{"id":24922,"date":"2022-06-05T15:30:28","date_gmt":"2022-06-05T10:00:28","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=24922"},"modified":"2022-05-23T15:58:35","modified_gmt":"2022-05-23T10:28:35","slug":"ncert-solutions-for-class-10-maths-chapter-1-ex-1-3","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-1-ex-1-3\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 1 Real Numbers Ex 1.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.3<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nProve that \\(\\sqrt{5}\\) is irrational.
\nSolution:
\nLet us assume, to the contrary, that \\(\\sqrt{5}\\) is irrational that is we can find integers a and b (b \u2260 0) such that \\(\\sqrt{5}\\) = \\(\\frac { a }{ b }\\). suppose a and b have a common factor other than 1 then we can divide by the common factor and assume that a and b are coprime.
\nSo b\\(\\sqrt{5}\\) = a
\nSquaring on both sides, and rearranging we get 5b\u00b2 = a\u00b3.
\nThus for a\u00b2 is divisible by 5, it follows that a is also divisible by 5.
\nSo, we can write a 5c for some integer c.
\nSubstituting for a, we get
\n5b\u00b2 = 25c\u00b2
\nb\u00b2 = 5c\u00b2
\nThis means that b2 is divisible by 5 and so b is also divisible by 5.
\nTherefore, a and b have at least 5 as common factor.
\nBut this contradicts the fact that a and b are coprime.
\nThis contradiction has arisen because of our incorrect assumption that \\(\\sqrt{5}\\) is irrational.
\nSo we conclude that \\(\\sqrt{5}\\) is irrational.<\/p>\n

Question 2.
\nProve that 3 + 2A\\(\\sqrt{5}\\) is irrational.
\nSolution:
\nLet us assume, to contrary, that 3 + 2\\(\\sqrt{5}\\) is rational.
\nThat is, we can find coprime a and b (b \u2260 0)
\nsuch that 3 + 2\\(\\sqrt{5}\\) = \\(\\frac { a }{ b }\\)
\nTherefore 3 – \\(\\frac { a }{ b }\\) = – 2\\(\\sqrt{5}\\)
\nRearranging this equation we get 2\\(\\sqrt{5}\\) = \\(\\frac { a }{ b }\\) – 3 = \\(\\frac { a – 3b }{ b }\\)
\nSince a and b are integers, \\(\\frac { a }{ b }\\) – 3 we get is
\nrational and so 2\\(\\sqrt{5}\\) is rational and so \\(\\sqrt{5}\\) is rational.
\nBut this contradicts the fact \\(\\sqrt{5}\\) is irrational.
\nThis contadiction has arise because of our incorrect assumption 3 + 2\\(\\sqrt{5}\\) is rational.
\nThus, we conclude that 3 + 2\\(\\sqrt{5}\\) is irrational.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nProve that the following are irrationals:
\n(i) \\(\\frac{1}{\\sqrt{2}}\\)
\n(ii) 7\\(\\sqrt{5}\\)
\n(iii) 6 + \\(\\sqrt{2}\\)
\nSolution:
\nLet us assume to the contrary, that \\(\\frac{1}{\\sqrt{2}}\\) is rational that is, we can find coprime a and b(b \u2260 0) such that = \\(\\frac{1}{\\sqrt{2}}\\) = \\(\\frac { a }{ b }\\)
\nSince a and b are integers so \\(\\frac { a }{ b }\\) is rational and so \\(\\sqrt{2}\\) is rational.
\nBut this contradicts the fact that \\(\\sqrt{2}\\) is irrational.
\nThus, we conclude that \\(\\frac{1}{\\sqrt{2}}\\) is irrational<\/p>\n

(ii) 7\\(\\sqrt{5}\\)
\nLet us assume, to the contrary that 7\\(\\sqrt{5}\\) is rational that is, we can find coprime a and b (\u2260 0)
\nsuch that 7\\(\\sqrt{5}\\) = \\(\\frac { a }{ b }\\) Rearranging, we get \\(\\sqrt{5}\\) = \\(\\frac { a }{ b }\\)
\nSince 7, a and b are integers, \\(\\frac { a }{ 7b }\\) is rational and so \\(\\sqrt{5}\\) is rational
\nBut this contradicts the fact that \\(\\sqrt{5}\\) is irrational. So, we conclude that 7\\(\\sqrt{5}\\) is irrational.<\/p>\n

(iii) 6 + \\(\\sqrt{2}\\)
\nLet us assume, to the contrary, that 6 + \\(\\sqrt{2}\\) is irrational.
\nThat is, we can find co prime a and b (* 0) such that 6 + \\(\\sqrt{2}\\) = 7 b
\nRearranging, we get \\(\\sqrt{2}\\) = \\(\\frac { a-6b }{ b }\\)
\nSince a, b and 6 are integers, so \\(\\frac { a-6b }{ b }\\) is rational and so is rational.
\nBut this contradicts the fact that \\(\\sqrt{2}\\) is
\nirrational. So we conclude that 6 + \\(\\sqrt{2}\\) is irrational.<\/p>\n

\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.3 Question 1. Prove that is irrational. Solution: Let us assume, to the contrary, that is irrational that is we …<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nNCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3 - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-1-ex-1-3\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3 Questions and Answers are prepared by our highly skilled subject experts. 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