ISC (Arts) Class 11CISCE

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Validation by Contradiction

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Here to check whether a statement p is true, we assume that p is not true i.e. ∼p is true. Then, we arrive at some result which contradicts our assumption. Therefore, we conclude that p is true.

Prove that : `sqrt 7` is irrational .
 In this method, we assume that the given statement is false. 
That is we assume that ` sqrt 7` is rational. 
This means that there exists positive integers a and b such that `sqrt 7 = a/b` , where a and b have no common factors. 
Squaring the equation , we get
`7 = a^2 / b^2 => a^2 = 7b^2 => 7` divides a. Therefore, there exists an integer c such  that a = 7c. 
Then  `a^2 = 49c^2 ` and  `a^2 = 7b^2` Hence, `7b^2 = 49c^2 ⇒ b^2 = 7c^2 `⇒ 7 divides b. 
But we have already shown that 7 divides a. This implies that 7 is a common factor of both of a and b which contradicts our earlier assumption that a and b have no common factors. This shows that the assumption `sqrt 7` is rational is wrong. Hence, the statement`sqrt 7` is irrational is true. | Validation by contradiction

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