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I'm not sure what you would want from such an exponentiation. (n/d)^x = (n^x) / (d^x) where x is an integer ? As a rule, we try to follow GMP's operator set, and as far as I known there is no rational exponentiation in GMP.
For now, we stick to GMP's set of rational operators. Maybe we could add more operators in the future, especially if we want to have the same features as Num's rationals.
I keep this as a possible enhancement, in case anyone is interested.
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