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If I am not wrong, rational powers of rational numbers can be factorized in an unique way as product of rational powers of different prime numbers:

  • $10^{1/2} = 2^{1/2} \cdot 5^{1/2}$
  • $(8/9)^{1/6} = 2^{1/2} \cdot 3^{-1/3}$
  • $\sqrt{6}/2 = (3/2)^{1/2} = 2^{-1/2} \cdot 3^{1/2}$

But such factorizations were removed from Wikipedia.

I'm almost sure somebody has already written about it. So I'd like to ask for a reference I will be able to use as source on Wikipedia.

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Could you explain a little more your factorization? What exactly do you mean by unique? – user37238 Jul 21 '14 at 9:59
There is one way to write them as product of rational powers of prime numbers. What is not clear? – BartekChom Jul 21 '14 at 10:06
It follows so quickly from unique factorization for integers that I wouldn't be surprised if no one wrote it down anywhere (not counting Wikipedia). – Gerry Myerson Jul 21 '14 at 10:14
Such factorizations are not unique, e.g. $\,2^{5/6} = 2^{1/2}\, 2^{1/3}\ \ $ – Bill Dubuque Jul 21 '14 at 13:51
Of course I'm talking about powers of different prime numbers. – BartekChom Jul 21 '14 at 18:51

If a real number can be written as the product of natural numbers raised to rational exponents, there is indeed a unique way to express it as a product of distinct primes raised to rational exponents. This follows from the unique factorization theorem for integers (as has been said in the comments) through elementary if slightly messy arguments. So I think the person who removed your "prime factorizations" from the pitch intervals page, on the ground that "(prime) factors not meaningful for non-rational intervals", is being pedantic. Whether or not the removed expressions (which can be easily derived from the equivalent ones that were left in the adjacent boxes) were likely to be useful, I don't really know, not being a musician.

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Prime number theorem. This is not what you mean. – Andrés E. Caicedo Jul 21 '14 at 19:10
of course - fixed - thanks. – Silvio Levy Jul 21 '14 at 19:17

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