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I have some expressions. For example:

 expr=f[x]^(7/3) + 2*f[x]^(2 + 2/(3*n)) + f[x]

where n is an integer.

I want to get the terms containing rational exponents to zero. For the example above; I want to reach f[x].

My try is

expr /. f[x]^Rational[_, _] -> 0

or

Assuming[expr /. f[x]^Rational[_, _] -> 0, n \[Element] Integers]
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2 Answers 2

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PossibleIntegerQ[a_, n] := 
  Resolve[Exists[n, Element[n, Integers], Element[a, Integers]]]

expr /. f[x]^a_. /; ! PossibleIntegerQ[a, n] -> 0
(*    f[x]    *)

The combination of Resolve and Exists determines for each term $f(x)^a$ whether there exists an integer $n$ that makes the exponent $a(n)$ integer-valued (i.e., whether or not $a$ is "possibly an integer"). If not, the term is set to zero.

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Rational numbers are atomic (i.e., cannot be divided into subexpressions)

AtomQ[3/4]

(* True *)

Consequently, rationals have no parts (although Numerator and Denominator act as if rationals had parts), and the pattern Rational[_, _] never occurs

expr = f[x]^(7/3) + 2*f[x]^(2 + 2/(3*n)) + f[x];

expr /. f[x]^(_?(! FreeQ[#, Rational] &)) :> 0

(* f[x] *)
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