Exercise 2.9

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

A function on the integers is said to be multiplicative if $f(ab) =f(a)f(b)$. whenever $(a, b) = 1$. Show that a multiplicative function is completely determined by its value on prime powers.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
Let the decomposition of $n$ in prime factors be $n =p_1^{k_1}\cdots p_t^{k_t}, p_1< \cdots<p_t$. As $p_i^{k_i} \wedge p_j^{k_j}=1$ for $i
e j,\ i,j = 1,\ldots,t$,
$$f(n) =f(p_1^{k_1}\cdots p_t^{k_t})=f(p_1^{k_1})\cdots f(p_t^{k_t})$$
(by induction on the number of prime factors.)

So $f(n)$ is completely determined by its value on prime powers.
\end{proof}

Exercise 2.9

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Exercise 2.9

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