- Chapter 1
- Relations and Functions
- 26 questions
Relations and Functions
Types of relations and the conditions under which a function is one-one or onto.
A relation is simply a rule that pairs elements of one set with elements of another. Class XII asks a sharper question than Class XI did: not merely what the pairing is, but what kind of pairing it is. Whether a relation is reflexive, symmetric or transitive decides whether it partitions a set into clean, non-overlapping classes, and whether a function is one-one and onto decides whether it can be reversed at all.
The chapter is short but it carries weight: the idea of an equivalence relation reappears throughout mathematics, and the one-one/onto test is what makes inverse trigonometric functions possible in the very next chapter.
What you should be able to do
- Decide whether a given relation is reflexive, symmetric or transitive, and justify each decision with a proof or a counter-example.
- Identify equivalence relations and describe the equivalence classes they produce.
- Determine whether a function is one-one (injective), onto (surjective), or both.
- Construct counter-examples to disprove a claimed property.
Topics
2 topics
- Types of Relations
Reflexive, symmetric, transitive and equivalence relations, with equivalence classes.
- Reflexive relations
- Symmetric relations
- Transitive relations
- Equivalence relations and equivalence classes
- Types of Functions
One-one (injective) and onto (surjective) functions, and bijections.
- One-one functions
- Onto functions
- Bijective functions
Formulas
11 items · All formulas
Relation as a subset
R \subseteq A \times BA relation from A to B is any set of ordered pairs (a,b) with a \in A and b \in B.
Reflexive relation
(a,a) \in R \quad \forall\, a \in AEvery element of the set is related to itself.
- Valid when
- Must hold for every element without exception.
Symmetric relation
(a,b) \in R \;\Rightarrow\; (b,a) \in RWhenever one element is related to another, the reverse also holds.
Transitive relation
(a,b) \in R \text{ and } (b,c) \in R \;\Rightarrow\; (a,c) \in RRelations chain: related through an intermediate means related directly.
Equivalence class
[a] = \{\, x \in A : (x,a) \in R \,\}The set of all elements related to a. Classes are disjoint and together cover A.
- Valid when
- Defined when R is an equivalence relation on A.
Number of relations on a finite set
2^{\,n^2}A set with n elements has n^2 ordered pairs, and a relation is any subset of them.
- Valid when
- |A| = n; counts relations from A to A.
Number of relations from $A$ to $B$
2^{\,mn}Where |A| = m and |B| = n.
One-one (injective) function
f(x_1) = f(x_2) \;\Rightarrow\; x_1 = x_2Distinct inputs are never sent to the same output.
Onto (surjective) function
\text{Range}(f) = BEvery element of the codomain is the image of at least one element of the domain.
Number of one-one functions
{}^{n}P_{m} = \dfrac{n!}{(n-m)!}One-one functions from a set of size m to a set of size n.
- Valid when
- Zero when m > n.
Number of functions from $A$ to $B$
n^{m}Each of the m elements of A may be sent to any of the n elements of B.
Key points
9 items · All key points
Reflexivity must hold for every element of the set — a single missing pair (a,a) destroys it.
This is the most commonly missed condition. On A = \{1,2,3\}, the relation \{(1,1),(2,2)\} is not reflexive because (3,3) is absent.
The empty relation on a non-empty set is symmetric and transitive, but never reflexive.
Symmetry and transitivity are conditional statements with no cases to check, so they hold vacuously. Reflexivity demands something exist, so it fails.
The universal relation A \times A is always an equivalence relation.
To show a relation is transitive you must check every chain; to show it is not, one broken chain suffices.
Write the counter-example explicitly: state (a,b) and (b,c) in R, and point out that (a,c) is not.
Equivalence classes are either identical or completely disjoint — they never partially overlap.
Whether a function is onto depends on the codomain you were given, not on the formula alone.
f(x) = x^2 is not onto from \mathbb{R} to \mathbb{R}, but is onto from \mathbb{R} to [0,\infty).
A strictly increasing or strictly decreasing function is automatically one-one.
This is often the quickest justification available: show f'(x) > 0 throughout and injectivity follows.
For a function from a finite set to itself, one-one and onto are equivalent. For infinite sets they are not.
f : \mathbb{N} \to \mathbb{N}, f(n) = n+1, is one-one but misses 1, so it is not onto.
A function is invertible exactly when it is bijective — this is what makes the next chapter possible.
Common mistakes
7 items · All common mistakes
- Mistake
Concluding a relation is reflexive after checking only the elements that happen to appear in R.
InsteadCheck every element of the underlying set A, including those that appear in no pair at all.
WhyStudents read the list of pairs rather than the set the relation is defined on, so elements absent from the list are never tested.
- Mistake
Treating symmetry and transitivity as though one implies the other.
InsteadThey are independent. R = \{(1,2),(2,1)\} on \{1,2\} is symmetric but not transitive, since (1,2) and (2,1) are present but (1,1) is not.
WhyBoth properties are about pairs pointing in more than one direction, which makes them feel related when they are not.
- Mistake
Claiming that a symmetric and transitive relation must be reflexive.
InsteadIt need not be. The empty relation on a non-empty set is the standard counter-example.
WhyThe tempting argument — (a,b) and (b,a) give (a,a) — assumes some pair involving a exists, which the definition does not guarantee.
- Mistake
Proving a property by verifying one or two numerical cases.
InsteadExamples can only disprove. To prove a property, argue with arbitrary a, b, c.
WhyChecking cases feels like evidence, but a relation can satisfy a property on many elements and still fail on one.
- Mistake
Checking ontoness against the range instead of the stated codomain.
InsteadCompare the range with the codomain given in the question. If they differ, the function is not onto.
WhyThe range is always covered by definition, so this reasoning makes every function look onto.
- Mistake
Deciding f(x) = x^2 is one-one because squaring 'looks like' a direct rule.
InsteadOn \mathbb{R} it is not: f(2) = f(-2) = 4. It becomes one-one only after the domain is restricted to [0,\infty) or (-\infty,0].
WhyEven powers collapse sign information, which is easy to overlook when reading the formula rather than the graph.
- Mistake
Assuming that a one-one function must also be onto because a finite example behaved that way.
InsteadThe equivalence holds only for functions from a finite set to itself. For infinite domains, check each property separately.