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mae that V6 is notrational number. |
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Answer» To prove: cubic root of 6 is irrationalSolution:Assume cube root 6 is rational. Then let cube root 6 = a/b here a and b are co prime numbers Cubing both sides : 6=a^3/b^3 a^3 = 6 b^3 a^3 = 2(3 b^3) Therefore, 2 divides a^3 or a^2 * a . By Euclid's Lemma if a prime number divides the product of two integers then it must divide one of the two integers Since all the terms here are the same we conclude that 2 divides a. Now there exists an integer k such that a=2k Substituting 2k in the above equation 8k^3 = 6b^3 b^3 = 2{(2k^3) / 3)} Here 2 divides b^3. thus 2 divides b. Hence 2 is common factor of both a & b. But this is a contradiction of the fact that a & b are co-prime. Therefore, the initial assumption is wrong. |
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