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I'm studying algebra, and I came across some questions on multiplicative functions (that should be number theory though?). One is: prove that mobius function is multiplicative. But I've not been given any clue about how to do such a proof. Can you help?

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  • $\begingroup$ @Fabian presumably the standard way $$\mu(n)=\begin{cases} 1 & n= 1 \\ 0 & n\text{ is not square free} \\ (-1)^k & n\text{ is square-free with }k\text{ prime factors}\end{cases}.$$ $\endgroup$ Aug 12, 2014 at 18:09

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I intend that you fill in the blanks. Suppose that $(n,m)=1$. If $m$ or $n$ is not squarefree, then $mn$ is not squarefree, and $\mu(mn)=\square$ and either one of $\mu(n)$ or $\mu(m)$ is $\square$, so equality holds. We may assume thus that $m$ and $n$ are squarefree. Then so is $mn$, and since $(m,n)=1$; the prime factors of $m$ and $n$ are $\square$. Thus if $m$ has $k$ (different) prime factors and $n$ has (different) $l$ prime factors, $mn$ has $\square$ (different) prime factors and $\mu(n)\mu(m)=(-1)^k(-1)^l=(-1)^{k+l}=\square$.

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Mobius function is not completely multiplicative but is multiplicative just like totient function i.e. when (m,n) == 1.

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