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Chapter 14 Class 11 Mathematical Reasoning
Serial order wise

Example 15 - Verify by contradiction: root 7 is irrational - Examples

Example 15 - Chapter 14 Class 11 Mathematical Reasoning - Part 2
Example 15 - Chapter 14 Class 11 Mathematical Reasoning - Part 3 Example 15 - Chapter 14 Class 11 Mathematical Reasoning - Part 4

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Transcript

Example 15 Verify by the method of contradiction. p: ﷐﷮7﷯is 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 ﷐𝑎﷮𝑏﷯ ( a & b are prove) Then there exist a, & b∈ Z Such that ﷐﷮7﷯ = ﷐𝑎﷮𝑏﷯ ( where a & b have no common factor) Squaring both sides (﷐﷮7﷯)2 = ﷐﷐𝑎﷮𝑏﷯﷯2 7 = ﷐𝑎2﷮𝑏2﷯ 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 = ﷐49﷮7﷯ 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 that﷐﷮7﷯ is rational Hence ﷐﷮7﷯ is irrational is true.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.