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For an integer raised to a rational number to give another integer, what quality must the rational possess?

What role does the prime factorization of the integer play?

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  • Anonymous
    9 years ago
    Favorite Answer

    All variables are assumed to be integers

    If n^(a/b) = x is an integer, and a/b are in lowest terms, then

    n^(1/b) must also be an integer, meaning n is a perfect bth power.

    If it's not, then n^(1/b) would be an irrational bth root, and raising it to the ath power would still be irrational, since a and b are relatively prime.

    There are some special cases to consider also. E.g., a/b might be an integer. You could have m/1 or 1/1, or -1/1.

    If b = 1, then n is always a "1st root" so that's not a problem, and n^(m/1) will always be an integer.

    If a/b = -1, n^-1 = 1/n, and n must be 1 or -1. This is the only case where a/b can be < 0.

    The requirements on n apply to each prime factor of n separately. A formal proof of all this would probably work with each prime factor separately, then apply them to all the factors generally.

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