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Math Help - [SOLVED] I do not understand one of the steps in this Dirichlet product proof.

  1. #1
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    [SOLVED] I do not understand one of the steps in this Dirichlet product proof.

    A note on notation:

    If f,g are arithmetic functions, then f*g is the Dirichlet product thereof, such that (f*g)(n)=\sum_{d|n}f(d)g\left(\frac{n}{d}\right).

    Theorem: If both g and f*g are multiplicative, then f is multiplicative.

    Proof: We shall prove by contradiction. Suppose f is not multiplicative. Let h=f*g. Since f is not multiplicative, there exist m,n,(m,n)=1 such that f(mn)\neq f(m)f(n). We choose mn as small as possible. If mn=1, then f(1)\neq f(1)f(1) so f(1)\neq 1. Since h(1)=f(1)g(1)=f(1)\neq1, h is not multiplicative, a contradiction. If mn>1, we have f(ab)=f(a)f(b) for all ab<mn and (a,b)=1. Now,

    h(mn)=f(mn)g(1)+\sum_{a|m,b|n,ab<mn}f(ab)g\left(\f  rac{mn}{ab}\right)

    =f(mn)+\sum_{a|m,b|n,ab<mn}f(a)f(b)g\left(\frac{m}  {a}\right)g\left(\frac{n}{b}\right)

    =f(mn)-f(m)f(n)+h(m)h(n).

    Since f(mn)\neq f(m)f(n), h(mn)\neq h(m)h(n). Therefore h is not multiplicative, a contradiction. \blacksquare

    Okay, here's my problem with this "proof": Why in the world do we have " f(ab)=f(a)f(b) for all ab<mn and (a,b)=1" if we are assuming that f is not multiplicative? As far as I can see, such a statement might be true or false for some f(mn)\neq f(m)f(n). Yet the author of the proof takes it to be true. Why?

    Thanks!
    Last edited by hatsoff; October 14th 2009 at 09:18 AM.
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  2. #2
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    Quote Originally Posted by hatsoff View Post
    theorem: if both f and f*g are multiplicative, then f is multiplicative.
    Is that right? Cuz if f is multiplicative, then obivously f is multiplicative, right?

    As for your question, the reason is if it's not multiplicative, it means that there exists at least one case where f(mn) \neq f(m)f(n). It doesn't mean this is true for all mn. The proof selects the smallest mn where this happens, meaning that before this, f(ab) = f(a)f(b) works. That's why they test when mn=1 and when mn>1. That's why the author did that ... at least that's how I've interpreted it ... I could be wrong
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  3. #3
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    Quote Originally Posted by Bingk View Post
    Is that right? Cuz if f is multiplicative, then obivously f is multiplicative, right?

    As for your question, the reason is if it's not multiplicative, it means that there exists at least one case where f(mn) \neq f(m)f(n). It doesn't mean this is true for all mn. The proof selects the smallest mn where this happens, meaning that before this, f(ab) = f(a)f(b) works. That's why they test when mn=1 and when mn>1. That's why the author did that ... at least that's how I've interpreted it ... I could be wrong
    Ah, okay. I'm surprised I missed that, yes.

    Thanks!
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