The question asks that if $f(n)$ is multiplicative to prove that $f(n)/n\qquad$ is also multiplicative.
This is what I have:
So, $f(n)\quad$ is multiplicative means that if $p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}\qquad$ 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\qquad $
then $g(n)=f(p_1^{e_1})f(p_2^{e_2})\cdots f(p_k^{e_k})/p_1^{e_1}p_2^{e_2}\cdots p_{k}^{e_{k}}\quad$. Which is by definition multiplicative since $gcd(p_1^{e_1},p_2^{e_2},\ldots,p_k^{e_k})=1\qquad$ 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!
Best Answer
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 $$g(ab) = \frac{f(ab)}{ab} = \frac{f(a)f(b)}{ab} = \frac{f(a)}{a}\frac{f(b)}{b} = g(a)g(b).$$