Relations

Two sets sit side by side — a list of book titles and a list of authors, or a set of countries and a set of capitals. A relation is simply the mathematical device for saying which items from the first set are linked to which items from the second: it is a chosen subset of ordered pairs picked out of the Cartesian product $A \times B$ . Once you can form $A \times B$ , a relation just filters that product down to the pairs that satisfy some rule or connection you care about — "is the square of," "divides," "is the capital of." This article builds the formal definition, shows how to read off a relation's domain, co-domain and range, and counts how many different relations can exist between two finite sets. Reflexive, symmetric and transitive relations are a Class 12 topic and are deliberately left out here.

The core insight

A relation $R$ from $A$ to $B$ is any subset of $A \times B$ , and it can always be described by roster form, set-builder form or an arrow diagram, with domain, co-domain and range read off from the ordered pairs it contains.

The core idea

Given two non-empty sets $A$ and $B$ , a relation $R$ from $A$ to $B$ is defined as any subset of the Cartesian product:

R \subseteq A \times B.

where: $A \times B$ is the set of all ordered pairs $(a,b)$ with $a \in A$ and $b \in B$ , and $R$ is whichever pairs we choose to keep. If $(a,b) \in R$ we say $a$ is related to $b$ under $R$ , written $a\,R\,b$ . A relation can be written in three equivalent ways. In roster form the pairs are listed explicitly, e.g. $R = \{(1,2),(2,4),(3,6)\}$ . In set-builder form the defining rule is stated instead of the pairs, e.g. $R = \{(x,y) : y = 2x,\ x \in A,\ y \in B\}$ . In an arrow diagram, $A$ and $B$ are drawn as two ovals and an arrow is drawn from each $a$ to every $b$ with $(a,b) \in R$ — a picture of the same information. Every relation carries three associated sets. The domain is the set of first components actually used:

\text{Domain}(R) = \{a : (a,b) \in R \text{ for some } b\}.

The co-domain is the whole of $B$ , fixed in advance regardless of what gets used. The range is the set of second components actually hit:

\text{Range}(R) = \{b : (a,b) \in R \text{ for some } a\}.

where: range is always a subset of co-domain, but the two need not be equal — co-domain is the declared target set $B$ , while range is only the part of $B$ that some pair actually reaches.

Figure

Two ovals labelled A and B with arrows from selected elements of A to selected elements of B, and the used elements of B marked as the range inside the larger co-domain oval

The number of possible relations from A to B

Since a relation is nothing more than a chosen subset of $A \times B$ , the goal here is to count, in one formula, how many distinct relations can be formed between two finite sets $A$ and $B$ purely from the sizes of the sets, using only the fact that $A \times B$ has $n(A)\cdot n(B)$ elements and that a set with $k$ elements has $2^k$ subsets in total.

Derivation ACounting the relations between two finite sets

The goal is to show that if $n(A) = p$ and $n(B) = q$ , the total number of relations that can be defined from $A$ to $B$ — including the empty relation and the full relation $R = A \times B$ — is exactly $2^{pq}$ , by combining the size of the Cartesian product with the subset-counting rule.

Step 1. Find the size of $A \times B$ . Every element of $A$ can be paired with every element of $B$ , so $A \times B$ contains exactly $n(A) \times n(B) = pq$ ordered pairs. This is the ground set from which a relation is drawn.

Step 2. Recall that a relation $R$ is, by definition, any subset of $A \times B$ — there is no restriction on which pairs may be included or excluded, so every subset of $A \times B$ is a legitimate relation, and every relation is some subset of $A \times B$ .

Step 3. Count the subsets of a set with $pq$ elements. A standard result (proved by choosing, independently, to include or exclude each of the $pq$ elements) is that a set with $k$ elements has exactly $2^k$ subsets. Applying this with $k = pq$ gives $2^{pq}$ subsets of $A \times B$ .

Step 4. Conclude. Since relations from $A$ to $B$ are exactly the subsets of $A \times B$ , the total number of possible relations from $A$ to $B$ is $2^{pq} = 2^{n(A)\cdot n(B)}$ . This count includes the empty relation ( $\varnothing$ , where nothing is related to anything) and the full relation ( $R = A \times B$ , where everything is related to everything).

Worked example

Let $A = \{1,2,3,4\}$ and $B = \{1,4,9,16,25\}$ , and define $R = \{(x,y) : y = x^2,\ x \in A,\ y \in B\}$ .

Step 1. Apply the rule to each element of $A$ . For $x=1$ , $y=1 \in B$ , so $(1,1) \in R$ . For $x=2$ , $y=4 \in B$ , so $(2,4) \in R$ . For $x=3$ , $y=9 \in B$ , so $(3,9) \in R$ . For $x=4$ , $y=16 \in B$ , so $(4,16) \in R$ .

Step 2. Write $R$ in roster form: $R = \{(1,1),(2,4),(3,9),(4,16)\}$ .

Step 3. Read off the three sets. Domain $= \{1,2,3,4\}$ , since every element of $A$ appeared as a first entry. Co-domain $= B = \{1,4,9,16,25\}$ , the full declared set. Range $= \{1,4,9,16\}$ — note $25$ sits in the co-domain but not the range, since no $x \in A$ satisfies $x^2 = 25$ .

Step 4. Apply the counting formula from the derivation as a check on scale. Here $n(A) = 4$ and $n(B) = 5$ , so $n(A\times B) = 20$ and the total number of possible relations from $A$ to $B$ is $2^{20} = 1{,}048{,}576$ — the single relation $R$ above is just one of over a million subsets of $A \times B$ .

The relation is $R = \{(1,1),(2,4),(3,9),(4,16)\}$ , with domain $\{1,2,3,4\}$ , co-domain $\{1,4,9,16,25\}$ and range $\{1,4,9,16\}$ .

Competitive stretch

Take $A = \{2,3,5\}$ , $B = \{4,6,9,10\}$ , and $R = \{(a,b) : a \text{ divides } b\}$ . Checking divisibility pair by pair gives $R = \{(2,4),(2,6),(2,10),(3,6),(3,9),(5,10)\}$ , so domain $= \{2,3,5\}$ and range $= \{4,6,9,10\}$ — here every element of $B$ is hit, so range equals co-domain. Since $n(A)=3$ and $n(B)=4$ , $n(A\times B)=12$ , so the total number of possible relations from $A$ to $B$ is $2^{12} = 4096$ .

Common traps

Confusing co-domain with range. Co-domain is the entire set $B$ as declared; range is only the part of $B$ actually reached by some pair. They coincide only when every element of $B$ is used.
Treating the rule as the relation. " $y = x^2$ " or " $a$ divides $b$ " is just a description of which pairs belong to $R$ ; the relation itself is the resulting set of ordered pairs, not the formula.
Applying reflexive, symmetric or transitive labels. These classifications belong to Class 12 — this syllabus only asks for domain, co-domain, range and the three representations.
Forgetting the $2^{pq}$ count includes the empty and full relations. Students often only count "interesting" relations and undercount, or forget that $R = \varnothing$ and $R = A \times B$ are both valid relations.

Ask the tutor: Still mixing up domain, co-domain and range, or want more practice converting a word rule into roster form? Ask the AI tutor for a fresh set of examples with step-by-step working.

Is every relation also a function

No. A function is a special kind of relation where every element of the domain set $A$ is related to exactly one element of $B$ — no element of $A$ may have two arrows going out or none at all. A general relation has no such restriction: some elements of $A$ can point to several elements of $B$ , and others can point to none. The Class 11 relations chapter only asks you to identify domain, co-domain and range for a given relation; the extra function condition and its vocabulary come right after this topic.

Why does range not always equal co-domain

Co-domain is simply the set $B$ you declared at the start, fixed before you even define $R$ . Range is worked out afterward, from whichever pairs actually satisfy the rule of $R$ . If some elements of $B$ never get paired with anything from $A$ — like $25$ in the squares example, since no element of $A = \{1,2,3,4\}$ squares to $25$ — those elements stay in the co-domain but are excluded from the range. Range equals co-domain only in the special case where every element of $B$ happens to be used.

How do I quickly count the number of relations possible between two sets

Multiply the sizes of the two sets to get $n(A \times B) = n(A) \cdot n(B)$ , then raise 2 to that power. This works because a relation is any subset of $A \times B$ , and a set with $k$ elements has exactly $2^k$ subsets. For example, with $n(A) = 3$ and $n(B) = 4$ , $n(A \times B) = 12$ , so there are $2^{12} = 4096$ possible relations, ranging from the empty relation to the full relation $A \times B$ itself.