CBSE Sample Paper 2022-23 · Section D · 5 marks

Q33: $(a, b)\,R\,(c, d) \iff ad = bc$ is an equivalence relation, or $A \subset B$ on $P(X)$

Sample 2022-23Ch 1 · Relations and Functions

The question

(a) Define the relation R in the set $N \times N$ as follows:

For (a, b), (c, d) $\in N \times N$ ⁠, (a, b) R (c, d) iff ad = bc. Prove that R is an equivalence relation in $N \times N$ ⁠.

(b) Given a non-empty set X, define the relation R in P(X) as follows:

For A, B $\in P(X)$ ⁠, $(A, B) \in R$ iff $A \subset B$ ⁠. Prove that R is reflexive, transitive and not symmetric.

Exam answer

(a) Let $(a, b) \in N \times N$ ⁠. Then $ab = ba$ (by the commutative property of multiplication of natural numbers), so $(a, b)\,R\,(a, b)$ ⁠.

Why?

Reflexive means every element is related to itself. Here the element is the pair $(a, b)$ ⁠, so we test $(a, b)\,R\,(a, b)$ ⁠.

The rule $(a, b)\,R\,(c, d) \iff ad = bc$ with $c = a$ ⁠, $d = b$ asks for $ab = ba$ ⁠, which is always true.

Hence, R is reflexive.

Let $(a, b), (c, d) \in N \times N$ such that $(a, b)\,R\,(c, d)$ ⁠. Then, by the commutative property of multiplication of natural numbers,

\begin{aligned} &ad = bc \\ \Rightarrow\ &cb = da \\ \Rightarrow\ &(c, d)\,R\,(a, b) \end{aligned}

Why?

Symmetric means: whenever $(a, b)\,R\,(c, d)$ ⁠, also $(c, d)\,R\,(a, b)$ ⁠.

For $(c, d)\,R\,(a, b)$ the rule needs (first of the first pair) $\times$ (second of the second pair) to equal (second of the first pair) $\times$ (first of the second pair), that is $c \times b = d \times a$ ⁠.

From $ad = bc$ ⁠, swap the two sides and the order of each product: $cb = da$ ⁠.

\begin{aligned} ad &= bc \\ bc &= ad \quad (\text{swap sides}) \\ cb &= da \quad (bc = cb,\ ad = da) \end{aligned}

Hence, R is symmetric.

Let $(a, b), (c, d), (e, f) \in N \times N$ such that $(a, b)\,R\,(c, d)$ and $(c, d)\,R\,(e, f)$ ⁠. Then

ad = bc \quad\text{and}\quad cf = de

\begin{aligned} \Rightarrow\ &adcf = bcde \\ \Rightarrow\ &af = be \quad [c, d \ne 0] \\ \Rightarrow\ &(a, b)\,R\,(e, f) \end{aligned}

Why?

Transitive means: if $(a, b)\,R\,(c, d)$ and $(c, d)\,R\,(e, f)$ ⁠, then $(a, b)\,R\,(e, f)$ ⁠, which needs $af = be$ ⁠.

Multiply the two equations side by side: $(ad)(cf) = (bc)(de)$ ⁠. Both sides contain the factor $cd$ ⁠. Since $c$ and $d$ are natural numbers, $cd \ne 0$ ⁠, so we may cancel it.

\begin{aligned} (ad)(cf) &= (bc)(de) \\ (cd)(af) &= (cd)(be) \\ \frac{(cd)(af)}{cd} &= \frac{(cd)(be)}{cd} \\ af &= be \end{aligned}

Watch out: cancelling is allowed only because $cd \ne 0$ ⁠. Write the reason beside the step.

Hence, R is transitive.

Since R is reflexive, symmetric and transitive, R is an equivalence relation on $N \times N$ ⁠.

Why?

An equivalence relation is one that is reflexive, symmetric and transitive. All three must be proved before this conclusion.

(b) Let $A \in P(X)$ ⁠. Then $A \subset A$ ⁠, so $(A, A) \in R$ ⁠.

Why?

$P(X)$ ⁠, the power set, is the set of all subsets of $X$ ⁠. Every set is a subset of itself, because each element of $A$ is an element of $A$ ⁠.

Hence, R is reflexive.

Let $A, B, C \in P(X)$ such that $(A, B) \in R$ and $(B, C) \in R$ ⁠. Then

\begin{aligned} &A \subset B,\ B \subset C \\ \Rightarrow\ &A \subset C \\ \Rightarrow\ &(A, C) \in R \end{aligned}

Why?

Every element of $A$ is in $B$ ⁠, and every element of $B$ is in $C$ ⁠. So every element of $A$ is in $C$ ⁠, which means $A \subset C$ ⁠.

Hence, R is transitive.

$\emptyset, X \in P(X)$ and $\emptyset \subset X$ ⁠. Hence, $(\emptyset, X) \in R$ ⁠. But $X \not\subset \emptyset$ (X is non-empty), which implies that $(X, \emptyset) \notin R$ ⁠.

Why?

To show that R is not symmetric, one counterexample is enough: a pair in R whose reverse is not in R.

The empty set is a subset of every set. $X$ is non-empty, so it has an element, and that element is not in $\emptyset$ ⁠. So $X \not\subset \emptyset$ ⁠.

Watch out: this is why the question says $X$ is non-empty. If $X = \emptyset$ ⁠, the pair would be $(\emptyset, \emptyset)$ ⁠, and its reverse is in R.

$\therefore$ R is not symmetric. Hence, R is reflexive and transitive but not symmetric.

Where this comes from

Chapter 1, Relations and Functions: Equivalence relation; reflexive, transitive only.

Question type: (a) Show that a relation is an equivalence relation; (b) A relation given by a rule

How to spot it: click to see

(a) Type 3. Show that a relation is an equivalence relation

Spot it: the words "is an equivalence relation", often with "2 divides $a - b$ " or " $|a - b|$ is even".

Here: The rule is $ad = bc$ on pairs. For transitive, multiply $ad = bc$ by $cf = de$ and cancel $cd$ ⁠, which is not 0 in $N$ ⁠, to get $af = be$ ⁠.

(b) Type 2. A relation given by a rule: letters to prove, numbers to disprove

Spot it: $R$ is given by a condition, such as $a \le b$ on $\R$ ⁠, and the set is too big to list every pair.

Here: The rule is $A \subset B$ on $P(X)$ ⁠. Reflexive and transitive hold for any sets; to break symmetry, take $\emptyset \subset X$ but $X \not\subset \emptyset$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Relations and Functions in past papers: question 14 of 18 · see them all

Q33: (a,b) R (c,d) ⟺ ad=bc(a, b)\,R\,(c, d) \iff ad = bc(a,b)R(c,d)⟺ad=bc is an equivalence relation, or A⊂BA \subset BA⊂B on P(X)P(X)P(X)

Q33: $(a, b)\,R\,(c, d) \iff ad = bc$ is an equivalence relation, or $A \subset B$ on $P(X)$