Proving Multiplicative Group of Positive Rationals Generated by Primes

futurebird
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From "The Theory of Groups" by Rotman

2.5. Prove that the multiplicative group of positive rationals is generated by all rationals of the form:
\frac{1}{p},
where p is prime.​

... um... no it's not. Right? How can I prove this when I don't even think it is true? I mean, for example take the positive rational number 75. How can I generate that using \frac{1}{p^{n}}, where p is prime, n \in \mathbb{N}? The largest possible value for \frac{1}{p^{n}} is 1/2...

I could see how this would be possible if we had addition as the operation for the generating set... but, then why is it the multiplicative group of positive rationals?

Please help me understand this question.
 
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futurebird said:
for example take the positive rational number 75. How can I generate that using \frac{1}{p^{n}}, where p is prime, n \in \mathbb{N}?
You can't, but why would you restrict yourself to positive exponents?
 
Hurkyl said:
You can't, but why would you restrict yourself to positive exponents?

Thanks!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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