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Proving the nth root of X is irrational
By Anonymous on Wednesday, July 12, 2000 - 09:01 pm:

Does anyone know of a proof for the following?

If the nth root of X (n and X are integers>1) is rational, then X = Yn (where Y is also an integer).

I am aware of the proof that sqrt(2) is irrational, but is there a general case for the nth root of X?

By Dan Goodman (Dfmg2) on Thursday, July 13, 2000 - 03:05 pm:

The appropriate theorem here is:

Theorem If a is a root of a monic polynomial with integer coefficients, then a is either irrational or an integer.

A monic polynomial is a polynomial whose leading coefficient (the coefficient of the highest power of x) is 1. This proves all of these sorts of problems. Specifically, in your case, consider the polynomial xn-X=0. This is a monic polynomial, so the roots must be either irrational or an integer, which is the proof you are looking for.

To prove the theorem above, suppose the solution is p/q in lowest terms, and q>1 (so that it is not an integer). If the polynomial is xn+c1xn-1+...+cn-1x+c n=0. Substitute p/q into this equation, and multiply by qn, to get pn+c1pn-1q+...+cn-1pq n-1+cnqn=0. q divides 0, and q divides c1pn-1q+...+cn-1pqn-1+c nqn, because there is a factor of q in each term. Therefore, q must also divide pn (because if q divides a+b and q divides b, q must also divide a). Therefore q divides p, but this contradicts p/q in lowest terms. Hence the result.

By Anonymous on Thursday, July 13, 2000 - 07:28 pm:

Thanks Dan, that is very helpful.

But is it not also true that if X is an integer then the nth root of X is also equal to Yn (where Y is an integer). So in other words the 5th root of 10 is irrational because 10 is not equal to Y5? Maybe the Theorem you gave demonstrates this and I am missing it, but from what I can tell it only says that the root can be any integer.

Thanks for your help, Joe

By Dan Goodman (Dfmg2) on Friday, July 14, 2000 - 01:52 pm:

Yup, the 5th root of 10 is irrational, because 10 is not equal to Y5 for any integer Y (15=1, 25=32, ...). And since x5-10=0 must have roots that are either integers or irrational, and it cannot have integer roots by the above, so it must have irrational roots. Furthermore, the 5th root of 10 is a root, and so must be irrational! I'm not entirely sure what you don't understand.

By Anonymous on Friday, July 14, 2000 - 02:52 pm:

Sorry, I got it after I posted the last message. I was slightly confused about xn-X=0 and if the second X was the same as the first x, but after working it out it makes sense. Thanks again for your help.