{"id":5151,"date":"2023-10-31T11:05:04","date_gmt":"2023-10-31T05:35:04","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=5151"},"modified":"2023-11-01T10:10:34","modified_gmt":"2023-11-01T04:40:34","slug":"ap-ssc-10th-class-maths-solutions-chapter-1-ex-1-4","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-ssc-10th-class-maths-solutions-chapter-1-ex-1-4\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.4"},"content":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 1 Real Numbers Ex 1.4 Textbook Questions and Answers.<\/p>\n

AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.4<\/h2>\n

10th Class Maths 1st Lesson Real Numbers Ex 1.4 Textbook Questions and Answers<\/h3>\n

Question 1.
\nProve that the following are irrational,
\ni) \\(\\frac{1}{\\sqrt{2}}\\)
\nii) \u221a3 + \u221a5
\niii) 6 + \u221a2
\niv) \u221a5
\nv) 3 + 2\u221a5
\nAnswer:
\ni) \\(\\frac{1}{\\sqrt{2}}\\)
\nOn the contrary, suppose that is a \\(\\frac{1}{\\sqrt{2}}\\) rational number;
\nthen \\(\\frac{1}{\\sqrt{2}}\\) is of the form where \\(\\frac{p}{q}\\) and q are integers.
\n\u2234 \\(\\frac{1}{\\sqrt{2}}\\) = \\(\\frac{p}{q}\\)
\n\u21d2 \\(\\frac{\\sqrt{2}}{1}\\) = \\(\\frac{q}{p}\\)
\n(i.e.,) \u221a2 is a rational number and it is a contradiction. This contradiction arised due to our supposition that \\(\\frac{1}{\\sqrt{2}}\\) is a rational number.
\nHence \\(\\frac{1}{\\sqrt{2}}\\) is an irrational number.<\/p>\n

\"AP<\/p>\n

ii) Suppose \u221a3 + \u221a5 is not an irrational number.
\nThen \u221a3 + \u221a5 must be a rational number.
\n\u221a3 + \u221a5 = \\(\\frac{p}{q}\\), q \u2260 0 and p, q \u2208 Z
\nSquaring on both sides
\n\"AP
\nbut \u221a15 is an irrational number.
\n\\(\\frac{p^{2}-8 q^{2}}{2 q^{2}}\\) is a rational number
\n(p2<\/sup> – 8q2<\/sup>, 2q2<\/sup> \u2208 Z, 2q2<\/sup> \u2260 0)
\nbut an irrational number can\u2019t be equal to a rational number, so our supposition that \u221a3 + \u221a5 is not an irrational number is false.
\n\u2234 \u221a3 + \u221a5 is an irrational number.<\/p>\n

iii) 6 + \u221a2
\nTo prove: 6 + \u221a2 is an irrational number.
\nLet us suppose that 6 + \u221a2 is a rational number.
\n\u2234 6 + \u221a2 = \\(\\frac{p}{q}\\), q \u2260 0
\n\u21d2 \u221a2 = \\(\\frac{p}{q}\\) – 6
\n\u21d2 \u221a2 = Difference of two rational numbers
\n\u21d2 \u221a2 = rational number But this contradicts the fact that \u221a2 is an irrational number.
\n\u2234 Our supposition is wrong.
\nHence the given statement is true.
\n6 + \u221a2 is an irrational number.<\/p>\n

iv) \u221a5
\nTo prove: \u221a5 is an irrational number.
\nOn the contrary, let us assume that \u221a5 is a rational number.
\n\u2234 \u221a5 = \\(\\frac{p}{q}\\), q \u2260 0
\nIf p, q have a common factor, on cancelling the common factor let it be
\nreduces to \\(\\frac{a}{b}\\) where a, b are co-primes.
\nNow \u221a5 = \\(\\frac{a}{b}\\), where HCF (a, b) = 1
\nSquaring on both sides we get
\n\u21d2 (\u221a5)2<\/sup> = \\(\\left(\\frac{a}{b}\\right)^{2}\\)
\n\u21d2 5 = \\(\\frac{\\mathrm{a}^{2}}{\\mathrm{~b}^{2}}\\)
\n\u21d2 5b2<\/sup> = a2<\/sup>
\n\u21d2 5 divides a2<\/sup> and thereby 5 divides 5
\nNow, take a = 5c
\nthen, a2<\/sup> = 25c2<\/sup>
\ni.e., 5b2<\/sup> = 25c2<\/sup>
\n\u21d2 b2<\/sup> = 5c2<\/sup>
\n\u21d2 5 divides b2<\/sup> and thereby b.
\n\u21d2 5 divides both b and a.
\nThis contradicts that a and b are co-primes.
\nThis contradiction arised due to our assumption that \u221a5 is a rational number.
\nHence our assumption is wrong and the given statement is true, i.e., \u221a5 is an irrational number,<\/p>\n

\"AP<\/p>\n

v) 3 + 2\u221a5
\nTo Prove: 3 + 2\u221a5 is an irrational.
\nOn the contrary, let us assume that 3 + 2\u221a5 is a rational number.
\n\"AP
\nHere p, q being integers we can say that \\(\\frac{p-3q}{2q}\\) is a rational number.
\nThis contradicts the fact that \u221a5 is an irrational number. This is due to our assumption “3 + 2\u221a5 is a rational number”.
\nHence our assumption is wrong.
\n\u2234 3 + 2\u221a5 is an irrational number.<\/p>\n

Question 2.
\nProve that \u221ap + \u221aq is an irrational, where p, q are primes.
\nAnswer:
\nGiven that p, q are primes.
\nHence fp and fq are irrationals.
\n[\u2235 p, q have no factors other than 1 \u2235 they are primes.]
\nNow \u221ap + \u221aq = sum of two irrational numbers = an irrational number
\nHence proved.<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers Ex 1.4 Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.4 10th Class Maths 1st Lesson Real Numbers Ex 1.4 Textbook Questions and Answers Question 1. Prove that the following are irrational, … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/5151"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=5151"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/5151\/revisions"}],"predecessor-version":[{"id":40605,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/5151\/revisions\/40605"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=5151"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=5151"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=5151"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}