A question concerning multiplicative function
Posted: Tue May 18, 2021 8:40 pm
The is a Mobius function in number theory which is stated as,
\[ \mu (n)= \begin{cases} 1 & \text{if n=1} \\ 0 & \text{if } p^2 | n \text{ for some prime } p>1 \\ (-1)^k & \text{if } n=p_1 \cdots p_k \text{ where } p_1,\cdots p_k \text{ are distinct primes} \\ \end{cases} \]
And it is also said that the mobius function is multiplicative meaning $\mu (mn)=\mu (m) \mu (n)$. But if we put $m=3$ and $n=2\times 3$ we get,
$\mu (m)=-1$
$\mu (n)=1$
$\mu (m) \mu (n) =-1$
But,
$\mu (mn)= \mu(18) = 0 \neq \mu (m) \mu (n)$.
I am not understanding why???
Proof that the Mobius function is multiplicative,
\[ \mu (n)= \begin{cases} 1 & \text{if n=1} \\ 0 & \text{if } p^2 | n \text{ for some prime } p>1 \\ (-1)^k & \text{if } n=p_1 \cdots p_k \text{ where } p_1,\cdots p_k \text{ are distinct primes} \\ \end{cases} \]
And it is also said that the mobius function is multiplicative meaning $\mu (mn)=\mu (m) \mu (n)$. But if we put $m=3$ and $n=2\times 3$ we get,
$\mu (m)=-1$
$\mu (n)=1$
$\mu (m) \mu (n) =-1$
But,
$\mu (mn)= \mu(18) = 0 \neq \mu (m) \mu (n)$.
I am not understanding why???
Proof that the Mobius function is multiplicative,