Prove (b^m)^(1/n) =(b^p)^(1/q) when (m/n)=(p/q) and hence prove b^(r+s)=b^r*b^s where r,s are rationals.
my working. (b^m)^(1/n)=b^(m/n)=b^(p/q)=(b^p)^(1/q).
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 $\displaystyle (b^m)^n = b^{mn}$ 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 $\displaystyle r = \frac{m}{n}, n > 0$ the definition of $\displaystyle b^r = (b^m)^{\frac1{n}}$.
[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 $\displaystyle b^r $ is well defined!]
Now to prove that for two rationals $\displaystyle r,s, b^r.b^s = b^{r+s}$, we have to assume $\displaystyle r = \frac{m}{n}, s = \frac{p}{q}, n,q > 0$ and show that
$\displaystyle (b^m)^{\frac1{n}}.(b^p)^{\frac1{q}} = (b^{mq + np})^{\frac1{nq}}$
So to prove the above statement define x and y so that:
$\displaystyle x = (b^m)^{\frac1{n}}, y = (b^p)^{\frac1{q}} z = (b^{mq + np})^{\frac1{nq}} \implies x^n = b^m, y^q = b^p, z^{nq} = b^{mq + np}$
Claim: $\displaystyle xy =z$
Proof:
Notice that $\displaystyle (xy)^{nq} = (x^n) ^q (y^q)^n = (b^m)^q (b^p)^n = b^{mq}b^{np}= b^{mq + np} = z^{nq}$
Thus $\displaystyle (xy)^{nq} = z^{nq}$. Since $\displaystyle nq$ 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 $\displaystyle xy = z$