Results 1 to 8 of 8

Math Help - Rudin's analysis

  1. #1
    Member
    Joined
    May 2011
    Posts
    169

    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).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    May 2011
    Posts
    169

    Re: Rudin's analysis

    i've done that part.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6

    Re: Rudin's analysis

    Quote Originally Posted by Duke View Post
    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    May 2011
    Posts
    169

    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6

    Re: Rudin's analysis

    Quote Originally Posted by Duke View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2011
    Posts
    169

    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
    Last edited by Duke; July 2nd 2011 at 02:20 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    May 2011
    Posts
    169

    Re: Rudin's analysis

    OK you're right. The question changes then lol. How do I prove (x^a)^b =x^ab
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6

    Re: Rudin's analysis

    Quote Originally Posted by Duke View Post
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Baby Rudin - ch 2, ex 27
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: November 21st 2010, 12:24 PM
  2. Analysis question from Baby Rudin (PMA), Chapter 1
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: September 14th 2010, 12:14 AM
  3. Rudin Analysis book:shrink nicely
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: January 6th 2010, 09:23 AM
  4. Principles of Mathematical Analysis(Rudin)#3
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: September 15th 2009, 03:02 PM
  5. Princple of mathematical analysis, by Walter rudin. Need help....
    Posted in the Differential Geometry Forum
    Replies: 10
    Last Post: March 19th 2009, 09:13 AM

Search Tags


/mathhelpforum @mathhelpforum