NCERT Solutions of Class 12 Maths | CBSE Textbook Solutions | Chapter 1 | Relations and Functions | Exercise 1.1 | Question 8 |

Question Details
BookNCERT Textbook
Chapter1 [Relations and Functions]
Question No.8
Question TypeExercise

Question: Show that the relation \(\mathrm{R}\) in the set \(\mathrm{A}=\{1,2,3,4,5\}\) given by \(\mathrm{R}=\{(a, b):|a-b|\) is even \(\}\), is an equivalence relation. Show that all the elements of \(\{1,3,5\}\) are related to each other and all the elements of \(\{2,4\}\) are related to each other. But no element of \(\{1,3,5\}\) is related to any element of \(\{2,4\}\).

Expert Answer

Given \(A=\{1,2,3,4,5\}\) and \(R=\{(a, b):|a-b|\) is even \(\}\)

(a) Reflexive : \(R=\{(a, a):|a-a|=0\) zero is an even no. \(\}\)
(b) Symmetric : If \(R=\left\{\left(a_1, a_2\right):\left|a_1-a_2\right|\right.\) is even is true \(\}\) then \(R=\left\{\left(a_2, a_1\right):\left|a_2-a_1\right|\right.\) will be even \(\}\)
So, symmetric.

(c) Transitive : If \(R=\left\{\left(a_1, a_2\right)\right.\) : \(\left|a_1-a_2\right|\) is even \(\}\) and \(R=\left\{\left(a_2, a_3\right):\left|a_2-a_3\right|\right.\) is even \(\}\) then \(R=\left\{\left(a_1, a_3\right)\right\}\) will also be even. So equivalence relation.
This is nothing but concept of equivalence class. \(\{1,3,5\}\) are odd numbers such that difference is always even. Thus \(\{(1,3)\}\), \(\{(1,5)\}\) will be even. So this is an equivalence class of 1 denoted by [1]. Now \(\{2,4\}\) are even numbers here also difference is even. These two are disjoint as well as \(\{1,3,5\} \cup\{2,4\} \equiv\{1,2,3,4,5\} \equiv A\).

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