# Math Help - Rudin's analysis

1. ## Rudin's analysis

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).

2. ## Re: Rudin's analysis

i've done that part.

3. ## Re: Rudin's analysis

Originally Posted by Duke
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

4. ## Re: Rudin's analysis

Yes I realised last night that the rule (x^a)^b=x^ab is using x^(a+b)=(x^a)(x^b). Please keep helping.

mq=np -> x^np =b^pm=y^mq. But surely this is still using the above rule. I am still using circular logic.

Thanks

5. ## Re: Rudin's analysis

Originally Posted by Duke
mq=np -> x^np =b^pm=y^mq. But surely this is still using the above rule. I am still using circular logic.
Thanks
No. That is not circular logic. I am assuming that you have proved the fact: Whenever m, n are integers, (b^m)(b^n) = b^(m+n); (b^m)^n = b^(mn) and you are using that fact in the above step.

6. ## Re: Rudin's analysis

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

7. ## Re: Rudin's analysis

OK you're right. The question changes then lol. How do I prove (x^a)^b =x^ab

8. ## Re: Rudin's analysis

Originally Posted by Duke
OK you're right. The question changes then lol. How do I prove (x^a)^b =x^ab
The case $(b^m)^n = b^{mn}$ 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 $r = \frac{m}{n}, n > 0$ the definition of $b^r = (b^m)^{\frac1{n}}$.
[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 $b^r$ is well defined!]

Now to prove that for two rationals $r,s, b^r.b^s = b^{r+s}$, we have to assume $r = \frac{m}{n}, s = \frac{p}{q}, n,q > 0$ and show that
$(b^m)^{\frac1{n}}.(b^p)^{\frac1{q}} = (b^{mq + np})^{\frac1{nq}}$

So to prove the above statement define x and y so that:
$x = (b^m)^{\frac1{n}}, y = (b^p)^{\frac1{q}} z = (b^{mq + np})^{\frac1{nq}} \implies x^n = b^m, y^q = b^p, z^{nq} = b^{mq + np}$

Claim: $xy =z$
Proof:
Notice that $(xy)^{nq} = (x^n) ^q (y^q)^n = (b^m)^q (b^p)^n = b^{mq}b^{np}= b^{mq + np} = z^{nq}$
Thus $(xy)^{nq} = z^{nq}$. Since $nq$ 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 $xy = z$