{"id":68976,"date":"2023-08-12T14:04:00","date_gmt":"2023-08-12T08:34:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=68976"},"modified":"2023-11-10T12:36:11","modified_gmt":"2023-11-10T07:06:11","slug":"rd-sharma-solutions-class-12-maths-chapter-1-exercise-1-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-1-exercise-1-2\/","title":{"rendered":"RD Sharma Solutions for Class 12 Maths Exercise 1.2 Chapter 1 Relation (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-117932\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-1-Exercise-1.2.png\" alt=\"RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-1-Exercise-1.2.png 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-1-Exercise-1.2-768x432.png 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2:<\/strong> Students will mostly gain conceptual knowledge of equivalence relations through this practice. If a relation is reflexive, symmetric, and transitive, it is said to be an equivalence relationship. Class 12 is a crucial year for students in terms of shaping and achieving their long-term objectives. These <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-1-relations\/\"><strong>RD Sharma Solutions Class 12 Maths Chapter 1<\/strong><\/a> Exercise 1.2 problems&#8217; solutions are created in a form of comprehensive manner, which develops students&#8217; problem-solving abilities.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d77dd14e355\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" 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src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Maths-Chapter-1-Ex-1.2-1.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Maths-Chapter-1-Ex-1.2-1.pdf\">RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-maths-rd-sharma-solutions-for-class-12-chapter-1-%e2%80%93-relations-exercise-12-question\"><\/span>Access answers to Maths RD Sharma Solutions For Class 12 Chapter 1 \u2013 Relations Exercise 1.2 Question<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. Show that the relation&nbsp;R&nbsp;defined by&nbsp;R&nbsp;= {(a, b):&nbsp;a \u2013 b&nbsp;is divisible by 3;&nbsp;a, b&nbsp;\u2208&nbsp;Z} is an equivalence relation.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given R&nbsp;= {(a, b):&nbsp;a \u2013 b&nbsp;is divisible by 3;&nbsp;a, b&nbsp;\u2208&nbsp;Z} is a relation<\/p>\n<p>To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.<\/p>\n<p>Let us check these properties on R.<\/p>\n<p>Reflexivity:&nbsp;<\/p>\n<p>Let&nbsp;a&nbsp;be&nbsp;an&nbsp;arbitrary&nbsp;element&nbsp;of&nbsp;R.&nbsp;<\/p>\n<p>Then, a \u2013 a = 0 = 0 \u00d7 3<\/p>\n<p>\u21d2&nbsp;a \u2212 a&nbsp;is&nbsp;divisible&nbsp;by&nbsp;3<\/p>\n<p>\u21d2&nbsp;(a,&nbsp;a)&nbsp;\u2208&nbsp;R&nbsp;for&nbsp;all&nbsp;a&nbsp;\u2208&nbsp;Z<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;reflexive&nbsp;on&nbsp;Z.<\/p>\n<p>Symmetry:<\/p>\n<p>Let&nbsp;(a,&nbsp;b)&nbsp;\u2208&nbsp;R<\/p>\n<p>\u21d2&nbsp;a \u2212 b&nbsp;is&nbsp;divisible&nbsp;by&nbsp;3<\/p>\n<p>\u21d2&nbsp;a \u2212 b&nbsp;= 3p&nbsp;for&nbsp;some&nbsp;p&nbsp;\u2208&nbsp;Z<\/p>\n<p>\u21d2&nbsp;b \u2212 a&nbsp;= 3&nbsp;(\u2212p)<\/p>\n<p>Here,&nbsp;\u2212p&nbsp;\u2208&nbsp;Z<\/p>\n<p>\u21d2&nbsp;b \u2212 a&nbsp;is&nbsp;divisible&nbsp;by&nbsp;3<\/p>\n<p>\u21d2&nbsp;(b,&nbsp;a)&nbsp;\u2208&nbsp;R&nbsp;for&nbsp;all&nbsp;a,&nbsp;b&nbsp;\u2208&nbsp;Z<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;symmetric&nbsp;on&nbsp;Z.<\/p>\n<p>Transitivity:<\/p>\n<p>Let&nbsp;(a,&nbsp;b)&nbsp;and&nbsp;(b,&nbsp;c)&nbsp;\u2208&nbsp;R<\/p>\n<p>\u21d2 a \u2212 b and b \u2212 c is divisible by 3<\/p>\n<p>\u21d2&nbsp;a \u2013 b = 3p&nbsp;for&nbsp;some&nbsp;p&nbsp;\u2208&nbsp;Z<\/p>\n<p>And&nbsp;b \u2212&nbsp;c&nbsp;=&nbsp;3q&nbsp;for&nbsp;some&nbsp;q&nbsp;\u2208&nbsp;Z<\/p>\n<p>Adding&nbsp;the&nbsp;above&nbsp;two equations,&nbsp;we&nbsp;get<\/p>\n<p>a&nbsp;\u2212&nbsp;b&nbsp;+&nbsp;b \u2013&nbsp;c =&nbsp;3p +&nbsp;3q<\/p>\n<p>\u21d2&nbsp;a \u2212 c&nbsp;= 3&nbsp;(p + q)<\/p>\n<p>Here,&nbsp;p + q&nbsp;\u2208&nbsp;Z<\/p>\n<p>\u21d2&nbsp;a \u2212 c&nbsp;is&nbsp;divisible&nbsp;by&nbsp;3<\/p>\n<p>\u21d2&nbsp;(a,&nbsp;c)&nbsp;\u2208&nbsp;R&nbsp;for&nbsp;all&nbsp;a,&nbsp;c&nbsp;\u2208&nbsp;Z<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;transitive&nbsp;on&nbsp;Z.<\/p>\n<p>Therefore R is reflexive, symmetric, and transitive.<\/p>\n<p>Hence, R is an equivalence relation to Z.<\/p>\n<p><strong>2. Show that the relation&nbsp;R&nbsp;on the set&nbsp;Z&nbsp;of integers, given by<br>R&nbsp;= {(a, b): 2 divides&nbsp;a \u2013 b}, is an equivalence relation.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given R&nbsp;= {(a, b): 2 divides&nbsp;a \u2013 b} is a relation defined on Z.<\/p>\n<p>To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.<\/p>\n<p>Let us check these properties on R.<\/p>\n<p>Reflexivity:<\/p>\n<p>Let&nbsp;a&nbsp;be&nbsp;an&nbsp;arbitrary&nbsp;element&nbsp;of&nbsp;the&nbsp;set&nbsp;Z.&nbsp;<\/p>\n<p>Then,&nbsp;a&nbsp;\u2208&nbsp;R<\/p>\n<p>\u21d2&nbsp;a \u2212 a&nbsp;=&nbsp;0&nbsp;=&nbsp;0&nbsp;\u00d7&nbsp;2<\/p>\n<p>\u21d2&nbsp;2&nbsp;divides&nbsp;a&nbsp;\u2212&nbsp;a<\/p>\n<p>\u21d2&nbsp;(a,&nbsp;a)&nbsp;\u2208&nbsp;R&nbsp;for&nbsp;all&nbsp;a&nbsp;\u2208&nbsp;Z<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;reflexive&nbsp;on&nbsp;Z.<\/p>\n<p>Symmetry:<\/p>\n<p>Let&nbsp;(a,&nbsp;b) \u2208&nbsp;R<\/p>\n<p>\u21d2&nbsp;2&nbsp;divides&nbsp;a \u2212 b<\/p>\n<p>\u21d2 (a-b)\/2 = p&nbsp;for&nbsp;some&nbsp;p&nbsp;\u2208&nbsp;Z<\/p>\n<p>&nbsp;\u21d2 (b-a)\/2 = \u2013 p<\/p>\n<p>Here,&nbsp;\u2212p&nbsp;\u2208&nbsp;Z<\/p>\n<p>\u21d2&nbsp;2&nbsp;divides&nbsp;b&nbsp;\u2212&nbsp;a<\/p>\n<p>\u21d2&nbsp;(b,&nbsp;a) \u2208&nbsp;R&nbsp;for&nbsp;all&nbsp;a,&nbsp;b&nbsp;\u2208&nbsp;Z<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;symmetric&nbsp;on&nbsp;Z<\/p>\n<p>Transitivity:<\/p>\n<p>Let&nbsp;(a,&nbsp;b)&nbsp;and&nbsp;(b,&nbsp;c)&nbsp;\u2208&nbsp;R<\/p>\n<p>\u21d2&nbsp;2&nbsp;divides&nbsp;a\u2212b&nbsp;and&nbsp;2&nbsp;divides&nbsp;b\u2212c<\/p>\n<p>\u21d2&nbsp;(a-b)\/2&nbsp;=&nbsp;p&nbsp;and (b-c)\/2 =&nbsp;q&nbsp;for&nbsp;some&nbsp;p,&nbsp;q&nbsp;\u2208&nbsp;Z<\/p>\n<p>Adding&nbsp;the&nbsp;above&nbsp;two equations,&nbsp;we&nbsp;get<\/p>\n<p>(a \u2013 b)\/2 + (b \u2013 c)\/2 = p + q<\/p>\n<p>\u21d2&nbsp;(a \u2013 c)\/2 =&nbsp;p +q<\/p>\n<p>Here,&nbsp;p+&nbsp;q&nbsp;\u2208&nbsp;Z<\/p>\n<p>\u21d2 2&nbsp;divides&nbsp;a&nbsp;\u2212&nbsp;c<\/p>\n<p>\u21d2&nbsp;(a,&nbsp;c) \u2208&nbsp;R&nbsp;for&nbsp;all&nbsp;a,&nbsp;c&nbsp;\u2208&nbsp;Z<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;transitive&nbsp;on&nbsp;Z.<\/p>\n<p>Therefore R is reflexive, symmetric, and transitive.<\/p>\n<p>Hence,&nbsp;R&nbsp;is an equivalence relation on&nbsp;Z.<\/p>\n<p><strong>3. Prove that the relation&nbsp;R&nbsp;on&nbsp;Z&nbsp;defined by (a,&nbsp;b) \u2208&nbsp;R&nbsp;\u21d4&nbsp;a&nbsp;\u2212&nbsp;b&nbsp;is divisible by 5 is an equivalence relation on&nbsp;Z.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given relation&nbsp;R&nbsp;on&nbsp;Z&nbsp;defined by (a,&nbsp;b) \u2208&nbsp;R&nbsp;\u21d4&nbsp;a&nbsp;\u2212&nbsp;b&nbsp;is divisible by 5<\/p>\n<p>To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.<\/p>\n<p>Let us check these properties on R.<\/p>\n<p>Reflexivity:<\/p>\n<p>Let&nbsp;a&nbsp;be&nbsp;an&nbsp;arbitrary&nbsp;element&nbsp;of&nbsp;R.&nbsp;Then,<\/p>\n<p>\u21d2&nbsp;a \u2212 a&nbsp;=&nbsp;0&nbsp;=&nbsp;0&nbsp;\u00d7&nbsp;5<\/p>\n<p>\u21d2&nbsp;a \u2212 a&nbsp;is&nbsp;divisible&nbsp;by&nbsp;5<\/p>\n<p>\u21d2&nbsp;(a,&nbsp;a)&nbsp;\u2208&nbsp;R&nbsp;for&nbsp;all&nbsp;a&nbsp;\u2208&nbsp;Z<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;reflexive&nbsp;on&nbsp;Z.<\/p>\n<p>Symmetry:<\/p>\n<p>Let&nbsp;(a,&nbsp;b)&nbsp;\u2208&nbsp;R<\/p>\n<p>\u21d2&nbsp;a \u2212 b&nbsp;is&nbsp;divisible&nbsp;by&nbsp;5<\/p>\n<p>\u21d2&nbsp;a \u2212 b&nbsp;=&nbsp;5p&nbsp;for&nbsp;some&nbsp;p&nbsp;\u2208&nbsp;Z<\/p>\n<p>\u21d2&nbsp;b \u2212 a&nbsp;=&nbsp;5&nbsp;(\u2212p)<\/p>\n<p>Here,&nbsp;\u2212p&nbsp;\u2208&nbsp;Z [Since&nbsp;p&nbsp;\u2208&nbsp;Z]<\/p>\n<p>\u21d2&nbsp;b \u2212 a&nbsp;is&nbsp;divisible&nbsp;by&nbsp;5<\/p>\n<p>\u21d2&nbsp;(b,&nbsp;a)&nbsp;\u2208&nbsp;R&nbsp;for&nbsp;all&nbsp;a,&nbsp;b&nbsp;\u2208&nbsp;Z<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;symmetric&nbsp;on&nbsp;Z.<\/p>\n<p>Transitivity:<\/p>\n<p>Let&nbsp;(a,&nbsp;b)&nbsp;and&nbsp;(b,&nbsp;c)&nbsp;\u2208&nbsp;R<\/p>\n<p>\u21d2&nbsp;a \u2212 b&nbsp;is&nbsp;divisible&nbsp;by&nbsp;5<\/p>\n<p>\u21d2&nbsp;a \u2212 b&nbsp;=&nbsp;5p&nbsp;for&nbsp;some&nbsp;Z<\/p>\n<p>Also,&nbsp;b \u2212 c&nbsp;is&nbsp;divisible&nbsp;by&nbsp;5<\/p>\n<p>\u21d2&nbsp;b \u2212 c&nbsp;=&nbsp;5q&nbsp;for&nbsp;some&nbsp;Z<\/p>\n<p>Adding&nbsp;the&nbsp;above&nbsp;two equations,&nbsp;we&nbsp;get<\/p>\n<p>a&nbsp;\u2212b&nbsp;+&nbsp;b \u2212 c&nbsp;=&nbsp;5p&nbsp;+&nbsp;5q<\/p>\n<p>\u21d2&nbsp;a \u2212 c&nbsp;=&nbsp;5&nbsp;(&nbsp;p&nbsp;+&nbsp;q&nbsp;)<\/p>\n<p>\u21d2&nbsp;a \u2212 c&nbsp;is&nbsp;divisible&nbsp;by&nbsp;5<\/p>\n<p>Here,&nbsp;p&nbsp;+&nbsp;q&nbsp;\u2208&nbsp;Z<\/p>\n<p>\u21d2&nbsp;(a,&nbsp;c)&nbsp;\u2208&nbsp;R&nbsp;for&nbsp;all&nbsp;a,&nbsp;c&nbsp;\u2208&nbsp;Z<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;transitive&nbsp;on&nbsp;Z.<\/p>\n<p>Therefore R is reflexive, symmetric, and transitive.<\/p>\n<p>Hence, R is an equivalence relation to Z.<\/p>\n<p><strong>4. Let&nbsp;n&nbsp;be a fixed positive integer. Define a relation&nbsp;R&nbsp;on&nbsp;Z&nbsp;as follows:<br>(a,&nbsp;b) \u2208&nbsp;R&nbsp;\u21d4&nbsp;a&nbsp;\u2212&nbsp;b&nbsp;is divisible by&nbsp;n.<br>Show that&nbsp;R&nbsp;is an equivalence relation on&nbsp;Z.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given (a,&nbsp;b) \u2208&nbsp;R&nbsp;\u21d4&nbsp;a&nbsp;\u2212&nbsp;b&nbsp;is divisible by&nbsp;n is a relation R defined on Z.<\/p>\n<p>To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.<\/p>\n<p>Let us check these properties on R.<\/p>\n<p>Reflexivity:<\/p>\n<p>Let&nbsp;a&nbsp;\u2208&nbsp;N<\/p>\n<p>Here, a&nbsp;\u2212&nbsp;a&nbsp;=&nbsp;0&nbsp;=&nbsp;0&nbsp;\u00d7&nbsp;n<\/p>\n<p>\u21d2&nbsp;a \u2212 a&nbsp;is&nbsp;divisible&nbsp;by&nbsp;n<\/p>\n<p>\u21d2&nbsp;(a,&nbsp;a)&nbsp;\u2208&nbsp;R<\/p>\n<p>\u21d2&nbsp;(a,&nbsp;a)&nbsp;\u2208&nbsp;R&nbsp;for&nbsp;all&nbsp;a&nbsp;\u2208&nbsp;Z<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;reflexive&nbsp;on&nbsp;Z.<\/p>\n<p>Symmetry:<\/p>\n<p>Let&nbsp;(a,&nbsp;b)&nbsp;\u2208&nbsp;R<\/p>\n<p>Here, a \u2212 b&nbsp;is&nbsp;divisible&nbsp;by&nbsp;n<\/p>\n<p>\u21d2&nbsp;a \u2212 b&nbsp;=&nbsp;n p&nbsp;for&nbsp;some&nbsp;p&nbsp;\u2208&nbsp;Z<\/p>\n<p>\u21d2&nbsp;b \u2212 a&nbsp;=&nbsp;n&nbsp;(\u2212p)<\/p>\n<p>\u21d2&nbsp;b \u2212 a&nbsp;is&nbsp;divisible&nbsp;by&nbsp;n [&nbsp;p&nbsp;\u2208&nbsp;Z\u21d2&nbsp;\u2212&nbsp;p&nbsp;\u2208&nbsp;Z]<\/p>\n<p>\u21d2&nbsp;(b,&nbsp;a)&nbsp;\u2208&nbsp;R&nbsp;<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;symmetric&nbsp;on&nbsp;Z.<\/p>\n<p>Transitivity:<\/p>\n<p>Let&nbsp;(a,&nbsp;b)&nbsp;and&nbsp;(b,&nbsp;c)&nbsp;\u2208&nbsp;R<\/p>\n<p>Here, a \u2212 b is divisible by n, and b \u2212 c is divisible by n.<\/p>\n<p>\u21d2&nbsp;a \u2212 b=&nbsp;n p&nbsp;for&nbsp;some&nbsp;p&nbsp;\u2208&nbsp;Z<\/p>\n<p>And&nbsp;b\u2212c&nbsp;=&nbsp;n q&nbsp;for&nbsp;some&nbsp;q&nbsp;\u2208&nbsp;Z<\/p>\n<p>a \u2013 b +&nbsp;b \u2212 c&nbsp;=&nbsp;n p&nbsp;+&nbsp;n q<\/p>\n<p>\u21d2&nbsp;a \u2212 c&nbsp;=&nbsp;n&nbsp;(p + q)<\/p>\n<p>\u21d2&nbsp;(a,&nbsp;c) \u2208&nbsp;R&nbsp;for&nbsp;all&nbsp;a,&nbsp;c&nbsp;\u2208&nbsp;Z<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;transitive&nbsp;on&nbsp;Z.<\/p>\n<p>Therefore R is reflexive, symmetric, and transitive.<\/p>\n<p>Hence, R is an equivalence relation to Z.<\/p>\n<p><strong>5. Let&nbsp;Z&nbsp;be the set of integers. Show that the relation R&nbsp;= {(a,&nbsp;b):&nbsp;a,&nbsp;b&nbsp;\u2208&nbsp;Z&nbsp;and&nbsp;a&nbsp;+&nbsp;b&nbsp;is even} is an equivalence relation on&nbsp;Z.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given R&nbsp;= {(a,&nbsp;b):&nbsp;a,&nbsp;b&nbsp;\u2208&nbsp;Z&nbsp;and&nbsp;a&nbsp;+&nbsp;b&nbsp;is even} is a relation defined on R.<\/p>\n<p>Also given that Z&nbsp;be the set of integers<\/p>\n<p>To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.<\/p>\n<p>Let us check these properties on R.<\/p>\n<p>Reflexivity:<\/p>\n<p>Let&nbsp;a&nbsp;be&nbsp;an&nbsp;arbitrary&nbsp;element&nbsp;of&nbsp;Z.&nbsp;<\/p>\n<p>Then, a&nbsp;\u2208&nbsp;R<\/p>\n<p>Clearly,&nbsp;a + a&nbsp;=&nbsp;2a&nbsp;is&nbsp;even&nbsp;for&nbsp;all&nbsp;a&nbsp;\u2208&nbsp;Z.<\/p>\n<p>\u21d2&nbsp;(a,&nbsp;a)&nbsp;\u2208&nbsp;R&nbsp;for&nbsp;all&nbsp;a&nbsp;\u2208&nbsp;Z<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;reflexive&nbsp;on&nbsp;Z.<\/p>\n<p>Symmetry:<\/p>\n<p>Let&nbsp;(a,&nbsp;b)&nbsp;\u2208&nbsp;R<\/p>\n<p>\u21d2&nbsp;a + b&nbsp;is&nbsp;even<\/p>\n<p>\u21d2&nbsp;b + a&nbsp;is&nbsp;even<\/p>\n<p>\u21d2&nbsp;(b,&nbsp;a)&nbsp;\u2208&nbsp;R&nbsp;for&nbsp;all&nbsp;a,&nbsp;b&nbsp;\u2208&nbsp;Z<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;symmetric&nbsp;on&nbsp;Z.<\/p>\n<p>Transitivity:<\/p>\n<p>Let&nbsp;(a,&nbsp;b)&nbsp;and&nbsp;(b,&nbsp;c)&nbsp;\u2208&nbsp;R<\/p>\n<p>\u21d2&nbsp;a + b&nbsp;and&nbsp;b + c&nbsp;are&nbsp;even<\/p>\n<p>Now,&nbsp;let&nbsp;a + b&nbsp;=&nbsp;2x&nbsp;for&nbsp;some&nbsp;x&nbsp;\u2208&nbsp;Z<\/p>\n<p>And&nbsp;b + c&nbsp;=&nbsp;2y&nbsp;for&nbsp;some&nbsp;y&nbsp;\u2208&nbsp;Z<\/p>\n<p>Adding&nbsp;the&nbsp;above&nbsp;two equations,&nbsp;we&nbsp;get<\/p>\n<p>A + 2b&nbsp;+ c&nbsp;=&nbsp;2x&nbsp;+&nbsp;2y<\/p>\n<p>\u21d2&nbsp;a + c&nbsp;=&nbsp;2&nbsp;(x + y \u2212 b),&nbsp;which&nbsp;is&nbsp;even&nbsp;for&nbsp;all&nbsp;x,&nbsp;y,&nbsp;b&nbsp;\u2208&nbsp;Z<\/p>\n<p>Thus,&nbsp;(a,&nbsp;c)&nbsp;\u2208&nbsp;R<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;transitive&nbsp;on&nbsp;Z.<\/p>\n<p>Therefore R is reflexive, symmetric, and transitive.<\/p>\n<p>Hence, R is an equivalence relation to Z<\/p>\n<p><strong>6. m&nbsp;is said to be related to&nbsp;n&nbsp;if&nbsp;m&nbsp;and&nbsp;n&nbsp;are integers and&nbsp;m&nbsp;\u2212&nbsp;n&nbsp;is divisible by 13. Does this define an equivalence relation?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given that m&nbsp;is said to be related to&nbsp;n&nbsp;if&nbsp;m&nbsp;and&nbsp;n&nbsp;are integers and&nbsp;m&nbsp;\u2212&nbsp;n&nbsp;is divisible by 13<\/p>\n<p>Now we have to check whether the given relation is equivalence or not.<\/p>\n<p>To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.<\/p>\n<p>Let&nbsp;R = {(m,&nbsp;n):&nbsp;m,&nbsp;n&nbsp;\u2208 Z&nbsp;:&nbsp;m \u2212 n&nbsp;is&nbsp;divisible&nbsp;by&nbsp;13}<\/p>\n<p>Let us check these properties on R.<\/p>\n<p>Reflexivity:&nbsp;<\/p>\n<p>Let&nbsp;m&nbsp;be&nbsp;an&nbsp;arbitrary&nbsp;element&nbsp;of&nbsp;Z.&nbsp;<\/p>\n<p>Then, m&nbsp;\u2208&nbsp;R<\/p>\n<p>\u21d2&nbsp;m \u2212 m&nbsp;=&nbsp;0&nbsp;=&nbsp;0&nbsp;\u00d7&nbsp;13<\/p>\n<p>\u21d2&nbsp;m \u2212 m&nbsp;is&nbsp;divisible&nbsp;by&nbsp;13<\/p>\n<p>\u21d2&nbsp;(m,&nbsp;m)&nbsp;is&nbsp;reflexive&nbsp;on&nbsp;Z.<\/p>\n<p>Symmetry:&nbsp;<\/p>\n<p>Let&nbsp;(m,&nbsp;n)&nbsp;\u2208&nbsp;R.&nbsp;<\/p>\n<p>Then, m \u2212 n&nbsp;is&nbsp;divisible&nbsp;by&nbsp;13<\/p>\n<p>\u21d2&nbsp;m \u2212 n&nbsp;=&nbsp;13p<\/p>\n<p>Here,&nbsp;p&nbsp;\u2208&nbsp;Z<\/p>\n<p>\u21d2&nbsp;n \u2013 m = 13&nbsp;(\u2212p)&nbsp;<\/p>\n<p>Here,&nbsp;\u2212p&nbsp;\u2208&nbsp;Z<\/p>\n<p>\u21d2&nbsp;n \u2212 m&nbsp;is&nbsp;divisible&nbsp;by&nbsp;13<\/p>\n<p>\u21d2&nbsp;(n,&nbsp;m) \u2208&nbsp;R&nbsp;for&nbsp;all&nbsp;m,&nbsp;n&nbsp;\u2208&nbsp;Z&nbsp;<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;symmetric&nbsp;on&nbsp;Z.<\/p>\n<p>Transitivity:&nbsp;<\/p>\n<p>Let&nbsp;(m,&nbsp;n)&nbsp;and&nbsp;(n,&nbsp;o) \u2208R<\/p>\n<p>\u21d2 m \u2212 n, and n \u2212 o are divisible by 13<\/p>\n<p>\u21d2&nbsp;m \u2013 n = 13p&nbsp;and&nbsp;n&nbsp;\u2212&nbsp;o&nbsp;= 13q&nbsp;for&nbsp;some&nbsp;p,&nbsp;q&nbsp;\u2208&nbsp;Z<\/p>\n<p>Adding&nbsp;the&nbsp;above&nbsp;two equations,&nbsp;we&nbsp;get<\/p>\n<p>m \u2013&nbsp;n + n \u2212 o&nbsp;= 13p&nbsp;+&nbsp;13q<\/p>\n<p>\u21d2&nbsp;m\u2212o&nbsp;= 13&nbsp;(p + q)<\/p>\n<p>Here,&nbsp;p + q&nbsp;\u2208&nbsp;Z<\/p>\n<p>\u21d2&nbsp;m \u2212 o&nbsp;is&nbsp;divisible&nbsp;by&nbsp;13<\/p>\n<p>\u21d2 (m,&nbsp;o)&nbsp;\u2208&nbsp;R&nbsp;for&nbsp;all&nbsp;m,&nbsp;o&nbsp;\u2208&nbsp;Z<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;transitive&nbsp;on&nbsp;Z.<\/p>\n<p>Therefore R is reflexive, symmetric, and transitive.<\/p>\n<p>Hence, R is an equivalence relation to Z.<\/p>\n<p><strong>7. Let&nbsp;R&nbsp;be a relation on the set&nbsp;A&nbsp;of ordered pair of integers defined by (x,&nbsp;y)&nbsp;R&nbsp;(u,&nbsp;v) if&nbsp;xv&nbsp;=&nbsp;y u. Show that&nbsp;R&nbsp;is an equivalence relation.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>First, let R be a relation on A<\/p>\n<p>It is given that set&nbsp;A&nbsp;of ordered pair of integers defined by (x,&nbsp;y)&nbsp;R&nbsp;(u,&nbsp;v) if&nbsp;xv&nbsp;=&nbsp;y u<\/p>\n<p>Now we have to check whether the given relation is equivalence or not.<\/p>\n<p>To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.<\/p>\n<p>Reflexivity:&nbsp;<\/p>\n<p>Let (a, b) be an arbitrary element of set A.&nbsp;<\/p>\n<p>Then, (a,&nbsp;b)&nbsp;\u2208&nbsp;A<\/p>\n<p>\u21d2&nbsp;a b&nbsp;=&nbsp;b a<\/p>\n<p>\u21d2&nbsp;(a,&nbsp;b)&nbsp;R&nbsp;(a,&nbsp;b)<\/p>\n<p>Thus,&nbsp;R&nbsp;is&nbsp;reflexive&nbsp;on&nbsp;A.<\/p>\n<p>Symmetry:&nbsp;<\/p>\n<p>Let&nbsp;(x,&nbsp;y)&nbsp;and&nbsp;(u,&nbsp;v) \u2208A&nbsp;such&nbsp;that&nbsp;(x,&nbsp;y)&nbsp;R&nbsp;(u,&nbsp;v).&nbsp;Then,<\/p>\n<p>&nbsp;x v = y u<\/p>\n<p>\u21d2&nbsp;v x = u y<\/p>\n<p>\u21d2&nbsp;u y = v x<\/p>\n<p>\u21d2&nbsp;(u,&nbsp;v)&nbsp;R&nbsp;(x,&nbsp;y)<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;symmetric&nbsp;on&nbsp;A.<\/p>\n<p>Transitivity:<\/p>\n<p>Let&nbsp;(x,&nbsp;y),&nbsp;(u,&nbsp;v)&nbsp;and&nbsp;(p,&nbsp;q) \u2208R&nbsp;such&nbsp;that&nbsp;(x,&nbsp;y)&nbsp;R&nbsp;(u,&nbsp;v)&nbsp;and&nbsp;(u,&nbsp;v)&nbsp;R&nbsp;(p,&nbsp;q)<\/p>\n<p>\u21d2&nbsp;x v&nbsp;=&nbsp;y u&nbsp;and&nbsp;u q&nbsp;=&nbsp;v p<\/p>\n<p>Multiplying&nbsp;the&nbsp;corresponding&nbsp;sides,&nbsp;we&nbsp;get<\/p>\n<p>x v&nbsp;\u00d7&nbsp;u q&nbsp;=&nbsp;y u&nbsp;\u00d7&nbsp;v p<\/p>\n<p>\u21d2&nbsp;x q&nbsp;=&nbsp;y p<\/p>\n<p>\u21d2&nbsp;(x,&nbsp;y)&nbsp;R&nbsp;(p,&nbsp;q)<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;transitive&nbsp;on&nbsp;A.<\/p>\n<p>Therefore R is reflexive, symmetric, and transitive.<\/p>\n<p>Hence, R is an equivalence relation to A.<\/p>\n<p><strong>8. Show that the relation&nbsp;R&nbsp;on the set&nbsp;A&nbsp;= {x&nbsp;\u2208&nbsp;Z; 0 \u2264&nbsp;x&nbsp;\u2264 12}, given by&nbsp;R&nbsp;= {(a,&nbsp;b):&nbsp;a&nbsp;=&nbsp;b}, is an equivalence relation. Find the set of all elements related to 1.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given set&nbsp;A&nbsp;= {x&nbsp;\u2208&nbsp;Z; 0 \u2264&nbsp;x&nbsp;\u2264 12}<\/p>\n<p>Also given that relation R&nbsp;= {(a,&nbsp;b):&nbsp;a&nbsp;=&nbsp;b} is defined on set A<\/p>\n<p>Now we have to check whether the given relation is equivalence or not.<\/p>\n<p>To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.<\/p>\n<p>Reflexivity:&nbsp;<\/p>\n<p>Let&nbsp;a&nbsp;be an arbitrary element of&nbsp;A.<\/p>\n<p>Then,&nbsp;a&nbsp;\u2208&nbsp;R<\/p>\n<p>\u21d2 a = a [Since every element is equal to itself]<\/p>\n<p>\u21d2&nbsp;(a,&nbsp;a)&nbsp;\u2208&nbsp;R&nbsp;for&nbsp;all&nbsp;a&nbsp;\u2208&nbsp;A<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;reflexive&nbsp;on&nbsp;A.<\/p>\n<p>Symmetry:&nbsp;<\/p>\n<p>Let&nbsp;(a,&nbsp;b)&nbsp;\u2208&nbsp;R<\/p>\n<p>\u21d2&nbsp;a&nbsp;b<\/p>\n<p>\u21d2&nbsp;b&nbsp;=&nbsp;a<\/p>\n<p>\u21d2&nbsp;(b,&nbsp;a)&nbsp;\u2208&nbsp;R&nbsp;for&nbsp;all&nbsp;a,&nbsp;b&nbsp;\u2208&nbsp;A<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;symmetric&nbsp;on&nbsp;A.<\/p>\n<p>Transitivity:&nbsp;<\/p>\n<p>Let&nbsp;(a,&nbsp;b)&nbsp;and&nbsp;(b,&nbsp;c)&nbsp;\u2208&nbsp;R<\/p>\n<p>\u21d2&nbsp;a&nbsp;=b&nbsp;and&nbsp;b&nbsp;=&nbsp;c<\/p>\n<p>\u21d2&nbsp;a&nbsp;=&nbsp;b&nbsp;c<\/p>\n<p>\u21d2&nbsp;a&nbsp;=&nbsp;c<\/p>\n<p>\u21d2&nbsp;(a,&nbsp;c)&nbsp;\u2208&nbsp;R<\/p>\n<p>So,&nbsp;R&nbsp;is&nbsp;transitive&nbsp;on&nbsp;A.<\/p>\n<p>Hence, R is an equivalence relation to A.<\/p>\n<p>Therefore R is reflexive, symmetric, and transitive.<\/p>\n<p>The set of all elements related to 1 is {1}.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-12-maths-chapter-1-exercise-12-important-topics-from-the-chapter\"><\/span>RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2: Important Topics From The Chapter<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Let us have a look at the important topics covered in this exercise.<\/p>\n<ul>\n<li>Types of relations\n<ul>\n<li>Void relation<\/li>\n<li>Universal relation<\/li>\n<li>Identity relation<\/li>\n<li>Reflexive relation<\/li>\n<li>Symmetric relation<\/li>\n<li>Transitive relation<\/li>\n<li>Antisymmetric relation<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<div>We have provided complete details of <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-12-solutions\/\"><strong>RD Sharma Solutions Class 12 Maths<\/strong><\/a> Chapter 1 Exercise 1.2. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>&nbsp;Class 12 Exam, feel free to ask us in the comment section below.<\/div>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-12-maths-chapter-1-exercise-12\"><\/span>FAQs on RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1629720081097\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-many-questions-are-there-in-rd-sharma-solutions-class-12-maths-chapter-1-exercise-12\"><\/span>How many questions are there in RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are a total of 8 questions in RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629720106829\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-rd-sharma-solutions-class-12-maths-chapter-1-exercise-12-free\"><\/span>Is RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2 free?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, You can get RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2 for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629720122916\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-download-rd-sharma-solutions-class-12-maths-chapter-1-exercise-12-free-pdf\"><\/span>Where can I download RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2 free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2: Students will mostly gain conceptual knowledge of equivalence relations through this practice. If a relation is reflexive, symmetric, and transitive, it is said to be an equivalence relationship. Class 12 is a crucial year for students in terms of shaping and achieving their long-term objectives. &#8230; <a title=\"RD Sharma Solutions for Class 12 Maths Exercise 1.2 Chapter 1 Relation (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-1-exercise-1-2\/\" aria-label=\"More on RD Sharma Solutions for Class 12 Maths Exercise 1.2 Chapter 1 Relation (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":117932,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2917,2985,73410],"tags":[3429,73223,73664],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68976"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=68976"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68976\/revisions"}],"predecessor-version":[{"id":505776,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68976\/revisions\/505776"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/117932"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=68976"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=68976"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=68976"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}