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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!

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    Ehr... break up the fraction?2012-07-03
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    @ArturoMagidin Thank you. so That is all that I am missing correct, everything else seems good?2012-07-03
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    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.2012-07-03
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    @ArturoMagidin I see that now. Thank you again, I really appreciate it.2012-07-03
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    Note that the same argument shows that if $f$ and $g$ are multiplicative, so is $fg$.2012-07-03
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    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.2012-07-03
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    @ArturoMagidin I see! That is good to know :D2012-07-03
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    @AndréNicolas Thanks for sharing that with me. You guys have been really helpful.2012-07-03

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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 $$g(ab) = \frac{f(ab)}{ab} = \frac{f(a)f(b)}{ab} = \frac{f(a)}{a}\frac{f(b)}{b} = g(a)g(b).$$

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    Thank you. I guess I am just missing the $\frac{f(p_{1}^{e_{1}})}{p_{1}^{e_{1}}}...\frac{f(p_{k}^{e_{k}})}{p_{k}^{e_{k}}}=g(p_{1}^{e_{1}})...g(p_{k}^{e_{k}})$.2012-07-03
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    @HowardRoark: Yes; that's *all* you are missing.2012-07-03