{"id":6004,"date":"2023-07-14T04:45:38","date_gmt":"2023-07-14T04:45:38","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6004"},"modified":"2023-07-23T05:41:47","modified_gmt":"2023-07-23T05:41:47","slug":"ncert-solutions-class-10-maths-chapter-1-real-numbers-ex-1-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-1-real-numbers-ex-1-3\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.3"},"content":{"rendered":"<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><br \/>\n<strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 1 (Real Numbers)\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 1 Real Numbers\u00a0 <\/strong>Exercise 1.3 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #ff0000;\"><strong>Chapter : 1 Real Numbers<\/strong><\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-1-real-numbers-ex-1-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 1.1<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-1-real-numbers-ex-1-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 1.2<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-1-real-numbers-ex-1-4\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 1.4<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 1.3\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. Prove that \u221a5<\/strong> <strong>is irrational.<br \/>\n<\/strong><strong>Solutions &#8211; <\/strong>Let \u221a5 is a rational number.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, we can find two integers <i>a<\/i>, <i>b<\/i> (<i>b<\/i> \u2260 0) such that \u221a5 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a}{b}\" alt=\"\\frac{a}{b}\" align=\"absmiddle\" \/><br \/>\nLet <i>a<\/i> and <i>b<\/i> have a common factor other than 1. Then we can divide them by the common factor, and assume that <i>a<\/i> and <i>b<\/i> are co-prime.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">a = \u221a5b<br \/>\na<sup>2 <\/sup>= 5b<sup>2<br \/>\n<\/sup>Therefore, <i>a<\/i><sup>2<\/sup> is divisible by 5 and it can be said that <i>a<\/i> is divisible by 5.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Let <i>a<\/i> = 5<i>k<\/i>, where <i>k<\/i> is an integer<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">(5k)<sup>2\u00a0<\/sup>= 5b<sup>2\u00a0\u00a0<\/sup>This means that <i>b<\/i><sup>2<\/sup> is divisible by 5 and hence, <i>b<\/i> is divisible by 5.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">b<sup>2\u00a0<\/sup>= 5k<sup>2\u00a0 <\/sup>This implies that <i>a<\/i> and <i>b<\/i> have 5 as a common factor.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">And this is a contradiction to the fact that <i>a<\/i> and <i>b<\/i> are co-prime.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence,\u221a5 cannot be expressed as <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{p}{q}\" alt=\"\\frac{p}{q}\" align=\"absmiddle\" \/> or it can be said that \u221a5 is irrational.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>2. Prove that 3 + 2\u221a5 + is irrational.<br \/>\n<\/strong><strong>Solutions &#8211; <\/strong>Let 3 + 2\u221a5 is rational.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, we can find two integers <i>a<\/i>, <i>b<\/i> (<i>b<\/i> \u2260 0) such that<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">3 + 2\u221a5 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a}{b}\" alt=\"\\frac{a}{b}\" align=\"absmiddle\" \/><br \/>\n2\u221a5 =\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a}{b}\" alt=\"\\frac{a}{b}\" align=\"absmiddle\" \/> &#8211; 3<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\">\u221a5 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{a}{b}&amp;space;-3\\right&amp;space;)\" alt=\"\\left ( \\frac{a}{b} -3\\right )\" align=\"absmiddle\" \/><br \/>\nSince <i>a<\/i> and <i>b<\/i> are integers, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{a}{b}&amp;space;-3\\right&amp;space;)\" alt=\"\\left ( \\frac{a}{b} -3\\right )\" align=\"absmiddle\" \/> will also be rational and therefore,\u221a5 is rational.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">This contradicts the fact that \u221a5 is irrational. Hence, our assumption that 3 + 2\u221a5 is rational is false. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, 3 + 2\u221a5\u00a0 is irrational.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>3. Prove that the following are irrationals:<br \/>\n<\/strong><strong>(i) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{\\sqrt{2}}}\" alt=\"\\mathbf{\\frac{1}{\\sqrt{2}}}\" align=\"absmiddle\" \/><br \/>\n<\/strong><strong>(ii) 7\u221a5<br \/>\n<\/strong><strong>(iii) 6 + <\/strong>\u221a<strong>2<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solutions &#8211;<br \/>\n<\/strong><strong>(i) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{\\sqrt{2}}}\" alt=\"\\mathbf{\\frac{1}{\\sqrt{2}}}\" align=\"absmiddle\" \/><br \/>\n<\/strong>Let <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{\\sqrt{2}}\" alt=\"\\frac{1}{\\sqrt{2}}\" align=\"absmiddle\" \/>\u00a0 is rational.<br \/>\nTherefore, we can find two integers <i>a<\/i>, <i>b<\/i> (<i>b<\/i> \u2260 0) such that<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{\\sqrt{2}}\" alt=\"\\frac{1}{\\sqrt{2}}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a}{b}\" alt=\"\\frac{a}{b}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\">\u221a2 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{b}{a}\" alt=\"\\frac{b}{a}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{b}{a}\" alt=\"\\frac{b}{a}\" align=\"absmiddle\" \/> is rational as <i>a<\/i> and <i>b<\/i> are integers. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, \u221a2 is rational which contradicts to the fact that \u221a2 is irrational.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, our assumption is false and <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{\\sqrt{2}}\" alt=\"\\frac{1}{\\sqrt{2}}\" align=\"absmiddle\" \/> is irrational.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) 7<\/strong>\u221a<strong>5<br \/>\n<\/strong>Let 7\u221a5 is a rational number.<br \/>\nTherefore, we can find two integers <i>a<\/i>,<i> b<\/i> (<i>b<\/i> \u2260 0) such that<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\">7\u221a5\u00a0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a}{b}\" alt=\"\\frac{a}{b}\" align=\"absmiddle\" \/> \u00a0for some integers <i>a<\/i> and <i>b<br \/>\n<\/i><br \/>\n\u2234 \u221a5 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a}{7b}\" alt=\"\\frac{a}{7b}\" align=\"absmiddle\" \/> <i><br \/>\n<\/i><\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"> <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a}{7b}\" alt=\"\\frac{a}{7b}\" align=\"absmiddle\" \/> is rational as <i>a<\/i> and <i>b<\/i> are integers.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, \u221a5 should be rational.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">This contradicts the fact that \u221a5 is irrational. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, our assumption that 7\u221a5 is rational is false. Hence, 7\u221a5 is irrational.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii) 6 + <\/strong>\u221a<strong>2<br \/>\n<\/strong>Let 6 + \u221a2 is a rational number.<br \/>\nTherefore, we can find two integers <i>a<\/i>, <i>b<\/i> (<i>b<\/i> \u2260 0) such that<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">6 + \u221a2 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a}{b}\" alt=\"\\frac{a}{b}\" align=\"absmiddle\" \/><br \/>\n\u221a2 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a}{b}\" alt=\"\\frac{a}{b}\" align=\"absmiddle\" \/> &#8211; 6<br \/>\nSince <i>a<\/i> and <i>b<\/i> are integers, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a}{b}\" alt=\"\\frac{a}{b}\" align=\"absmiddle\" \/> &#8211; 6 is also rational and hence, \u221a2 should be rational. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">This contradicts the fact that \u221a2 is irrational. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, our assumption is false and hence, 6 + \u221a2 is irrational.<\/span><\/p>\n<p><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\"><b><i>Go Back To Chapters<\/i><\/b><\/a><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 1 (Real Numbers)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 1 Real Numbers\u00a0 Exercise 1.3 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 1 Real Numbers NCERT Class 10 Maths Solution Ex &#8211; 1.1 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1046,1051,1044,1049,1048],"class_list":["post-6004","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-1-real-numbers-solutions","tag-ncert-class-10-mathematics-exercise-1-3-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - 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