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

• Oct 14th 2009, 08:49 AM
hatsoff
[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 and $(a,b)=1$. Now,

$h(mn)=f(mn)g(1)+\sum_{a|m,b|n,ab

$=f(mn)+\sum_{a|m,b|n,ab

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

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 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!
• Oct 14th 2009, 09:17 AM
Bingk
Quote:

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 :)
• Oct 14th 2009, 09:18 AM
hatsoff
Quote:

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!