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Taking $\nu_p(n)$ to be the p-adic valuation (returning the highest power of $p$ that divides $n$),

How would I go about proving that this function is completely additive for a fixed prime? I've done proofs for additive and multiplicative functions before but not sure how to go about this one since it has both a $p$ and $n$ element. Maybe I'm just complicating and confusing things in my head?

raznbagel
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  • You have $p$ fixed, so it doesn't matter. You need to show that, for a fixed prime $p$, $f(n):=\nu_p(n)$ is completely additive. – learner Mar 20 '19 at 16:10

1 Answers1

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Write $n_1 = k_1p^{a_1}$ and $n_2 = k_2p^{a_2}$ where the $k_i$ are prime to $p$.

Then $$\nu_p(n_1n_2) = \nu_p(k_1k_2p^{a_1+a_2}) = a_1+a_2 = \nu_p(n_1) + \nu_p(n_2)$$

since $\,p\nmid k_1,k_2\,\Rightarrow\, p\nmid k_1 k_2\,$ by Euclid's Lemma.

Bill Dubuque
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user113102
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