Show that an n-th root of an integer is either an integer or irrational.
arithmetic/rational.roots New
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Problem
Show that an n-th root of an integer is either an integer or irrational. [ Editor: problem statement reconstructed from the posting, which generalizes the irrationality of the square root of two. ]
Show solution
One easy theorem that generalizes the result about the irrationality of
the square root of two is:
n n-1
If the polynomial x + a x + . . . + a x + a with integer
n-1 1 0
coefficients has a rational root, then the rational root must be an
integer. For let the rational root be p/q with p and q relatively
n
prime. Substituting x=p/q, multiplying by q and rearranging terms
gives
n n-1 n
p = -(a p q + . . . + a q )
n-1 0
Thus q divides p and so q = +/- 1.
[ Editor: the equation shows that q divides p^n; since p and q are
relatively prime, q = +/- 1. ]
n
Applying this to the polynomial x - k we see that an n-th root of an
integer is either an integer (i.e. k is an n-th power) or is
irrational.