Exercise 2.11

Proof. Let’s verify that $\mu$ is a multiplicative function.

If $n\wedge m=1$, then $n=p_1^{a_1}\cdots p_l^{a_l},m=q_1^{b_1}\cdots q_r^{b_r}$, where $p_1,\dots ,p_l,q_1,\dots q_r$ are distinct primes. Then the decomposition in prime factors of $\mathit{nm}$ is $\mathit{nm}=p_1^{a_1}\cdots p_l^{a_l}{q}_1^{b_1}\cdots q_r^{b_r}$. If one of the $a_i$ or one of the $b_j$ is greater than 1, then $\mu (\mathit{nm})=0=\mu (n)\mu (m)$. Otherwise, $n=p_1\cdots p_l,m=q_1\cdots q_r,\mathit{nm}=p_1\cdots p_l{q}_1\cdots q_r$, and $\mu (\mathit{nm})={(-1)}^{l+r}={(-1)}^l{(-1)}^r=\mu (n)\mu (m)$. So

that is, $n\mapsto \frac{\mu (n)}{n}$ is a multiplicative function.

From Ex.2.10, $n\mapsto {\sum <!--\; FUNCTION\; APPLICATION\; -->}_{d\mid n}\frac{\mu (d)}{d}$ is also a multiplicative function, and so is $\psi$, where $\psi$ is defined by

To verify the equality $\varphi =\psi$, it is sufficient from Ex. 2.9 to verify $\varphi (p^k)=\psi (p^k)$ for all prime powers $p^k,k\ge 1$ ( $\varphi (1)=\psi (1)=1$).

(The other terms are null.)

So

Thus $\varphi =\psi$ : for all $n\ge 1$,

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