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Verify by the method of contradiction that √2 is irrational |
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Answer» If possible Let √2 is rational ∴ √2 = p/q, p, q ∈ z, q ≠ 0 We assume that p and q do not have any common factor p = √2q ⇒ p = 2q2 p2 is a multiple of 2 ⇒ ∴ p is a multiple of 2 ∴ p = 2k, k ∈ z ⇒ p2 = 4k 2q2 = 4k2 ⇒ q2 = 2k2 q2 = 2k2 q2 is a multiple of 2 ⇒ ∴ q is a multiple of z ∴ p and q are both multiple of 2 and hence has a common factor z which is a contradiction ∴ our assumption is wron ∴ √2 is irrational |
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