MonkiezDevil
Theorem : the nth root of an integer
is either an integer or irrational.
Prove the more general result:
Theorem: If x is a root of the polynomial xm + c1 xm-1 + c2 xm-2+ ... + cm = 0,
(with the coefficients c1, c2, ... , cm all integers,
of course), then x is either an integer or irrational.
(Hint: suppose a/b is a root of the polynomial, and a and
b are relatively prime. Plug into the equation above and multiply
both sides by bm. Show that b must equal 1.)

Incidentally, this result naturally leads to the
definition of a quadratic integer .

Please teach me . Thank you very much
fivedoughnut
How do you prove a question?.... only an answer surely!
AlphaNumeric
Let the n'th root of an integer, k, would be r_i, so (r_i)^n = k, for n different r_i.

Therefore you are solving the polynomial x^n = k.

Suppose x = (a/b) where a and b are COPRIME. You have (a/b)^n = k, so a^n = k(b^n).

Consider prime factorisation, there MUST be the same prime factorisaton on both sides of that equation or they aren't equal. Since a and b are coprime, a contains none of the prime factors b does. But that means they can't possibly be equal unless b contains the only factor it could (ie it has no prime factorisation) is 1 (1 is not a prime!)

Therefore b=1 [i]provided[/b] x is rational. Therefore the root must be integer if rational. If it's not an integer (b not 1), then you've just shown that it can't be rational (A -> B has the equivalent logic statement ¬B -> ¬A). Therefore it's irratinal.

Can you extend this method to the second part of the question?
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