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Let $\zeta(s)$ be the Riemann zeta function, that is, $$\zeta(s) = \sum_{n=1}^{\infty}{\frac{1}{n^s}}.$$

A function $g$ is said to be multiplicative if, whenever $\gcd(x,y)=1$, we have $$g(xy) = g(x)g(y).$$ Examples of multiplicative (arithmetic) functions include the divisor-sum $\sigma(z)=\sigma_{1}(z)$ and the abundancy index $I(z)=\sigma(z)/z$.

It is known that, if a function $f$ is multiplicative (and not identically zero), then $f(1)=1$.

Proof: Suppose that $f$ is multiplicative (and not identically zero). Since $\gcd(m,1)=1$ and $f \neq 0$ is multiplicative, then $$f(m)=f(m\cdot{1})=f(m)f(1)$$ which implies that $f(1)=1$.

Since we know that $\zeta(1) \neq 1$ (in fact, the sum $$\zeta(1) = \sum_{n=1}^{\infty}{\frac{1}{n}}$$ is known to diverge), then can we already conclude that the Riemann zeta function $\zeta$ is not multiplicative?

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    $\begingroup$ The term "multiplicative" is for functions defined on the positive integers. Zeta isn't defined at $1.$ $\endgroup$ – coffeemath Jan 8 at 12:02
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    $\begingroup$ @coffeemath, ohh yeah right! Can you write your last comment as an actual answer, so that I would be able to accept it? Thanks! $\endgroup$ – Jose Arnaldo Bebita-Dris Jan 8 at 12:06
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The term "multiplicative" is used for a function $f$ defined on the positive integers, for which $f(mn)=f(m)f(n)$ whenever $\gcd(m,n)=1.$ The Riemann zeta function $\zeta$ isn't even defined at $1$ so is not a candidate to be called multiplicative. Even ignoring the problem at $1$ it could be checked numerically that $\zeta(2)\zeta(3) \neq \zeta(6)$.

Added: No need for a numerical check. Exact values for $\zeta(2k)$ are known, and $\zeta(6)/\zeta(2)<0.7.$ So it can't equal $\zeta(3),$ which is greater than $1.$

Another note: $\zeta(n)$is strictly monotone decreasing as $n>1$ increases. So if for $a,b$ with $1 <a,b$ we had $\zeta(ab)=\zeta(a)\zeta(b),$ then since $ab>a$ it would follow that $\zeta(ab)/\zeta(a)<1,$ which cannot be $\zeta(b)>1.$ So $\zeta(n)$ is not multiplicative for any values of $a,b>1,$ coprime or not.

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    $\begingroup$ Thank you for your answer. I hope you don't mind my minor edit. $\endgroup$ – Jose Arnaldo Bebita-Dris Jan 8 at 12:18

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