i've done that part.
If by that, you mean your working in the first post, then the proof is not right. In the second equality, you claim b^(m/n) = b^(p/q). But that is precisely what you have to prove! You cannot assume that!
Anyway, the second part of the problem is proved by the same method. Assume r = m/n and s = p/q, then define x and y such that b^m = x^n and b^p = y^q. Now try to continue the proof; I think I have given enough hints
That is what I am trying to prove. That (b^m)^1/n =(b^p)^1/q. This is surely going to be achieved by proving that (b^m)^1/n=b^(m/n). Thanks. Sorry for the confusion. I should also say b>1
The case when m,n are integers is easy to prove (induction?) and that is all we will need.
This time I will be more explicit.
For a rational the definition of .
[Note:Just writing the above definition assumes that you have proved the existence and uniqueness of nth root of a positive real number. My first post in this thread is the outline of the proof that is well defined!]
Now to prove that for two rationals , we have to assume and show that
So to prove the above statement define x and y so that:
Claim:
Proof:
Notice that
Thus . Since is positive and there is a result that says mth root of a positive real number is unique for a natural number 'm'.
We can take nq-th root on both sides and obtain