Board 2025 Class 12 Maths Q37 Solution: Case study: five relations on {1, 2, 3}

The question

A class-room teacher is keen to assess the learning of her students the concept of “relations” taught to them. She writes the following five relations each defined on the set A = $\{1, 2, 3\}$ :

\begin{aligned} R_1 &= \{(2, 3), (3, 2)\} \\ R_2 &= \{(1, 2), (1, 3), (3, 2)\} \\ R_3 &= \{(1, 2), (2, 1), (1, 1)\} \\ R_4 &= \{(1, 1), (1, 2), (3, 3), (2, 2)\} \\ R_5 &= \{(1, 1), (1, 2), (3, 3), (2, 2), \\ &\qquad (2, 1), (2, 3), (3, 2)\} \end{aligned}

The students are asked to answer the following questions about the above relations :

(i) Identify the relation which is reflexive, transitive but not symmetric. (1 mark)

(ii) Identify the relation which is reflexive and symmetric but not transitive. (1 mark)

(iii) (a) Identify the relations which are symmetric but neither reflexive nor transitive. (2 marks)

OR

(iii) (b) What pairs should be added to the relation $R_2$ to make it an equivalence relation ? (2 marks)

Exam answer

(i) $R_4$ is reflexive, as $(1, 1), (2, 2), (3, 3) \in R_4$ ⁠. It is not symmetric, as $(1, 2) \in R_4$ but $(2, 1) \notin R_4$ ⁠. It is transitive: its only pair with unequal entries is $(1, 2)$ ⁠, and the chains $(1, 1), (1, 2)$ and $(1, 2), (2, 2)$ both give $(1, 2) \in R_4$ ⁠.

Why?

The three tests on a set A:

Reflexive: $(a, a) \in R$ for every $a$ in A.

Symmetric: whenever $(a, b) \in R$ ⁠, also $(b, a) \in R$ ⁠.

Transitive: whenever $(a, b), (b, c) \in R$ ⁠, also $(a, c) \in R$ ⁠.

One pair that breaks a rule is enough to say "not". To say "yes" for transitivity, every chain must work.

The relation is $R_4$ ⁠.

Why?

No other relation fits. $R_1$ ⁠, $R_2$ and $R_3$ are not reflexive ( $(1, 1) \notin R_1$ ⁠, $(1, 1) \notin R_2$ ⁠, $(2, 2) \notin R_3$ ⁠), and $R_5$ is symmetric.

(ii) $R_5$ is reflexive, as $(1, 1), (2, 2), (3, 3) \in R_5$ ⁠. It is symmetric, as $(1, 2), (2, 1)$ and $(2, 3), (3, 2)$ are in $R_5$ ⁠. It is not transitive, as $(1, 2), (2, 3) \in R_5$ but $(1, 3) \notin R_5$ ⁠.

Why?

For symmetry, each pair with unequal entries must come with its mirror image: $(1, 2)$ with $(2, 1)$ ⁠, and $(2, 3)$ with $(3, 2)$ ⁠.

One broken chain is enough to show it is not transitive: $1 \to 2 \to 3$ needs $(1, 3)$ ⁠, which is missing.

The relation is $R_5$ ⁠.

Why?

Of the other relations, only $R_4$ is reflexive, and it is not symmetric.

(iii) (a) $R_1$ is symmetric. It is not reflexive, as $(1, 1) \notin R_1$ ⁠, and not transitive, as $(2, 3), (3, 2) \in R_1$ but $(2, 2) \notin R_1$ ⁠.

Why?

$R_1$ is symmetric because its two pairs are mirror images of each other.

In the transitive test, $c$ may equal $a$ ⁠. So $(2, 3), (3, 2)$ requires $(2, 2)$ ⁠.

Watch out: students often miss this chain and wrongly call $R_1$ transitive.

$R_3$ is symmetric. It is not reflexive, as $(2, 2) \notin R_3$ ⁠, and not transitive, as $(2, 1), (1, 2) \in R_3$ but $(2, 2) \notin R_3$ ⁠.

Why?

In $R_3$ ⁠, $(1, 2)$ and $(2, 1)$ are mirror images, and $(1, 1)$ is its own mirror image. So $R_3$ is symmetric.

The chain $2 \to 1 \to 2$ needs $(2, 2)$ ⁠, which is missing.

( $R_2$ and $R_4$ are not symmetric, and $R_5$ is reflexive.)

The relations are $R_1$ and $R_3$ ⁠.

(iii) (b) For reflexivity, $R_2$ needs $(1, 1), (2, 2), (3, 3)$ ⁠. For symmetry, it needs $(2, 1), (3, 1), (2, 3)$ ⁠.

Why?

An equivalence relation is reflexive, symmetric and transitive. Add pairs one property at a time.

Reflexive: all three pairs $(a, a)$ are missing. Symmetric: the mirror images of $(1, 2), (1, 3), (3, 2)$ are $(2, 1), (3, 1), (2, 3)$ ⁠.

Then $R_2$ becomes $A \times A$ ⁠, which is also transitive.

Why?

Now $R_2$ has $3 + 6 = 9$ pairs, which is every pair of $A \times A$ ⁠. Then every $(a, c)$ is in it, so it is transitive too.

None of the six pairs can be left out: each one is needed for reflexivity or for symmetry.

$\therefore$ The pairs to be added are $(1, 1), (2, 2), (3, 3)$ ⁠, $(2, 1), (3, 1)$ and $(2, 3)$ ⁠.

Where this comes from

Chapter 1, Relations and Functions: Case study: reflexive, symmetric, transitive.

Question type: A relation given as a list of pairs

How to spot it: click to see

Type 1. A relation given as a list of pairs: check each property

Spot it: $R$ is written out pair by pair, on a small set like $\{1, 2, 3\}$ ⁠.

Here: All five relations are listed pair by pair on $\{1, 2, 3\}$ ⁠, so test each property on the pairs: for example $(1, 2), (2, 3) \in R_5$ but $(1, 3) \notin R_5$ ⁠. In (iii)(b), add the missing $(a, a)$ pairs and the reverses of $(1, 2), (1, 3), (3, 2)$ ⁠.

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Q37: Case study: five relations on {1,2,3}\{1, 2, 3\}{1,2,3}

Q37: Case study: five relations on $\{1, 2, 3\}$