Will every n t h root of 2 be an irrational number? If yes, how can I prove that?

Question

Will every       n          t      h       root of   2 be an irrational number? If yes, how can I prove that?

stormsteghj · Accepted Answer

Yes. In fact, for every integer   k and every   n  &amp;gt;  1, the   n root of   k is either an integer or irrational.One way to prove it is to use exactly the same idea as for proving the square root of   2 is irrational: suppose       k    n    =      p    q  , with   p and   q integers, relatively prime. Then       q    n    k  =      p    n  . Now think about the prime factorizations: every prime that divides   q must divide   p, but p and q are relatively prime, so   q  =  1. That means that you must have   k  =      p    n   with   p an integer. That is, the only way for the   n root of   k to be a rational is if   k is an nth power of an integer.Or you can use the Rational Root Test: an   n root of   k is a root of the polynomial       x    n    −  k. But a rational root of a polynomial with integer coefficients that is written in lowest term       p    q   must have denominator   q that divides the leading coefficient and numerator   q that divides the constant coefficient. So any rational root of       x    n    −  k must be an integer.Getting this back to your question, since   2 is not an   n power of an integer for any   n  &amp;gt;  1,       2    n   is not a rational for any integer   n  &amp;gt;  1.

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