**Example 15**

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 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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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.