# Example 15 - Chapter 14 Class 11 Mathematical Reasoning

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 15 Verify by the method of contradiction. p: 7is irrational p : 7 is irrational. By method of contradiction We assume p is net true i.e. ∼ p is true & we arrive at some result which Contradiction our assumption ,we conclude that p is true We assume that given statement is false i.e. 7 is not irrational. i.e. 7 is rational. if 7 is rational then 7 must be in the from of 𝑎𝑏 ( a & b are prove) Then there exist a, & b∈ Z Such that 7 = 𝑎𝑏 ( where a & b have no common factor) Squaring both sides (7)2 = 𝑎𝑏2 7 = 𝑎2𝑏2 7b2 = a2 ⇒ a2 = 7b2 ⇒ 7 Divide a2 ⇒ Therefore there exist on integer c s.t a = 7C Squaring both side (a)2 = 49C2 From (1) & (2) a2 = 7b2 a2 = 49C2 ⇒ 7b2 = 49C2 ⇒ b2 = 497 C2 ⇒ b2 = 7C2 ⇒ 7 Divide b2 ⇒ 7 Divide b But we already show that 7 Divide a But we already show that 7 Divide a This implies 7 is a common factor of a & b but this contradict our assumption that a & b have no common factor. Hence our assumption is wrong that7 is rational Hence 7 is irrational is true.

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Chapter 14 Class 11 Mathematical Reasoning

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.