
Problem: Rational or Irrational
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Let
and assume that x is rational,
however it is not an integer. There should exist a minimal positive
integer n such that xn is an integer. Consider
. Since the fractional part of x, ![]()
Note that mis an integer for m=nx-n[x] which is an integer.
Also,
which is also an integer. Due to
the minimality of n, m=0. In other words, x =[x] and is an integer
in contradiction to our assumption.
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