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Prove that√6 is irrational |
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Answer» Let √6 be a rational number , then √6 = p÷q , where p,q are integers , q not = 0 and p,q have no common factors ( except 1 )=> 6 = p² ÷ q² => p² = 2q² ................(i)As 2 divides 6q² , so 2 divides p² but 2 is a prime number > 2 divides p.....Let p = 2m , where m is an integer .......Substituting this value of p in (i) , we get ,......(2m)² = 6q² .......=> 2m² = 3q² .......As 2 divides 2m² , 2 divides 3q² .......=> 2 divides 3 or 2 divide q² ......But 2 does not divide 3 , therefore , 2 divides q²........=> 2 divides q .......Thus , p and q have a common factor 2 . This contradicts that p and q have no common factors ( except 1 ).......Hence , our supposition is wrong . Therefore , √6 is an irrational number.... Square root hai cube root hai 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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