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Verify by the method of contradiction that √2 is irrational 

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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 = 2k

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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