{"id":68975,"date":"2021-08-23T13:06:00","date_gmt":"2021-08-23T07:36:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=68975"},"modified":"2021-08-31T18:25:49","modified_gmt":"2021-08-31T12:55:49","slug":"rd-sharma-solutions-class-12-maths-chapter-1-exercise-1-1","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-1-exercise-1-1\/","title":{"rendered":"RD Sharma Solutions for Class 12 Maths Exercise 1.1 Chapter 1 Relation (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-117929\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-1-Exercise-1.1.png\" alt=\"RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-1-Exercise-1.1.png 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-1-Exercise-1.1-768x432.png 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1:<\/strong> The <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-12-solutions\/\"><strong>RD Sharma Class 12 Solution<\/strong><\/a> PDF of Chapter 1 Exercise 1.1 Relations can be downloaded from the link given. Solutions for correct answer practice are prepared by experts in the best possible way and are easily understood by the students. In this practice, students will gain knowledge about relations and types. <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 1.1 are detailed and detailed to make learning easier for students. In this exercise, there are two levels according to the increasing order of difficulty.<\/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-69e7f01db1998\" 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\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-1-exercise-1-1\/#is-rd-sharma-solutions-class-12-maths-chapter-1-exercise-11-for-free\" title=\"Is RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1 for free?\">Is RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1 for free?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-1-exercise-1-1\/#where-can-i-download-rd-sharma-solutions-class-12-maths-chapter-1-exercise-11-free-pdf\" title=\"Where can I download RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1 free PDF?\">Where can I download RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1 free PDF?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-12-maths-chapter-1-exercise-11-pdf\"><\/span>Download RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1 PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script 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-1.1-2.pdf\", \"#example1\");<\/script><\/p>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Maths-Chapter-1-Ex-1.1-2.pdf\">RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-12-maths-chapter-1-exercise-11\"><\/span>Access answers to RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. Let\u00a0<em>A<\/em> be the set of all human beings in a town at a particular time. Determine whether the following relation is reflexive, symmetric, and transitive:<\/strong><\/p>\n<p><strong>(i)\u00a0R<em>\u00a0<\/em>= {(x,\u00a0y):\u00a0x\u00a0and\u00a0y\u00a0work at the same place}<\/strong><\/p>\n<p><strong>(ii) R<em>\u00a0<\/em>= {(x,\u00a0y):\u00a0x\u00a0and\u00a0y\u00a0live in the same locality}<\/strong><\/p>\n<p><strong>(iii) R<em>\u00a0<\/em>= {(x,\u00a0y):\u00a0x\u00a0is wife of y}<\/strong><\/p>\n<p><strong>(iv) R<em>\u00a0<\/em>= {(x,\u00a0y):\u00a0x\u00a0is father of y}<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Given R<em>\u00a0<\/em>= {(x,\u00a0y):\u00a0x\u00a0and\u00a0y\u00a0work at the same place}<\/p>\n<p>Now we have to check whether the relation is reflexive:<\/p>\n<p>Let\u00a0x\u00a0be\u00a0an\u00a0arbitrary\u00a0element\u00a0of\u00a0R.\u00a0<\/p>\n<p>Then, x \u2208R\u00a0\u00a0\u00a0<\/p>\n<p>\u21d2 x and x work at the same place is true since they are the same.<\/p>\n<p>\u21d2(x,\u00a0x)\u00a0\u2208R [condition for reflexive relation]<\/p>\n<p>So,\u00a0R\u00a0is\u00a0a\u00a0reflexive\u00a0relation.<\/p>\n<p>Now let us check the Symmetric relation:<\/p>\n<p>Let\u00a0(x,\u00a0y) \u2208R<\/p>\n<p>\u21d2x\u00a0and\u00a0y\u00a0work\u00a0at\u00a0the\u00a0same\u00a0place\u00a0[given]<\/p>\n<p>\u21d2y\u00a0and\u00a0x\u00a0work\u00a0at\u00a0the\u00a0same\u00a0place<\/p>\n<p>\u21d2(y,\u00a0x) \u2208R<\/p>\n<p>So,\u00a0R\u00a0is\u00a0a\u00a0symmetric\u00a0relation.<\/p>\n<p>Transitive relation:<\/p>\n<p>Let\u00a0(x,\u00a0y) \u2208R\u00a0and\u00a0(y,\u00a0z) \u2208R.\u00a0<\/p>\n<p>Then, x\u00a0and\u00a0y\u00a0work\u00a0at\u00a0the\u00a0same\u00a0place. [Given]<\/p>\n<p>y\u00a0and\u00a0z\u00a0also\u00a0work\u00a0at\u00a0the\u00a0same\u00a0place. [(y, z) \u2208R]<\/p>\n<p>\u21d2 x, y, and z all work at the same place.<\/p>\n<p>\u21d2x\u00a0and\u00a0z\u00a0work\u00a0at\u00a0the\u00a0same\u00a0place.<\/p>\n<p>\u21d2\u00a0(x,\u00a0z) \u2208R<\/p>\n<p>So,\u00a0R\u00a0is\u00a0a\u00a0transitive\u00a0relation.<\/p>\n<p>Hence R is reflexive, symmetric, and transitive.<\/p>\n<p>(ii) Given R<em>\u00a0<\/em>= {(x,\u00a0y):\u00a0x\u00a0and\u00a0y\u00a0live in the same locality}<\/p>\n<p>Now we have to check whether the relation R is reflexive, symmetric, and transitive.<\/p>\n<p>Let\u00a0x\u00a0be\u00a0an\u00a0arbitrary\u00a0element\u00a0of\u00a0R.\u00a0<\/p>\n<p>Then, x \u2208R<\/p>\n<p>It is given that x\u00a0and\u00a0x\u00a0live\u00a0in\u00a0the\u00a0same\u00a0locality\u00a0is\u00a0true\u00a0since\u00a0they\u00a0are\u00a0the\u00a0same.<\/p>\n<p>So,\u00a0R\u00a0is\u00a0a\u00a0reflexive\u00a0relation.<\/p>\n<p>Symmetry:<\/p>\n<p>Let (x, y) \u2208 R<\/p>\n<p>\u21d2\u00a0x\u00a0and\u00a0y\u00a0live\u00a0in\u00a0the\u00a0same\u00a0locality [given]<\/p>\n<p>\u21d2\u00a0y\u00a0and\u00a0x\u00a0live\u00a0in\u00a0the\u00a0same\u00a0locality<\/p>\n<p>\u21d2 (y, x) \u2208 R\u00a0<\/p>\n<p>So,\u00a0R\u00a0is\u00a0a\u00a0symmetric\u00a0relation.<\/p>\n<p>Transitivity:<\/p>\n<p>Let\u00a0(x,\u00a0y) \u2208R\u00a0and\u00a0(y,\u00a0z) \u2208R.\u00a0<\/p>\n<p>Then,<\/p>\n<p>x\u00a0and\u00a0y\u00a0live\u00a0in\u00a0the\u00a0same\u00a0locality\u00a0and\u00a0y\u00a0and\u00a0z\u00a0live\u00a0in\u00a0the\u00a0same\u00a0locality<\/p>\n<p>\u21d2 x, y, and z all live in the same locality<\/p>\n<p>\u21d2\u00a0x\u00a0and\u00a0z\u00a0live\u00a0in\u00a0the\u00a0same\u00a0locality\u00a0<\/p>\n<p>\u21d2 (x, z) \u2208 R<\/p>\n<p>So, R is a transitive relation.<\/p>\n<p>Hence R is reflexive, symmetric, and transitive.<\/p>\n<p>(iii) Given R<em>\u00a0<\/em>= {(x,\u00a0y):\u00a0x\u00a0is wife of y}<\/p>\n<p>Now we have to check whether the relation R is reflexive, symmetric, and transitive.<\/p>\n<p>First, let us check whether the relation is reflexive:<\/p>\n<p>Let\u00a0x\u00a0be\u00a0an\u00a0element\u00a0of\u00a0R.\u00a0<\/p>\n<p>Then, x is the wife of x cannot be true.<\/p>\n<p>\u21d2 (x,\u00a0x) \u2209R<\/p>\n<p>So,\u00a0R\u00a0is\u00a0not\u00a0a\u00a0reflexive\u00a0relation.<\/p>\n<p>Symmetric relation:<\/p>\n<p>Let\u00a0(x,\u00a0y) \u2208R<\/p>\n<p>\u21d2 x is the wife of y\u00a0<\/p>\n<p>\u21d2 x\u00a0is\u00a0female\u00a0and\u00a0y\u00a0is\u00a0male<\/p>\n<p>\u21d2 y cannot be the wife of x as y is the husband of x<\/p>\n<p>\u21d2 (y,\u00a0x) \u2209R\u00a0\u00a0<\/p>\n<p>So,\u00a0R\u00a0is\u00a0not\u00a0a\u00a0symmetric\u00a0relation.<\/p>\n<p>Transitive relation:<\/p>\n<p>Let\u00a0(x,\u00a0y) \u2208R,\u00a0but\u00a0(y,\u00a0z) \u2209R<\/p>\n<p>Since x is the wife of y, but y cannot be the wife of z, y is the husband of x.<\/p>\n<p>\u21d2 x\u00a0is\u00a0not\u00a0the\u00a0wife\u00a0of\u00a0z<\/p>\n<p>\u21d2(x,\u00a0z) \u2208R<\/p>\n<p>So,\u00a0R\u00a0is\u00a0a\u00a0transitive\u00a0relation.<\/p>\n<p>(iv) Given R<em>\u00a0<\/em>= {(x,\u00a0y):\u00a0x\u00a0is father of y}<\/p>\n<p>Now we have to check whether the relation R is reflexive, symmetric, and transitive.<\/p>\n<p>Reflexivity:<\/p>\n<p>Let\u00a0x\u00a0be\u00a0an\u00a0arbitrary\u00a0element\u00a0of\u00a0R.\u00a0<\/p>\n<p>Then, x is the father of x cannot be true since no one can be the father of himself.<\/p>\n<p>So,\u00a0R\u00a0is\u00a0not\u00a0a\u00a0reflexive\u00a0relation.<\/p>\n<p>Symmetry:\u00a0<\/p>\n<p>Let\u00a0(x,\u00a0y) \u2208R<\/p>\n<p>\u21d2 x is a father of y<\/p>\n<p>\u21d2 y is the son\/daughter of x<\/p>\n<p>\u21d2 (y,\u00a0x) \u2209R\u00a0<\/p>\n<p>So,\u00a0R\u00a0is\u00a0not\u00a0a\u00a0symmetric\u00a0relation.<\/p>\n<p>Transitivity:<\/p>\n<p>Let\u00a0(x,\u00a0y) \u2208R\u00a0and\u00a0(y,\u00a0z) \u2208R.\u00a0<\/p>\n<p>Then, x is a father of y and y is a father of z<\/p>\n<p>\u21d2 x is the grandfather of z<\/p>\n<p>\u21d2 (x,\u00a0z) \u2209R<\/p>\n<p>So,\u00a0R\u00a0is\u00a0not\u00a0a\u00a0transitive\u00a0relation.<\/p>\n<p><strong>2. Three relations\u00a0R<sub>1<\/sub>,<em>\u00a0<\/em>R<sub>2,<\/sub>\u00a0and\u00a0R<sub>3<\/sub>\u00a0are defined on a set\u00a0A\u00a0= {a<em>,\u00a0<\/em>b<em>,\u00a0<\/em>c} as follows:<br \/>R<sub>1<\/sub>\u00a0= {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}<br \/>R<sub>2<\/sub>\u00a0= {(a, a)}<br \/>R<sub>3<\/sub>\u00a0= {(b, c)}<br \/>R<sub>4<\/sub>\u00a0= {(a, b), (b, c), (c, a)}.<\/strong><\/p>\n<p><strong>Find whether or not each of the relations\u00a0R<sub>1<\/sub>,\u00a0R<sub>2<\/sub>,\u00a0R<sub>3<\/sub>,\u00a0R<sub>4<\/sub>\u00a0on\u00a0<em>A\u00a0<\/em>is (i) reflexive (ii) symmetric, and (iii) transitive.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Consider R<sub>1<\/sub><br \/><br \/>Given R<sub>1<\/sub>\u00a0= {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}<\/p>\n<p>Now we have check R<sub>1<\/sub>\u00a0is reflexive, symmetric and transitive<\/p>\n<p>Reflexive:<\/p>\n<p>Given (a, a), (b, b) and (c, c) \u2208 R<sub>1<\/sub><\/p>\n<p>So, R<sub>1<\/sub>\u00a0is reflexive.<\/p>\n<p>Symmetric:<br \/><br \/>We see that the ordered pairs obtained by interchanging the components of R<sub>1<\/sub>\u00a0are also in R<sub>1<\/sub>.<br \/><br \/>So, R<sub>1<\/sub>\u00a0is symmetric.<\/p>\n<p>Transitive:<br \/><br \/>Here, (a,\u00a0b) \u2208R<sub>1<\/sub>,\u00a0(b,\u00a0c) \u2208R<sub>1,<\/sub>\u00a0and\u00a0also\u00a0(a,\u00a0c) \u2208R<sub>1<\/sub><\/p>\n<p>So, R<sub>1<\/sub>\u00a0is transitive.<\/p>\n<p>(ii) Consider R<sub>2<\/sub><\/p>\n<p>Given R<sub>2<\/sub>\u00a0= {(a, a)}<\/p>\n<p>Reflexive:\u00a0<\/p>\n<p>Clearly\u00a0(a, a)\u00a0\u2208R<sub>2<\/sub>.<\/p>\n<p>So, R<sub>2<\/sub>\u00a0is reflexive.<\/p>\n<p>Symmetric:\u00a0<\/p>\n<p>Clearly\u00a0(a, a)\u00a0\u2208R\u00a0<\/p>\n<p>\u21d2\u00a0(a, a)\u00a0\u2208R.<\/p>\n<p>So, R<sub>2<\/sub>\u00a0is symmetric.<\/p>\n<p>Transitive:\u00a0<\/p>\n<p>R<sub>2<\/sub> is clearly a transitive relation since there is only one element in it.<\/p>\n<p>(iii) Consider R<sub>3<\/sub><br \/><br \/>Given R<sub>3<\/sub>\u00a0= {(b, c)}<br \/><br \/>Reflexive:<br \/><br \/>Here,(b,\u00a0b)\u2209\u00a0R<sub>3<\/sub>\u00a0neither\u00a0(c,\u00a0c)\u00a0\u2209\u00a0R<sub>3<\/sub><\/p>\n<p>So, R<sub>3<\/sub>\u00a0is not reflexive.<\/p>\n<p>Symmetric:<br \/><br \/>Here, (b,\u00a0c)\u00a0\u2208R<sub>3<\/sub>,\u00a0but\u00a0(c, b)\u00a0\u2209R<sub>3<\/sub><\/p>\n<p>So, R<sub>3\u00a0<\/sub>is not symmetric.<\/p>\n<p>Transitive:<br \/><br \/>Here, R<sub>3<\/sub>\u00a0has only two elements.<\/p>\n<p>Hence, R<sub>3<\/sub>\u00a0is transitive.<\/p>\n<p>(iv) Consider R<sub>4<\/sub><br \/><br \/>Given R<sub>4<\/sub>\u00a0= {(a, b), (b, c), (c, a)}.<\/p>\n<p>Reflexive:<br \/><br \/>Here, (a,\u00a0a)\u00a0\u2209\u00a0R<sub>4<\/sub>,\u00a0(b,\u00a0b) \u2209\u00a0R<sub>4<\/sub>\u00a0(c,\u00a0c) \u2209\u00a0R<sub>4<\/sub><\/p>\n<p>So,\u00a0R<sub>4<\/sub>\u00a0is\u00a0not\u00a0reflexive.<\/p>\n<p>Symmetric:<br \/><br \/>Here, (a,\u00a0b) \u2208\u00a0R<sub>4<\/sub>,\u00a0but\u00a0(b, a)\u00a0\u2209\u00a0R<sub>4<\/sub>.<\/p>\n<p>So,\u00a0R<sub>4<\/sub>\u00a0is\u00a0not\u00a0symmetric<\/p>\n<p>Transitive:<br \/><br \/>Here, (a,\u00a0b) \u2208R<sub>4<\/sub>,\u00a0(b,\u00a0c) \u2208R<sub>4<\/sub>,\u00a0but\u00a0(a,\u00a0c) \u2209R<sub>4<\/sub><\/p>\n<p>So,\u00a0R<sub>4<\/sub>\u00a0is\u00a0not\u00a0transitive.<\/p>\n<p><strong>3. \u00a0Test whether the following relation\u00a0R<sub>1<\/sub>, R<sub>2<\/sub>, and R<sub>3\u00a0<\/sub>are (i) reflexive (ii) symmetric, and (iii) transitive:<\/strong><\/p>\n<p><strong>(i) R<sub>1<\/sub>\u00a0on\u00a0Q<sub>0<\/sub>\u00a0defined by (a, b) \u2208 R<sub>1<\/sub>\u00a0\u21d4\u00a0a\u00a0= 1\/b.<\/strong><\/p>\n<p><strong>(ii) R<sub>2<\/sub>\u00a0on\u00a0Z\u00a0defined by (a, b) \u2208 R<sub>2<\/sub>\u00a0\u21d4 |a \u2013 b| \u2264 5<\/strong><\/p>\n<p><strong>(iii) R<sub>3\u00a0<\/sub>on\u00a0R\u00a0defined by (a, b) \u2208\u00a0R<sub>3<\/sub>\u00a0\u21d4\u00a0a<sup>2<\/sup><sub>\u00a0<\/sub>\u2013 4ab\u00a0+ 3b<sup>2<\/sup><sub>\u00a0<\/sub>= 0.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Given R<sub>1<\/sub>\u00a0on\u00a0Q<sub>0<\/sub>\u00a0defined by (a, b) \u2208 R<sub>1<\/sub>\u00a0\u21d4\u00a0a\u00a0= 1\/b.<\/p>\n<p>Reflexivity:<br \/><br \/>Let\u00a0a\u00a0be an arbitrary element of\u00a0R<sub>1<\/sub>.<\/p>\n<p>Then, a\u00a0\u2208\u00a0R<sub>1<\/sub><\/p>\n<p>\u21d2\u00a0a\u00a0\u22601\/a\u00a0for\u00a0all\u00a0a\u00a0\u2208\u00a0Q<sub>0<\/sub><\/p>\n<p>So,\u00a0R<sub>1<\/sub>\u00a0is\u00a0not\u00a0reflexive.<\/p>\n<p>Symmetry:<br \/><br \/>Let (a,\u00a0b)\u00a0\u2208\u00a0R<sub>1<\/sub>\u00a0<\/p>\n<p>Then, (a,\u00a0b)\u00a0\u2208\u00a0R<sub>1<\/sub><\/p>\n<p>Therefore we can write \u2018a\u2019 as a =1\/b<\/p>\n<p>\u21d2 b = 1\/a<\/p>\n<p>\u21d2 (b, a)\u00a0\u2208 R<sub>1<\/sub><\/p>\n<p>So,\u00a0R<sub>1<\/sub>\u00a0is\u00a0symmetric.<\/p>\n<p>Transitivity:<br \/><br \/>Here, (a,\u00a0b)\u00a0\u2208R<sub>1<\/sub>\u00a0and\u00a0(b,\u00a0c)\u00a0\u2208R<sub>2<\/sub><\/p>\n<p>\u21d2 a = 1\/b and b = 1\/c<\/p>\n<p>\u21d2 a = 1\/ (1\/c) = c<\/p>\n<p>\u21d2 a\u00a0\u2260 1\/c<\/p>\n<p>\u21d2 (a, c)\u00a0\u2209 R<sub>1<\/sub><\/p>\n<p>So,\u00a0R<sub>1<\/sub>\u00a0is\u00a0not\u00a0transitive.<\/p>\n<p>\u00a0<\/p>\n<p>(ii) Given R<sub>2<\/sub>\u00a0on\u00a0Z\u00a0defined by (a, b) \u2208 R<sub>2<\/sub>\u00a0\u21d4 |a \u2013 b| \u2264 5<\/p>\n<p>Now we have to check whether R<sub>2<\/sub> is reflexive, symmetric, and transitive.<\/p>\n<p>Reflexivity:<\/p>\n<p>Let\u00a0a\u00a0be an arbitrary element of\u00a0R<sub>2<\/sub>.<\/p>\n<p>Then, a\u00a0\u2208\u00a0R<sub>2<\/sub><\/p>\n<p>On applying the given condition we get,<\/p>\n<p>\u21d2\u00a0|\u00a0a\u2212a\u00a0|\u00a0=\u00a00\u00a0\u2264\u00a05<\/p>\n<p>So,\u00a0R<sub>1<\/sub>\u00a0is\u00a0reflexive.<\/p>\n<p>Symmetry:<\/p>\n<p>Let\u00a0(a,\u00a0b) \u2208\u00a0R<sub>2<\/sub><\/p>\n<p>\u21d2\u00a0|a\u2212b|\u00a0\u2264\u00a05\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0[Since,\u00a0|a\u2212b|\u00a0=\u00a0|b\u2212a|]<\/p>\n<p>\u21d2\u00a0|b\u2212a|\u00a0\u2264\u00a05<\/p>\n<p>\u21d2\u00a0(b,\u00a0a)\u00a0\u2208\u00a0R<sub>2<\/sub><\/p>\n<p>So,\u00a0R2\u00a0is\u00a0symmetric.<\/p>\n<p>Transitivity:<\/p>\n<p>Let\u00a0(1,\u00a03)\u00a0\u2208\u00a0R<sub>2<\/sub>\u00a0and\u00a0(3,\u00a07)\u00a0\u2208R<sub>2<\/sub><\/p>\n<p>\u21d2|1\u22123|\u22645\u00a0and\u00a0|3\u22127|\u22645<\/p>\n<p>But\u00a0|1\u22127|\u00a0\u22705\u00a0<\/p>\n<p>\u21d2\u00a0(1, 7)\u00a0\u2209\u00a0R<sub>2<\/sub><\/p>\n<p>So,\u00a0R<sub>2<\/sub>\u00a0is\u00a0not\u00a0transitive.<\/p>\n<p>(iii) Given R<sub>3\u00a0<\/sub>on\u00a0R\u00a0defined by (a, b) \u2208\u00a0R<sub>3<\/sub>\u00a0\u21d4\u00a0a<sup>2<\/sup><sub>\u00a0<\/sub>\u2013 4ab\u00a0+ 3b<sup>2<\/sup><sub>\u00a0<\/sub>= 0.<\/p>\n<p>Now we have check whether R<sub>2<\/sub>\u00a0is reflexive, symmetric and transitive.<\/p>\n<p>Reflexivity:<\/p>\n<p>Let a\u00a0be an arbitrary element of\u00a0R<sub>3<\/sub>.<\/p>\n<p>Then, a\u00a0\u2208\u00a0R<sub>3<\/sub><\/p>\n<p>\u21d2\u00a0a<sup>2\u00a0<\/sup>\u2212\u00a04a\u00a0\u00d7\u00a0a+\u00a03a<sup>2<\/sup>=\u00a00\u00a0<\/p>\n<p>So,\u00a0R<sub>3<\/sub>\u00a0is\u00a0reflexive<\/p>\n<p>Symmetry:<\/p>\n<p>Let\u00a0(a,\u00a0b)\u00a0\u2208\u00a0R<sub>3<\/sub><\/p>\n<p>\u21d2\u00a0a<sup>2<\/sup>\u22124ab+3b<sup>2<\/sup>=0<\/p>\n<p>But\u00a0b<sup>2<\/sup>\u22124ba+3a<sup>2<\/sup>\u22600\u00a0for\u00a0all\u00a0a,\u00a0b\u00a0\u2208\u00a0R<\/p>\n<p>So,\u00a0R<sub>3<\/sub>\u00a0is\u00a0not\u00a0symmetric.<\/p>\n<p>Transitivity:<\/p>\n<p>Let (1,\u00a02)\u00a0\u2208\u00a0R<sub>3<\/sub>\u00a0and\u00a0(2,\u00a03)\u00a0\u2208\u00a0R<sub>3<\/sub><\/p>\n<p>\u21d2\u00a01 \u2212 8 + 6 = 0\u00a0and\u00a04 \u2013 24 + 27 = 0<\/p>\n<p>But\u00a01\u00a0\u2013\u00a012 + 9 \u2260 0<\/p>\n<p>So,\u00a0R<sub>3<\/sub>\u00a0is\u00a0not\u00a0transitive.<\/p>\n<p><strong>4. Let\u00a0A\u00a0= {1, 2, 3}, and let\u00a0R<sub>1<\/sub>\u00a0= {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)},\u00a0R<sub>2<\/sub>\u00a0= {(2, 2), (3, 1), (1, 3)},\u00a0R<sub>3<\/sub>\u00a0= {(1, 3), (3, 3)}. Find whether or not each of the relations\u00a0R<sub>1<\/sub>,\u00a0R<sub>2<\/sub>,\u00a0R<sub>3<\/sub>\u00a0on A is (i) reflexive (ii) symmetric (iii) transitive.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Consider R<sub>1<\/sub><br \/><br \/>Given R<sub>1<\/sub>\u00a0= {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}<\/p>\n<p>Reflexivity:<br \/><br \/>Here, (1,\u00a01),\u00a0(2,\u00a02),\u00a0(3,\u00a03)\u00a0\u2208R<\/p>\n<p>So,\u00a0R<sub>1<\/sub>\u00a0is\u00a0reflexive.<\/p>\n<p>Symmetry:<\/p>\n<p>Here, (2, 1)\u00a0\u2208\u00a0R<sub>1<\/sub>,<\/p>\n<p>But\u00a0(1, 2)\u00a0\u2209\u00a0R<sub>1<\/sub><\/p>\n<p>So,\u00a0R<sub>1<\/sub>\u00a0is\u00a0not\u00a0symmetric.<\/p>\n<p>Transitivity:<\/p>\n<p>Here,\u00a0(2,\u00a01)\u00a0\u2208R<sub>1<\/sub>\u00a0and\u00a0(1,\u00a03) \u2208R<sub>1<\/sub>,\u00a0<\/p>\n<p>But\u00a0(2,\u00a03) \u2209R<sub>1<\/sub>\u00a0<\/p>\n<p>So,\u00a0R<sub>1<\/sub>\u00a0is\u00a0not\u00a0transitive.<\/p>\n<p>Now consider R<sub>2<\/sub><br \/><br \/>Given R<sub>2<\/sub>\u00a0= {(2, 2), (3, 1), (1, 3)}<\/p>\n<p>Reflexivity:<\/p>\n<p>Clearly,\u00a0(1,\u00a01)\u00a0and\u00a0(3,\u00a03) \u2209R<sub>2<\/sub>\u00a0<\/p>\n<p>So,\u00a0R<sub>2<\/sub>\u00a0is\u00a0not\u00a0reflexive.<\/p>\n<p>Symmetry:<\/p>\n<p>Here,\u00a0(1,\u00a03)\u00a0\u2208\u00a0R<sub>2<\/sub>\u00a0and\u00a0(3,\u00a01)\u00a0\u2208\u00a0R<sub>2<\/sub><\/p>\n<p>So,\u00a0R<sub>2<\/sub>\u00a0is\u00a0symmetric.<\/p>\n<p>Transitivity:<\/p>\n<p>Here,\u00a0(1, 3)\u00a0\u2208\u00a0R<sub>2<\/sub>\u00a0and\u00a0(3, 1)\u00a0\u2208\u00a0R<sub>2\u00a0<\/sub><\/p>\n<p>But\u00a0(3,\u00a03) \u2209R<sub>2<\/sub><\/p>\n<p>So,\u00a0R<sub>2<\/sub>\u00a0is\u00a0not\u00a0transitive.<\/p>\n<p>Consider R<sub>3<\/sub><br \/><br \/>Given R<sub>3<\/sub>\u00a0= {(1, 3), (3, 3)}<\/p>\n<p>Reflexivity:<\/p>\n<p>Clearly,\u00a0(1, 1)\u00a0\u2209\u00a0R<sub>3<\/sub><\/p>\n<p>So,\u00a0R<sub>3<\/sub>\u00a0is\u00a0not\u00a0reflexive.<\/p>\n<p>Symmetry:<\/p>\n<p>Here,\u00a0(1,\u00a03)\u00a0\u2208\u00a0R<sub>3<\/sub>,\u00a0but\u00a0(3,\u00a01)\u00a0\u2209\u00a0R<sub>3<\/sub><\/p>\n<p>So,\u00a0R<sub>3<\/sub>\u00a0is\u00a0not\u00a0symmetric.<\/p>\n<p>Transitivity:<\/p>\n<p>Here,\u00a0(1,\u00a03)\u00a0\u2208\u00a0R<sub>3<\/sub>\u00a0and\u00a0(3,\u00a03)\u00a0\u2208\u00a0R<sub>3<\/sub>\u00a0<\/p>\n<p>Also,\u00a0(1,\u00a03)\u00a0\u2208\u00a0R<sub>3<\/sub><\/p>\n<p>So,\u00a0R<sub>3<\/sub>\u00a0is\u00a0transitive.<\/p>\n<p><strong>5. The following relation is defined on the set of real numbers.<br \/>(i) aRb\u00a0if\u00a0a\u00a0\u2013\u00a0b\u00a0&gt; 0<\/strong><\/p>\n<p><strong>(ii) aRb iff 1 + a b &gt; 0<\/strong><\/p>\n<p><strong>(iii) aRb if |a| \u2264 b.<\/strong><\/p>\n<p><strong>Find whether a relation is reflexive, symmetric, or transitive.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Consider aRb\u00a0if\u00a0a\u00a0\u2013\u00a0b\u00a0&gt; 0<\/p>\n<p>Now for this relation, we have to check whether it is reflexive, transitive, and symmetric.<\/p>\n<p>Reflexivity:<\/p>\n<p>Let\u00a0a\u00a0be an arbitrary element of\u00a0R.<\/p>\n<p>Then, a\u00a0\u2208\u00a0R<\/p>\n<p>But\u00a0a \u2212 a\u00a0=\u00a00\u00a0\u226f\u00a00<\/p>\n<p>So,\u00a0this\u00a0relation\u00a0is\u00a0not\u00a0reflexive.<\/p>\n<p>Symmetry:<\/p>\n<p>Let\u00a0(a,\u00a0b)\u00a0\u2208\u00a0R<\/p>\n<p>\u21d2\u00a0a \u2212 b\u00a0&gt;\u00a00<\/p>\n<p>\u21d2\u00a0\u2212 (b \u2212 a)\u00a0&gt;0<\/p>\n<p>\u21d2\u00a0b \u2212 a\u00a0&lt;\u00a00<\/p>\n<p>So,\u00a0the\u00a0given\u00a0relation\u00a0is\u00a0not\u00a0symmetric.<\/p>\n<p>Transitivity:<\/p>\n<p>Let\u00a0(a,\u00a0b) \u2208R\u00a0and\u00a0(b,\u00a0c) \u2208R.\u00a0<\/p>\n<p>Then, a \u2212 b\u00a0&gt;\u00a00\u00a0and\u00a0b \u2212 c\u00a0&gt; 0<\/p>\n<p>Adding\u00a0the\u00a0two,\u00a0we\u00a0get<\/p>\n<p>a\u00a0\u2013\u00a0b + b\u00a0\u2212\u00a0c\u00a0&gt;\u00a00<\/p>\n<p>\u21d2\u00a0a\u00a0\u2013\u00a0c &gt;\u00a00\u00a0<\/p>\n<p>\u21d2\u00a0(a,\u00a0c)\u00a0\u2208\u00a0R.<\/p>\n<p>So,\u00a0the\u00a0given\u00a0relation\u00a0is\u00a0transitive.<\/p>\n<p>(ii) Consider aRb iff 1 + a b &gt; 0<\/p>\n<p>Now for this relation, we have to check whether it is reflexive, transitive, and symmetric.<\/p>\n<p>Reflexivity:<\/p>\n<p>Let\u00a0a\u00a0be an arbitrary element of\u00a0R.<\/p>\n<p>Then, a\u00a0\u2208\u00a0R<\/p>\n<p>\u21d2 1\u00a0+\u00a0a\u00a0\u00d7\u00a0a\u00a0&gt;\u00a00<\/p>\n<p>i.e.\u00a01\u00a0+\u00a0a<sup>2<\/sup>\u00a0&gt;\u00a00\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Since,\u00a0square\u00a0of\u00a0any\u00a0number\u00a0is\u00a0positive]<\/p>\n<p>So,\u00a0the\u00a0given\u00a0relation\u00a0is\u00a0reflexive.<\/p>\n<p>Symmetry:<\/p>\n<p>Let\u00a0(a,\u00a0b)\u00a0\u2208\u00a0R<\/p>\n<p>\u21d2\u00a01 +\u00a0a b\u00a0&gt;\u00a00<\/p>\n<p>\u21d2\u00a01 + b a\u00a0&gt;\u00a00<\/p>\n<p>\u21d2\u00a0(b,\u00a0a)\u00a0\u2208\u00a0R<\/p>\n<p>So,\u00a0the\u00a0given\u00a0relation\u00a0is\u00a0symmetric.<\/p>\n<p>Transitivity:<\/p>\n<p>Let\u00a0(a,\u00a0b) \u2208R\u00a0and\u00a0(b,\u00a0c) \u2208R<\/p>\n<p>\u21d21 +\u00a0a b\u00a0&gt;\u00a00\u00a0and\u00a01 +\u00a0b c\u00a0&gt;0<\/p>\n<p>But\u00a01+\u00a0ac\u00a0\u226f\u00a00<\/p>\n<p>\u21d2\u00a0(a,\u00a0c)\u00a0\u2209\u00a0R<\/p>\n<p>So,\u00a0the\u00a0given\u00a0relation\u00a0is\u00a0not\u00a0transitive.<\/p>\n<p>(iii) Consider aRb if |a| \u2264 b.<\/p>\n<p>Now for this relation, we have to check whether it is reflexive, transitive, and symmetric.<\/p>\n<p>Reflexivity:<\/p>\n<p>Let\u00a0a\u00a0be an arbitrary element of\u00a0R.<\/p>\n<p>Then,\u00a0a\u00a0\u2208\u00a0R\u00a0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0[Since,\u00a0|a|=a]<\/p>\n<p>\u21d2\u00a0|a|\u226e\u00a0a<\/p>\n<p>So,\u00a0R\u00a0is\u00a0not\u00a0reflexive.<\/p>\n<p>Symmetry:<\/p>\n<p>Let\u00a0(a,\u00a0b)\u00a0\u2208\u00a0R<\/p>\n<p>\u21d2\u00a0|a|\u00a0\u2264\u00a0b\u00a0<\/p>\n<p>\u21d2\u00a0|b|\u00a0\u2270\u00a0a\u00a0for\u00a0all\u00a0a,\u00a0b\u00a0\u2208\u00a0R<\/p>\n<p>\u21d2\u00a0(b,\u00a0a)\u00a0\u2209\u00a0R\u00a0<\/p>\n<p>So,\u00a0R\u00a0is\u00a0not\u00a0symmetric.<\/p>\n<p>Transitivity:<\/p>\n<p>Let\u00a0(a,\u00a0b)\u00a0\u2208\u00a0R\u00a0and\u00a0(b,\u00a0c)\u00a0\u2208\u00a0R<\/p>\n<p>\u21d2\u00a0|a|\u00a0\u2264\u00a0b\u00a0and\u00a0|b|\u00a0\u2264\u00a0c<\/p>\n<p>Multiplying\u00a0the\u00a0corresponding\u00a0sides,\u00a0we\u00a0get<\/p>\n<p>|a|\u00a0\u00d7\u00a0|b|\u00a0\u2264\u00a0b c<\/p>\n<p>\u21d2\u00a0|a|\u00a0\u2264\u00a0c<\/p>\n<p>\u21d2\u00a0(a,\u00a0c)\u00a0\u2208\u00a0R<\/p>\n<p>Thus,\u00a0R\u00a0is\u00a0transitive.\u00a0<\/p>\n<p><strong>6. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as\u00a0R = {(a,\u00a0b):\u00a0b\u00a0=\u00a0a\u00a0+ 1} is reflexive, symmetric or transitive.<\/strong><\/p>\n<p><strong>\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given R = {(a,\u00a0b):\u00a0b\u00a0=\u00a0a\u00a0+ 1}<\/p>\n<p>Now for this relation, we have to check whether it is reflexive, transitive, and symmetric Reflexivity:<\/p>\n<p>Let a be an arbitrary element of\u00a0R.<\/p>\n<p>Then, a\u00a0=\u00a0a\u00a0+\u00a01\u00a0cannot\u00a0be\u00a0true\u00a0for\u00a0all\u00a0a\u00a0\u2208\u00a0A.<\/p>\n<p>\u21d2\u00a0(a,\u00a0a)\u00a0\u2209\u00a0R\u00a0<\/p>\n<p>So,\u00a0R\u00a0is\u00a0not\u00a0reflexive\u00a0on\u00a0A.<\/p>\n<p>Symmetry:<\/p>\n<p>Let\u00a0(a,\u00a0b)\u00a0\u2208\u00a0R<\/p>\n<p>\u21d2\u00a0b\u00a0=\u00a0a\u00a0+\u00a01<\/p>\n<p>\u21d2\u00a0\u2212a\u00a0=\u00a0\u2212b\u00a0+\u00a01<\/p>\n<p>\u21d2\u00a0a\u00a0=\u00a0b\u00a0\u2212\u00a01<\/p>\n<p>Thus,\u00a0(b,\u00a0a)\u00a0\u2209\u00a0R<\/p>\n<p>So,\u00a0R\u00a0is\u00a0not\u00a0symmetric\u00a0on\u00a0A.<\/p>\n<p>Transitivity:\u00a0<\/p>\n<p>Let\u00a0(1,\u00a02)\u00a0and\u00a0(2,\u00a03)\u00a0\u2208\u00a0R<\/p>\n<p>\u21d2\u00a02\u00a0=\u00a01\u00a0+\u00a01\u00a0and\u00a03\u00a0<\/p>\n<p>2\u00a0+\u00a01\u00a0\u00a0is\u00a0true.<\/p>\n<p>But\u00a03\u00a0\u2260\u00a01+1<\/p>\n<p>\u21d2\u00a0(1,\u00a03)\u00a0\u2209\u00a0R<\/p>\n<p>So,\u00a0R\u00a0is\u00a0not\u00a0transitive\u00a0on\u00a0A.<\/p>\n<p><strong>7. Check whether the relation R on\u00a0R\u00a0defined as R = {(<em>a<\/em>,\u00a0<em>b<\/em>):\u00a0<em>a<\/em>\u00a0\u2264\u00a0<em>b<\/em><sup>3<\/sup>} is reflexive, symmetric, or transitive.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given R = {(<em>a<\/em>,\u00a0<em>b<\/em>):\u00a0<em>a\u00a0<\/em>\u2264\u00a0<em>b<\/em><sup>3<\/sup>}<\/p>\n<p>It is observed that (1\/2, 1\/2) in R as 1\/2 &gt; (1\/2)<sup>3<\/sup>\u00a0= 1\/8<\/p>\n<p>\u2234 R is not reflexive.<\/p>\n<p>Now,<\/p>\n<p>(1, 2) \u2208 R (as 1 &lt; 2<sup>3<\/sup>\u00a0= 8)<\/p>\n<p>But,<\/p>\n<p>(2, 1) \u2209 R (as 2 &gt; 1<sup>3<\/sup>\u00a0= 1)<\/p>\n<p>\u2234 R is not symmetric.<\/p>\n<p>We have (3, 3\/2), (3\/2,\u00a06\/5) in \u201cR as\u201d\u00a03 &lt; (3\/2)<sup>3<\/sup>\u00a0and 3\/2 &lt; (6\/5)<sup>3<\/sup><\/p>\n<p>But\u00a0(3, 6\/5)\u00a0\u2209 R as 3 &gt; (6\/5)<sup>3<\/sup><\/p>\n<p>\u2234 R is not transitive.<\/p>\n<p>Hence, R is\u00a0neither reflexive, nor symmetric, nor transitive.<\/p>\n<p>\u00a0<\/p>\n<p><strong>8. Prove that every identity relation on a set is reflexive, but the converse is not necessarily true.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u00a0A\u00a0be a set.<\/p>\n<p>Then, Identity\u00a0relation\u00a0IA=I<sub>A<\/sub>\u00a0is\u00a0reflexive,\u00a0since\u00a0(a,\u00a0a)\u00a0\u2208 A \u2200a<\/p>\n<p>The converse of it need not be necessarily true.<br \/><br \/>Consider the set\u00a0A\u00a0= {1, 2, 3}<\/p>\n<p>Here,<br \/><br \/>Relation\u00a0R\u00a0= {(1, 1), (2, 2) , (3, 3), (2, 1), (1, 3)} is reflexive on\u00a0A.<br \/><br \/>However,\u00a0R\u00a0is not an identity relation.<\/p>\n<p><strong>9. If\u00a0A\u00a0= {1, 2, 3, 4} define relations on\u00a0A\u00a0which have properties of being<\/strong><\/p>\n<p><strong>(i) Reflexive, transitive but not symmetric<\/strong><\/p>\n<p><strong>(ii) Symmetric but neither reflexive nor transitive.<\/strong><\/p>\n<p><strong>(iii) Reflexive, symmetric and transitive.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) The relation on\u00a0A\u00a0having properties of being reflexive, transitive, but not symmetric is<br \/><br \/>R\u00a0= {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)}<\/p>\n<p>Relation\u00a0R\u00a0satisfies\u00a0reflexivity\u00a0and\u00a0transitivity.<\/p>\n<p>\u21d2 (1,\u00a01),\u00a0(2,\u00a02),\u00a0(3,\u00a03)\u00a0\u2208\u00a0R\u00a0<\/p>\n<p>And\u00a0(1,\u00a01),\u00a0(2,\u00a01)\u00a0\u2208\u00a0R\u00a0\u21d2 (1,\u00a01)\u00a0\u2208\u00a0R<\/p>\n<p>However,\u00a0(2,\u00a01)\u00a0\u2208\u00a0R,\u00a0but\u00a0(1,\u00a02)\u00a0\u2209\u00a0R<\/p>\n<p>(ii) \u00a0The relation on\u00a0A\u00a0having properties of being reflexive, transitive, but not symmetric is<br \/><br \/>R\u00a0= {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)}<\/p>\n<p>Relation\u00a0R\u00a0satisfies\u00a0reflexivity\u00a0and\u00a0transitivity.<\/p>\n<p>\u21d2 (1,\u00a01),\u00a0(2,\u00a02),\u00a0(3,\u00a03)\u00a0\u2208\u00a0R\u00a0<\/p>\n<p>And\u00a0(1,\u00a01),\u00a0(2,\u00a01)\u00a0\u2208\u00a0R\u00a0\u21d2 (1,\u00a01)\u00a0\u2208\u00a0R<\/p>\n<p>However,\u00a0(2,\u00a01)\u00a0\u2208\u00a0R,\u00a0but\u00a0(1,\u00a02)\u00a0\u2209\u00a0R<\/p>\n<p>(iii) The relation on A having properties of being symmetric, reflexive, and transitive is<br \/><br \/>R\u00a0= {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1)}<br \/><br \/>The relation\u00a0R\u00a0is an equivalence relation on\u00a0A.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-12-maths-chapter-1-exercise-11-important-topics\"><\/span>RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1: Important Topics<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Let us have a look at 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 RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a> 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-11\"><\/span><strong>FAQs on RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1<\/strong><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-1629718051264\" 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-11\"><\/span>How many questions are there in RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are a total of 9 questions in RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629718115751\" 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-11-for-free\"><\/span>Is RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1 for 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 1.1 for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629718141754\" 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-11-free-pdf\"><\/span>Where can I download RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1 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 1.1 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 1.1: The RD Sharma Class 12 Solution PDF of Chapter 1 Exercise 1.1 Relations can be downloaded from the link given. Solutions for correct answer practice are prepared by experts in the best possible way and are easily understood by the students. In this practice, students &#8230; <a title=\"RD Sharma Solutions for Class 12 Maths Exercise 1.1 Chapter 1 Relation (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-1-exercise-1-1\/\" aria-label=\"More on RD Sharma Solutions for Class 12 Maths Exercise 1.1 Chapter 1 Relation (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":117929,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[2985,2917,73716,73411,73410],"tags":[3429,73223,73664,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68975"}],"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=68975"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68975\/revisions"}],"predecessor-version":[{"id":121948,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68975\/revisions\/121948"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/117929"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=68975"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=68975"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=68975"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}