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Math Help - Multiplicative Proof

  1. #1
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    Multiplicative Proof

    I'm definitely overthinking this...

    Show that if f and g are multiplicative functions, then fg is also multiplicative, where (fg)(n) = f(n)g(n) for every positive integer n.

    Here's what I'm thinking:

    If n is prime, then n = 1*n, and f(1*n) = f(1)f(n), and same for g.... 1 and n are clearly relatively prime, so the definition of multiplicative applies and I can finish it out.

    If n is composite, n =p1p2....pk for k distinct primes.
    f(n) = f(p1)f(p2)f(p3)....etc... and same for g since all the pi's are relatively prime to each other, thus letting the definition for multiplicative work.

    So no matter if n is composite or prime, each time it works.
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  2. #2
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    Quote Originally Posted by Janu42 View Post
    I'm definitely overthinking this...

    Show that if f and g are multiplicative functions, then fg is also multiplicative, where (fg)(n) = f(n)g(n) for every positive integer n.

    Here's what I'm thinking:

    If n is prime, then n = 1*n, and f(1*n) = f(1)f(n), and same for g.... 1 and n are clearly relatively prime, so the definition of multiplicative applies and I can finish it out.

    If n is composite, n =p1p2....pk for k distinct primes.
    f(n) = f(p1)f(p2)f(p3)....etc... and same for g since all the pi's are relatively prime to each other, thus letting the definition for multiplicative work.

    So no matter if n is composite or prime, each time it works.
    Doesn't this work?

    Let a,b be coprime. So f(ab)=f(a)f(b) and g(ab)=g(a)g(b).

    So (fg)(ab) = f(ab)g(ab) = f(a)f(b)g(a)f(b) = f(a)g(a)f(b)g(b) = (fg)(a)(fg)(b).

    We just use multiplicative commutativity of integers.
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  3. #3
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    Quote Originally Posted by Janu42 View Post
    I'm definitely overthinking this...

    Show that if f and g are multiplicative functions, then fg is also multiplicative, where (fg)(n) = f(n)g(n) for every positive integer n.

    Here's what I'm thinking:

    If n is prime, then n = 1*n, and f(1*n) = f(1)f(n), and same for g.... 1 and n are clearly relatively prime, so the definition of multiplicative applies and I can finish it out.

    If n is composite, n =p1p2....pk for k distinct primes.
    f(n) = f(p1)f(p2)f(p3)....etc... and same for g since all the pi's are relatively prime to each other, thus letting the definition for multiplicative work.

    So no matter if n is composite or prime, each time it works.

    I think you're definitely overthinking it:

    (m,n)=1\Longrightarrow (fg)(mn):= f(mn)g(mn)=f(m)f(n)g(m)g(n) = f(m)g(m)f(n)g(n) = (fg)(m)(fg)(n)

    Tonio
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