Exercise 2.15

Proof. Here $\nu ={\sigma }_0,\sigma ={\sigma }_1$.

(a)

From the Möbius Inversion Theorem, as $\nu (n)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{d\mid n}1={\sum }_{d\mid n}I(d)$, where $I(n)=1$ for all $n\ge 1$, $$1=I(n)=\underset{d\mid n}{\sum }\mu (n∕d)\nu (d).$$

(b)

From the same theorem, as $\sigma (n)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{d\mid n}d={\sum }_{d\mid n}\mathrm{Id}(d)$, where $\mathrm{Id}(n)=n$ for all $n\ge 1$, $$n=\mathrm{Id}(n)=\underset{d\mid n}{\sum }\mu (n∕d)\sigma (d).$$