# if $f(n)$ is multiplicative prove that $f(n)/n$ is also multiplicative.

The question asks that if $f(n)$ is multiplicative to prove that $f(n)/n$ is also multiplicative.

This is what I have:

So, $f(n)$ is multiplicative means that if $p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ is the prime-power decomposition of $n$, then

$f(n)=f(p_1^{e_1})f(p_2^{e_2})\cdots f(p_k^{e_k})$.

Now I say, let $g(n)=f(n)/n$ then $g(n)=f(p_1^{e_1})f(p_2^{e_2})\cdots f(p_k^{e_k})/p_1^{e_1}p_2^{e_2}\cdots p_{k}^{e_{k}}$. Which is by definition multiplicative since $gcd(p_1^{e_1},p_2^{e_2},\ldots,p_k^{e_k})=1$ and $g(p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}) = f(p_1^{e_1}) f(p_2^{e_2}) \cdots f(p_k^{e_k})/p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$.

I can't think of any other thing to do. Thank you in advance!

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Ehr... break up the fraction? – Arturo Magidin Jul 3 '12 at 5:35
Correct, but too much work. An arithmetical function is multiplicative if and only if whenever $\gcd(a,b)=1$, $g(ab)=g(a)g(b)$. So you don't have to go all the way to prime factorizations. – Arturo Magidin Jul 3 '12 at 5:41
Note that the same argument shows that if $f$ and $g$ are multiplicative, so is $fg$. – André Nicolas Jul 3 '12 at 5:44
And the claim at hand would follow from @AndréNicolas' statement, by letting $g(n) = \frac{1}{n}$, which is easily seen to be not just multiplicative, but completely multiplicative. – Arturo Magidin Jul 3 '12 at 5:46
@AndréNicolas Thanks for sharing that with me. You guys have been really helpful. – HowardRoark Jul 3 '12 at 5:49

You have it! Just break up the fraction to recognize it as the product of values of $g$ at the prime powers.
Of course, you don't have to go all the way to prime factorizations. Let $g(n) = f(n)/n$. We need to show that if $\gcd(a,b)=1$, then $g(ab) = g(a)g(b)$. We know that $f(ab)=f(a)f(b)$. So $$g(ab) = \frac{f(ab)}{ab} = \frac{f(a)f(b)}{ab} = \frac{f(a)}{a}\frac{f(b)}{b} = g(a)g(b).$$
Thank you. I guess I am just missing the $\frac{f(p_{1}^{e_{1}})}{p_{1}^{e_{1}}}...\frac{f(p_{k}^{e_{k}})}{p_{k}^{e_{k}}}‌​=g(p_{1}^{e_{1}})...g(p_{k}^{e_{k}})$. – HowardRoark Jul 3 '12 at 5:39