Key Takeaways
• State the underlying sets before testing a relation or function. • Equivalence relations combine reflexivity, symmetry and transitivity, and their classes partition the set. • Injectivity controls collisions; surjectivity controls coverage of the codomain; bijectivity combines the two. • Composition follows successive stages, while inverse functions reverse bijections and satisfy identities in both directions. • Finite counting arguments require their stated size assumptions, and relation counts require all forced pairs. • Use definitions, complete proofs and explicit counterexamples to connect numerical, geometric and abstract examples.
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Composition, Inverses and Connected Applications
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