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

**hatsoff** · October 14th 2009, 07:49 AM

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!

**Bingk** · October 14th 2009, 08:17 AM

Originally Posted by hatsoff

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

**hatsoff** · October 14th 2009, 08:18 AM

Originally Posted by Bingk

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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