# Factorization of rational powers of rational numbers

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.

• 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