Let N be the set of all natural numbers and R be a relation on N× N defined by (a,b)R(c,d) ad=bc for all (a,b),(c,d) N× N. Show that R is an equivalence relation on N× N. Also, find the equivalence class of (2,6), i.e., [(2,6)].

R is an equivalence relation; [(2,6)]=\(x,3x):x N\=\(1,3),(2,6),(3,9),…\

MathsCBSE Class 12 · Sample paper, session 2023-24 · 5 marks · Long answer

CBSE Class 12 Mathematics sample paper 2023-24: question 33(a)

From the official CBSE sample question paper for 2023-24, section D, question 33(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(a) · 5 marks · Long answerRelations and Functions

Let $N$ be the set of all natural numbers and $R$ be a relation on $N \times N$ defined by $(a, b) R (c, d) \Leftrightarrow a d = b c$ for all $(a, b), (c, d) \in N \times N$ . Show that $R$ is an equivalence relation on $N \times N$ . Also, find the equivalence class of $(2, 6)$ , i.e., $[(2, 6)]$ .

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

Open one hint at a time and try again before the next.

Hint 1

Check reflexive, symmetric and transitive one by one, using the condition $a d = b c$ .

Hint 2

For transitivity, multiply $a d = b c$ and $c f = d e$ and cancel $c d$ . For the class, solve $x \cdot 6 = y \cdot 2$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Reflexive: For any $(a, b) \in N \times N$ , $ab = ba$ because multiplication is commutative. So $(a, b) R (a, b)$ , and $R$ is reflexive.
1 mark
Symmetric: Suppose $(a, b) R (c, d)$ . Then $a d = b c$ , so $c b = d a$ . This means $(c, d) R (a, b)$ , and $R$ is symmetric.
1 mark
Transitive: Suppose $(a, b) R (c, d)$ and $(c, d) R (e, f)$ . Then $a d = b c$ and $c f = d e$ . Multiplying, $(a d) (c f) = (b c) (d e)$ , so $a f \cdot c d = b e \cdot c d$ . Since $c, d \in N$ , $c d \neq = 0$ , which gives $a f = b e$ . So $(a, b) R (e, f)$ , and $R$ is transitive.
1 mark
Since $R$ is reflexive, symmetric and transitive, $R$ is an equivalence relation on $N \times N$ .
½ mark
$[(2, 6)] = {(x, y) \in N \times N : (x, y) R (2, 6)} = {(x, y) : 6 x = 2 y} = {(x, y) : y = 3 x}$ .
½ mark
$[(2, 6)] = {(x, 3 x) : x \in N} = {(1, 3), (2, 6), (3, 9), \dots}$ .
1 mark

Answer: $R$ is an equivalence relation; $[(2, 6)] = {(x, 3 x) : x \in N} = {(1, 3), (2, 6), (3, 9), \dots}$

Where the marks are

Proves reflexivity (1 mark)
Proves symmetry (1 mark)
Proves transitivity, cancelling $c d \neq = 0$ (1 mark)
Concludes that $R$ is an equivalence relation (½ mark)
Sets up the condition $6 x = 2 y$ for the class (½ mark)
Writes $[(2, 6)] = {(x, 3 x) : x \in N}$ (1 mark)

Revise the chapter

Chapter: Relations and Functions (notes, formulas, where marks are lost and every past question)
Topic: Equivalence Relations and Equivalence Classes

Practise this chapterBoard-style questions from Relations and Functions, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2023-24: every question, by section.

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CBSE Class 12 Mathematics sample paper 2023-24: question 33(a)

Hints

Step-by-step solution

Revise the chapter

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