CBSE Class 12 Mathematics

Relations and their representation

The Cartesian product $A \times B$ of two sets is the set of all ordered pairs $(a, b)$ with $a \in A$ and $b \in B$. A relation $R$ from $A$ to $B$ is any subset of $A \times B$, and we write $a\,R\,b$ when $(a, b) \in R$. A relation on a set $A$ is a relation from $A$ to $A$, so it is a subset of $A \times A$.

If $A$ has $n$ elements then $A \times A$ has $n^2$ ordered pairs, and each pair is either in a relation or out of it, so the number of relations on $A$ is $2^{n^2}$. For $A = \{1, 2\}$ that is $2^4 = 16$.

Two extreme relations are named. The empty relation has no pairs at all, so $R = \varnothing$. The universal relation contains every pair, so $R = A \times A$. A relation on a small set can be written as a list of pairs, drawn as arrows between the elements, or read from a rule such as "$a\,R\,b$ if $a - b$ is even".

Reflexive, symmetric and transitive relations

A relation $R$ on a set $A$ is

• reflexive if $(a, a) \in R$ for every $a \in A$;
• symmetric if $(a, b) \in R$ implies $(b, a) \in R$;
• transitive if $(a, b) \in R$ and $(b, c) \in R$ together imply $(a, c) \in R$.

Each property is a statement about all elements, so one failing case is enough to show that it does not hold, and to show that it does hold you must argue for a general element. A single missing pair decides reflexivity: on $\{1, 2, 3\}$ the relation must contain $(1,1)$, $(2,2)$ and $(3,3)$.

For a relation given as a list of pairs, check symmetry by looking at each pair and its reverse. Check transitivity by taking every pair $(a, b)$ and every pair that starts with $b$, and confirming that the pair $(a, c)$ is present. A pair such as $(a, a)$ followed by $(a, b)$ gives back $(a, b)$, so pairs with equal entries rarely cause trouble; the trouble comes from chains between different elements.

Take $A = \{1, 2, 3\}$ and $R = \{(1, 2), (2, 1)\}$. The relation is symmetric, because $(1, 2)$ and $(2, 1)$ are both present. It is not reflexive, because $(1, 1)$ is missing. It is not transitive either: $(1, 2)$ and $(2, 1)$ together would need $(1, 1)$. So a relation can be symmetric without being either of the other two.

The empty relation on a non-empty set is symmetric and transitive, because there is no pair that could break either condition, and it is not reflexive. The universal relation has all three properties.

Relations defined by a rule

To test a rule-based relation, work with general elements.

On $\mathbb{R}$, let $a\,R\,b$ mean $a \le b$. It is reflexive, since $a \le a$. It is transitive, since $a \le b$ and $b \le c$ give $a \le c$. It is not symmetric, because $2 \le 5$ but $5 \not\le 2$.

On $\mathbb{N}$, let $a\,R\,b$ mean that $a$ divides $b$. It is reflexive because $a \mid a$, and transitive because $b = ka$ and $c = lb$ give $c = (kl)a$. It is not symmetric, because $3 \mid 12$ but $12 \nmid 3$.

On the set of all straight lines in a plane, let $L_1\,R\,L_2$ mean that $L_1$ is perpendicular to $L_2$. It is symmetric, since perpendicularity does not depend on the order. It is not reflexive, because no line is perpendicular to itself, and it is not transitive, because if $L_1 \perp L_2$ and $L_2 \perp L_3$ then $L_1$ and $L_3$ are parallel.

On $\mathbb{Z}$, let $a\,R\,b$ mean $a < b$. It is transitive and nothing else: $a < a$ is false, and $a < b$ rules out $b < a$.

Equivalence relations

A relation is an equivalence relation if it is reflexive, symmetric and transitive. The idea is that it groups together things that are alike in some chosen way, and "alike" must pass all three tests.

For $A = \{1, 2, 3\}$, take

It contains every $(a, a)$, so it is reflexive. The only pairs with different entries are $(1, 2)$ and $(2, 1)$, each the reverse of the other, so it is symmetric. For transitivity the chains to check are $(1,2)$ then $(2,1)$, which needs $(1,1)$; $(2,1)$ then $(1,2)$, which needs $(2,2)$; and $(1,2)$ then $(2,2)$, which needs $(1,2)$. All are present. So $R$ is an equivalence relation.

Proving a rule is an equivalence relation

On $\mathbb{Z}$ let $a\,R\,b$ mean that $a - b$ is divisible by $5$. Write the proof in three lines, one for each property.

For reflexivity, $a - a = 0 = 5 \times 0$, so $a\,R\,a$.

For symmetry, suppose $a - b = 5k$ for an integer $k$. Then $b - a = 5(-k)$, which is also a multiple of $5$, so $b\,R\,a$.

For transitivity, suppose $a - b = 5k$ and $b - c = 5m$. Adding gives $a - c = 5(k + m)$, so $a\,R\,c$.

So $R$ is an equivalence relation. Another example is the relation "$|a| = |b|$" on $\mathbb{R}$: equality of a quantity always gives an equivalence relation, because equality is itself reflexive, symmetric and transitive. Having the same birthday, or the same remainder on division by a fixed number, works in the same way.

A relation that looks like an equivalence relation can fail. On $\mathbb{R}$, "$|a - b| \le 1$" is reflexive and symmetric, but $0$ and $1$ are related, $1$ and $2$ are related, and $0$ and $2$ are not, so it is not transitive.

Equivalence classes

For an equivalence relation $R$ on $A$, the equivalence class of an element $a$ is the set $[a] = \{x \in A : x\,R\,a\}$ of everything related to $a$. Because $R$ is reflexive, $a \in [a]$, so no class is empty. Because $R$ is symmetric and transitive, two classes are either identical or have nothing in common. The classes therefore divide $A$ into separate pieces that together cover $A$, which is called a partition of $A$. Conversely, any partition defines an equivalence relation, in which two elements are related when they lie in the same piece.

In the example above, $[1] = [2] = \{1, 2\}$ and $[3] = \{3\}$.

For "$a - b$ is divisible by $5$" on $\mathbb{Z}$ there are exactly five classes, one for each remainder on division by $5$:

and likewise $[2]$, $[3]$ and $[4]$. The number $13$ lies in $[3]$, since $13 - 3 = 10$, and the number $-1$ lies in $[4]$, since $-1 - 4 = -5$.

The class of an element is found by asking what it is related to, so for "$|a| = |b|$" on $\mathbb{R}$ the class of $3$ is $\{3, -3\}$ and the class of $0$ is $\{0\}$.

Checking a relation step by step

1. Write out what each property asks for, using the rule of the relation.
2. For reflexivity, substitute $b = a$ and see whether the condition is always true.
3. For symmetry, assume $a\,R\,b$ and work out whether $b\,R\,a$ follows.
4. For transitivity, assume $a\,R\,b$ and $b\,R\,c$ and work out whether $a\,R\,c$ follows.
5. To show a property fails, give one specific counterexample with actual numbers.

As a last example, on $\mathbb{N}$ let $a\,R\,b$ mean that $a + b$ is even. Reflexive: $a + a = 2a$ is even. Symmetric: $a + b = b + a$. Transitive: if $a + b$ and $b + c$ are even then $a$, $b$ and $c$ all have the same parity, so $a + c$ is even. The relation is an equivalence relation with two classes, the odd numbers and the even numbers.