{"id":125039,"date":"2023-09-10T06:49:00","date_gmt":"2023-09-10T01:19:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125039"},"modified":"2023-11-03T10:40:43","modified_gmt":"2023-11-03T05:10:43","slug":"rd-sharma-class-10-solutions-chapter-1-exercise-1-5","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-1-exercise-1-5\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125048\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-1-Exercise-1.5.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-1-Exercise-1.5.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-1-Exercise-1.5-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5:&nbsp;<\/strong>Numbers that cannot be stated in the usual form of p\/q are referred to as irrational numbers. This exercise explains how to prove the irrationality of numbers. Kopykitab&#8217;s subject experts have prepared the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Solutions Class 10<\/strong><\/a> to help students understand how to do workout problems correctly. If you need help with any of the questions in this exercise, you can use the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-1-real-numbers\/\"><strong>RD Sharma Solutions for Class 10 Maths Chapter 1 Real Numbers<\/strong><\/a> Exercise 1.5, which is available in PDF format below.<\/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-69d073c3244a7\" 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\" 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ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-1-exercise-1-5\/#how-much-does-it-cost-to-download-rd-sharma-class-10-solutions-chapter-1-exercise-15-pdf\" title=\"How much does it cost to download RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5 PDF?\">How much does it cost to download RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5 PDF?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-1-exercise-15-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/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\/09\/RD-Sharma-Solution-Class-10-Chapter-1-Exercise-1.5.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-1-Exercise-1.5.pdf\">RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5 PDF<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-1-exercise-15-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 1 Exercise 1.5- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Show that the following numbers are irrational.<\/strong><\/p>\n<p><strong>(i) 1\/\u221a2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Consider 1\/\u221a2 is a rational number<\/p>\n<p>Let us assume 1\/\u221a2 = r where r is a rational number<\/p>\n<p>On further calculation, we get<\/p>\n<p>1\/r = \u221a2<\/p>\n<p>Since r is a rational number, 1\/r = \u221a2 is also a rational number<\/p>\n<p>But we know that \u221a2 is an irrational number<\/p>\n<p>So, our supposition is wrong.<\/p>\n<p>Hence, 1\/\u221a2 is an irrational number.<\/p>\n<p><strong>(ii) 7\u221a5<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume, on the contrary, that 7\u221a5 is a rational number. Then, there exist positive integers a and b such that<\/p>\n<p>7\u221a5 = a\/b, where a and b are co-primes<\/p>\n<p>\u21d2 \u221a5 = a\/7b<\/p>\n<p>\u21d2 \u221a5 is rational [\u2235 7, a and b are integers \u2234 a\/7b is a rational number]<\/p>\n<p>This contradicts the fact that \u221a5 is irrational. So, our assumption is incorrect.<\/p>\n<p>Hence, 7\u221a5 is an irrational number.<\/p>\n<p><strong>(iii) 6 + \u221a2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume, on the contrary, that 6+\u221a2 is a rational number. Then, there exist coprime positive integers a and b such that<\/p>\n<p>6 + \u221a2 = a\/b<\/p>\n<p>\u21d2 \u221a2 = a\/b \u2013 6<\/p>\n<p>\u21d2 \u221a2 = (a \u2013 6b)\/b<\/p>\n<p>\u21d2 \u221a2 is rational [\u2235 a and b are integers \u2234 (a-6b)\/b is a rational number]<\/p>\n<p>This contradicts the fact that \u221a2 is irrational. So, our assumption is incorrect.<\/p>\n<p>Hence, 6 + \u221a2 is an irrational number.<\/p>\n<p><strong>(iv)<\/strong>&nbsp;<strong>3 \u2212 \u221a5<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume, on the contrary, that 3-\u221a5 is a rational number. Then, there exist coprime positive integers a and b such that<\/p>\n<p>3-\u221a5 = a\/b<\/p>\n<p>\u21d2 \u221a5 = a\/b + 3<\/p>\n<p>\u21d2 \u221a5 = (a + 3b)\/b<\/p>\n<p>\u21d2 \u221a5 is rational [\u2235 a and b are integers \u2234 (a+3b)\/b is a rational number]<\/p>\n<p>This contradicts the fact that \u221a5 is irrational. So, our assumption is incorrect.<\/p>\n<p>Hence, 3-\u221a5 is an irrational number.<\/p>\n<p><strong>2. Prove that the following numbers are irrationals.<\/strong><\/p>\n<p><strong>(i) 2\/\u221a7<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume, on the contrary, that 2\/\u221a7 is a rational number. Then, there exist coprime positive integers a and b such that<\/p>\n<p>2\/\u221a7 = a\/b<\/p>\n<p>\u21d2 \u221a7 = 2b\/a<\/p>\n<p>\u21d2 \u221a7 is rational [\u2235 2, a and b are integers \u2234 2b\/a is a rational number]<\/p>\n<p>This contradicts the fact that \u221a7 is irrational. So, our assumption is incorrect.<\/p>\n<p>Hence, 2\/\u221a7 is an irrational number.<\/p>\n<p><strong>(ii) 3\/(2\u221a5)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume, on the contrary, that 3\/(2\u221a5) is a rational number. Then, there exist coprime positive integers a and b such that<\/p>\n<p>3\/(2\u221a5) = a\/b<\/p>\n<p>\u21d2 \u221a5 = 3b\/2a<\/p>\n<p>\u21d2 \u221a5 is rational [\u2235 3, 2, a and b are integers \u2234 3b\/2a is a rational number]<\/p>\n<p>This contradicts the fact that \u221a5 is irrational. So, our assumption is incorrect.<\/p>\n<p>Hence, 3\/(2\u221a5) is an irrational number.<\/p>\n<p><strong>(iii) 4 + \u221a2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume, on the contrary, that 4 + \u221a2 is a rational number. Then, there exist coprime positive integers a and b such that<\/p>\n<p>4 + \u221a2 = a\/b<\/p>\n<p>\u21d2 \u221a2 = a\/b \u2013 4<\/p>\n<p>\u21d2 \u221a2 = (a \u2013 4b)\/b<\/p>\n<p>\u21d2 \u221a2 is rational [\u2235 a and b are integers \u2234 (a \u2013 4b)\/b is a rational number]<\/p>\n<p>This contradicts the fact that \u221a2 is irrational. So, our assumption is incorrect.<\/p>\n<p>Hence, 4 + \u221a2 is an irrational number.<\/p>\n<p><strong>(iv) 5\u221a2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume, on the contrary, that 5\u221a2 is a rational number. Then, there exist positive integers a and b such that<\/p>\n<p>5\u221a2 = a\/b, where a and b are co-primes<\/p>\n<p>\u21d2 \u221a2 = a\/5b<\/p>\n<p>\u21d2 \u221a2 is rational [\u2235 a and b are integers \u2234 a\/5b is a rational number]<\/p>\n<p>This contradicts the fact that \u221a2 is irrational. So, our assumption is incorrect.<\/p>\n<p>Hence, 5\u221a2 is an irrational number.<\/p>\n<p><strong>3. Show that&nbsp;2 \u2212 \u221a3&nbsp;is an irrational number.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume, on the contrary, that 2 \u2013 \u221a3 is a rational number. Then, there exist coprime positive integers a and b such that<\/p>\n<p>2 \u2013 \u221a3= a\/b<\/p>\n<p>\u21d2 \u221a3 = 2 \u2013 a\/b<\/p>\n<p>\u21d2 \u221a3 = (2b \u2013 a)\/b<\/p>\n<p>\u21d2 \u221a3 is rational [\u2235 a and b are integers \u2234 (2b \u2013 a)\/b is a rational number]<\/p>\n<p>This contradicts the fact that \u221a3 is irrational. So, our assumption is incorrect.<\/p>\n<p>Hence, 2 \u2013 \u221a3 is an irrational number.<\/p>\n<p><strong>4. Show that&nbsp;3 + \u221a2&nbsp;is an irrational number<\/strong>.<\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume, on the contrary, that 3 + \u221a2 is a rational number. Then, there exist coprime positive integers a and b such that<\/p>\n<p>3 + \u221a2= a\/b<\/p>\n<p>\u21d2 \u221a2 = a\/b \u2013 3<\/p>\n<p>\u21d2 \u221a2 = (a \u2013 3b)\/b<\/p>\n<p>\u21d2 \u221a2 is rational [\u2235 a and b are integers \u2234 (a \u2013 3b)\/b is a rational number]<\/p>\n<p>This contradicts the fact that \u221a2 is irrational. So, our assumption is incorrect.<\/p>\n<p>Hence, 3 + \u221a2 is an irrational number.<\/p>\n<p><strong>5. Prove that&nbsp;4 \u2212 5\u221a2&nbsp;is an irrational number.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume, on the contrary, that 4 \u2013 5\u221a2 is a rational number. Then, there exist coprime positive integers a and b such that<\/p>\n<p>4 \u2013 5\u221a2 = a\/b<\/p>\n<p>\u21d2 5\u221a2 = 4 \u2013 a\/b<\/p>\n<p>\u21d2 \u221a2 = (4b \u2013 a)\/(5b)<\/p>\n<p>\u21d2 \u221a2 is rational [\u2235 5, a and b are integers \u2234 (4b \u2013 a)\/5b is a rational number]<\/p>\n<p>This contradicts the fact that \u221a2 is irrational. So, our assumption is incorrect.<\/p>\n<p>Hence, 4 \u2013 5\u221a2 is an irrational number.<\/p>\n<p><strong>6. Show that&nbsp;5 \u2212 2\u221a3&nbsp;is an irrational number.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume, on the contrary, that 5 \u2013 2\u221a3 is a rational number. Then, there exist coprime positive integers a and b such that<\/p>\n<p>5 \u2013 2\u221a3 = a\/b<\/p>\n<p>\u21d2 2\u221a3 = 5 \u2013 a\/b<\/p>\n<p>\u21d2 \u221a3 = (5b \u2013 a)\/(2b)<\/p>\n<p>\u21d2 \u221a3 is rational [\u2235 2, a and b are integers \u2234 (5b \u2013 a)\/2b is a rational number]<\/p>\n<p>This contradicts the fact that \u221a3 is irrational. So, our assumption is incorrect.<\/p>\n<p>Hence, 5 \u2013 2\u221a3 is an irrational number.<\/p>\n<p><strong>7. Prove that&nbsp;2\u221a3 \u2212 1&nbsp;is an irrational number.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume, on the contrary, that 2\u221a3 \u2013 1 is a rational number. Then, there exist coprime positive integers a and b such that<\/p>\n<p>2\u221a3 \u2013 1 = a\/b<\/p>\n<p>\u21d2 2\u221a3 = a\/b + 1<\/p>\n<p>\u21d2 \u221a3 = (a + b)\/(2b)<\/p>\n<p>\u21d2 \u221a3 is rational [\u2235 2, a and b are integers \u2234 (a + b)\/2b is a rational number]<\/p>\n<p>This contradicts the fact that \u221a3 is irrational. So, our assumption is incorrect.<\/p>\n<p>Hence, 2\u221a3 \u2013 1 is an irrational number.<\/p>\n<p><strong>8. Prove that&nbsp;2 \u2212 3\u221a5&nbsp;is an irrational number.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume, on the contrary, that 2 \u2013 3\u221a5 is a rational number. Then, there exist co-prime positive integers a and b such that<\/p>\n<p>2 \u2013 3\u221a5 = a\/b<\/p>\n<p>\u21d2 3\u221a5 = 2 \u2013 a\/b<\/p>\n<p>\u21d2 \u221a5 = (2b \u2013 a)\/(3b)<\/p>\n<p>\u21d2 \u221a5 is rational [\u2235 3, a and b are integers \u2234 (2b \u2013 a)\/3b is a rational number]<\/p>\n<p>This contradicts the fact that \u221a5 is irrational. So, our assumption is incorrect.<\/p>\n<p>Hence, 2 \u2013 3\u221a5 is an irrational number.<\/p>\n<p><strong>9. Prove that&nbsp;<\/strong>\u221a5 + \u221a3<strong>&nbsp;is irrational.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume, on the contrary, that \u221a5 + \u221a3 is a rational number. Then, there exist coprime positive integers a and b such that<\/p>\n<p>\u221a5 + \u221a3 = a\/b<\/p>\n<p>\u21d2 \u221a5 = (a\/b) \u2013 \u221a3<\/p>\n<p>\u21d2 (\u221a5)<sup>2<\/sup>&nbsp;= ((a\/b) \u2013 \u221a3)<sup>2<\/sup>&nbsp;[Squaring on both sides]<\/p>\n<p>\u21d2 5 = (a<sup>2<\/sup>\/b<sup>2<\/sup>) + 3 \u2013 (2\u221a3a\/b)<\/p>\n<p>\u21d2 (a<sup>2<\/sup>\/b<sup>2<\/sup>) \u2013 2 = (2\u221a3a\/b)<\/p>\n<p>\u21d2 (a\/b) \u2013 (2b\/a) = 2\u221a3<\/p>\n<p>\u21d2 (a<sup>2<\/sup>&nbsp;\u2013 2b<sup>2<\/sup>)\/2ab = \u221a3<\/p>\n<p>\u21d2 \u221a3 is rational [\u2235 a and b are integers \u2234 (a<sup>2<\/sup>&nbsp;\u2013 2b<sup>2<\/sup>)\/2ab is rational]<\/p>\n<p>This contradicts the fact that \u221a3 is irrational. So, our assumption is incorrect.<\/p>\n<p>Hence, \u221a5 + \u221a3 is an irrational number.<\/p>\n<p><strong>10. Prove that&nbsp;\u221a2 + \u221a3&nbsp;is irrational.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume, on the contrary, that \u221a2 + \u221a3 is a rational number. Then, there exist coprime positive integers a and b such that<\/p>\n<p>\u221a2 + \u221a3 = a\/b<\/p>\n<p>\u21d2 \u221a2 = (a\/b) \u2013 \u221a3<\/p>\n<p>\u21d2 (\u221a2)<sup>2<\/sup>&nbsp;= ((a\/b) \u2013 \u221a3)<sup>2<\/sup>&nbsp;[Squaring on both sides]<\/p>\n<p>\u21d2 2 = (a<sup>2<\/sup>\/b<sup>2<\/sup>) + 3 \u2013 (2\u221a3a\/b)<\/p>\n<p>\u21d2 (a<sup>2<\/sup>\/b<sup>2<\/sup>) + 1 = (2\u221a3a\/b)<\/p>\n<p>\u21d2 (a\/b) + (b\/a) = 2\u221a3<\/p>\n<p>\u21d2 (a<sup>2<\/sup>&nbsp;+ b<sup>2<\/sup>)\/2ab = \u221a3<\/p>\n<p>\u21d2 \u221a3 is rational [\u2235 a and b are integers \u2234 (a<sup>2<\/sup>&nbsp;+ 2b<sup>2<\/sup>)\/2ab is rational]<\/p>\n<p>This contradicts the fact that \u221a3 is irrational. So, our assumption is incorrect.<\/p>\n<p>Hence, \u221a2 + \u221a3 is an irrational number.<\/p>\n<p><strong>11.&nbsp;Prove that for any prime positive integer p,&nbsp;<\/strong>\u221ap<strong>&nbsp;is an irrational number.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Consider \u221ap as a rational number<\/p>\n<p>Assume \u221ap = a\/b where a and b are integers and b \u2260 0<\/p>\n<p>By squaring on both sides<\/p>\n<p>p = a<sup>2<\/sup>\/b<sup>2<\/sup><\/p>\n<p>pb = a<sup>2<\/sup>\/b<\/p>\n<p>p and b are integers pb= a<sup>2<\/sup>\/b will also be an integer<\/p>\n<p>But we know that a<sup>2<\/sup>\/b is a rational number, so our supposition is wrong<\/p>\n<p>Therefore, \u221ap is an irrational number.<\/p>\n<p><strong>12. If p and q are prime positive integers, prove that&nbsp;<\/strong>\u221ap + \u221aq<strong>&nbsp;is an irrational number.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume, on the contrary, that \u221ap + \u221aq is a rational number. Then, there exist coprime positive integers a and b such that<\/p>\n<p>\u221ap + \u221aq = a\/b<\/p>\n<p>\u21d2 \u221ap = (a\/b) \u2013 \u221aq<\/p>\n<p>\u21d2 (\u221ap)<sup>2<\/sup>&nbsp;= ((a\/b) \u2013 \u221aq)<sup>2<\/sup>&nbsp;[Squaring on both sides]<\/p>\n<p>\u21d2 p = (a<sup>2<\/sup>\/b<sup>2<\/sup>) + q \u2013 (2\u221aq a\/b)<\/p>\n<p>\u21d2 (a<sup>2<\/sup>\/b<sup>2<\/sup>) \u2013 (p+q) = (2\u221aq a\/b)<\/p>\n<p>\u21d2 (a\/b) \u2013 ((p+q)b\/a) = 2\u221aq<\/p>\n<p>\u21d2 (a<sup>2<\/sup>&nbsp;\u2013 b<sup>2<\/sup>(p+q))\/2ab = \u221aq<\/p>\n<p>\u21d2 \u221aq is rational [\u2235 a and b are integers \u2234 (a<sup>2<\/sup>&nbsp;\u2013 b<sup>2<\/sup>(p+q))\/2ab is rational]<\/p>\n<p>This contradicts the fact that \u221aq is irrational. So, our assumption is incorrect.<\/p>\n<p>Hence, \u221ap + \u221aq is an irrational number.<\/p>\n<p><br>We have provided complete details of RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a> Class 10, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-1-exercise-15\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5<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-1631019496476\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-download-rd-sharma-class-10-solutions-chapter-1-exercise-15-free-pdf\"><\/span>Where can I download RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5 free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631019656980\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-it-required-to-remember-all-of-the-questions-in-chapter-1-exercise-15-of-rd-sharma-solutions-for-class-10-maths\"><\/span>Is it required to remember all of the questions in Chapter 1 Exercise 1.5 of RD Sharma Solutions for Class 10 Maths?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, all of the questions in RD Sharma Solutions for Class 10 Maths Chapter 1 Exercise 1.5 must be learned. These questions may appear on both board exams and class tests. Students will be prepared for their board exams if they learn these questions.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631019784714\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-1-exercise-15\"><\/span>What are the benefits of using RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>1. Correct answers according to the last CBSE guidelines and syllabus.<br \/>2. The solutions are written in simple language to assist students in their board examination, &amp; competitive examination preparation.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1681197469070\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-rd-sharma-class-10-solutions-chapter-1-exercise-15-pdf\"><\/span>How much does it cost to download RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5 PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5 PDF for free.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5:&nbsp;Numbers that cannot be stated in the usual form of p\/q are referred to as irrational numbers. This exercise explains how to prove the irrationality of numbers. Kopykitab&#8217;s subject experts have prepared the RD Sharma Solutions Class 10 to help students understand how to do workout problems &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-1-exercise-1-5\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":125048,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3243,9206,73520,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125039"}],"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=125039"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125039\/revisions"}],"predecessor-version":[{"id":501549,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125039\/revisions\/501549"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125048"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125039"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125039"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125039"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}