Proove that cuberoot 6 is a rational or irrational?

1.	Proove that cuberoot 6 is a rational or irrational?
Answer» Assume cube root 6 is rational. Then let cube root 6 = a/b ( a & b are co-prime and b not = 0)\xa0Cubing both sides : 6=a^3/b^3\xa0a^3 = 6b^3\xa0a^3 = 2(3b^3)\xa0Therefore, 2 divides a^3 or a^2 * a . By Euclid\'s Lemma if a prime number\xa0divides the product of two integers then it must divide one of the two integers\xa0Since all the terms here are the same we conclude that 2 divides a.\xa0Now there exists an integer k such that a=2k\xa0Substituting 2k in the above equation\xa08k^3 = 6b^3\xa0b^3 = 2{(2k^3) / 3)}\xa0Therefore, 2 divides b^3. Using the same logic as above. 2 divides b.\xa0Hence 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.\xa0cube root 6 is irrational

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