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Verify by the method of contradiction that “√7 is irrational.” |
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Answer» Prove that √7 is irrational Assume √7. is an rational ∴ √7 = p/q Squaring on b.s. ∴ 7 = p2/q2 ∴ 7q2 = p2 i.e., p in a multiple of again put p = 7k (k+1) : ∴ 7q2 = (7k)2 ⇒ 7q2 = 49k2 ⇒ q2 = 7k2 ⇒ q is also a multiple of q ∴ both “p’ and ‘q’ are the factor of 7 ∴ √7 ≠ p/q (q = 0) ∴ our assumption that √7 is an partial is wrong. ∴ √7 is an irrational. |
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