# If $f$ is an arithmetic function with $f(1)=1$ then $f$ is multiplicative

I'm studying analytic number theory for undergraduates and I read this theorem in Tom Apostol's book on the second chapter:

Theorem 2.12. If $f$ is multiplicative then $f(1)=1$

And under need there is a note that says that $\Lambda (n)$ is not multiplicative because $\Lambda (1) = 0$. So I wonder if the converse of the theorem is true.

• You mean $\Lambda(1)=0,$ I think. – saulspatz Mar 20 '18 at 3:26
• You are right, I fixed it. Thanks @salpatz – Luis Victoria Mar 20 '18 at 3:33

The arithmetic function $$f(x)=\begin{cases}1 & x = 1 \\ 2 & x > 1\end{cases}$$ is not multiplicative even though $f(1)=1$.
(But when the book says that $\Lambda(n)$ is not multiplicative because $\Lambda(1)=0$, it's using the contrapositive, not the converse...)
Consider $f(x)=2x-1$. $f(1)=1$ but $f(6)=11 \neq f(2)\cdot f(3)=3\cdot 5=15$