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Prove that $f(n) = 2^{\omega(n)}$ is multiplicative where $\omega(n)$ is the number of distinct primes.

My attempt:

Let $a = p_1p_2\cdots p_k$ and $b = q_1q_2\cdots q_t$ where $p_i$ and $q_j$ are prime factors, and $p_i \neq q_j$ for all $1 \leq i \leq k$ and $1 \leq j \leq t$. We will show that $2^{\omega(ab)} = 2^{\omega(a)} \times 2^{\omega(b)}$

Indeed, $\omega(a) = k$ and $\omega(b) = t$. Then $2^{\omega(a)} \times 2^{\omega(b)} = 2^{k + t}$
Where $2^{\omega(ab)} = 2^{k + t}$
$\therefore 2^{\omega(ab)} = 2^{\omega(a)} \times 2^{\omega(b)}$

Am I in the right track?

Thanks,

  • 0
    $\omega(n)$ is the number of distinct prime factors of $n$.2011-08-24

2 Answers 2

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Hint: Notice $2^{\omega(a)}\times 2^{\omega(b)}=2^{\omega(a)+\omega(b)}$ so try to relate $\omega(a)+\omega(b)$ and $\omega(ab)$.

To simplify things, you need only show it holds for $a=p^r$, $b=q^t$ where $q,p$ are prime numbers.

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    Done, hope that helps.2011-03-01
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Note that if $f : \mathbb{N} \to \mathbb{C}$ is additive:

$ \gcd(m,n) = 1 \quad \Longrightarrow \quad f(mn) = f(m) + f(n), $

then for any $t > 0$, $g(n) := t^{f(n)}$ is multiplicative:

$ \gcd(m,n) = 1 \quad \Longrightarrow \quad g(mn) = g(m) g(n), $

since

$ g(mn) = t^{f(mn)} = t^{f(m) + f(n)} = t^{f(m)}t^{f(n)} = g(m) g(n). $

Note also that this $g$ will never be the zero function. Therefore, it suffices to show that $\omega(n)$ is additive (which has indeed already been shown).