Determining the irrationality of a quotient

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Determine whether (7)1/2/(15)1/3 is either rational or irrational and prove your answer is correct.

So I know that (7)1/2 is irrational from previous theorems since it is a prime, I also split up (15)1/3 into (5)1/3 times (3)1/3. I previously had shown that (5)1/3 is also irrational. Doing this question I showed that (3)1/3 is irrational by the uniqueness of prime factorization.

But my problem lies in showing that this whole quotient is irrational. I don't know where to start. Maybe assume that the quotient is rational and obtain a contradiction? If so how could I start it?
 
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trap101 said:
Determine whether (7)1/2/(15)1/3 is either rational or irrational and prove your answer is correct.

So I know that (7)1/2 is irrational from previous theorems since it is a prime, I also split up (15)1/3 into (5)1/3 times (3)1/3. I previously had shown that (5)1/3 is also irrational. Doing this question I showed that (3)1/3 is irrational by the uniqueness of prime factorization.

But my problem lies in showing that this whole quotient is irrational. I don't know where to start. Maybe assume that the quotient is rational and obtain a contradiction? If so how could I start it?

Yes, assume it's equal to a rational and assume its in lowest terms. Then take the sixth power of both sides. Then make the usual sort of arguments about prime divisibility.
 
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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