Since $\zeta(1) \neq 1$, does this mean that $\zeta$ is not multiplicative?

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?

• The term "multiplicative" is for functions defined on the positive integers. Zeta isn't defined at $1.$ – coffeemath Jan 8 at 12:02
• @coffeemath, ohh yeah right! Can you write your last comment as an actual answer, so that I would be able to accept it? Thanks! – Jose Arnaldo Bebita-Dris Jan 8 at 12:06

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 $$1 we had $$\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.