TITLE: If $f$ is a group homomorphism from $(\mathbb{Z},+)$ to $(\mathbb{Q}-\{0\},.)$ such that $f(2)=\frac{1}{3}$, then find $f(-8)$.
QUESTION [6 upvotes]: I came across the following question:
If $f$ is a group homomorphism from $(\mathbb{Z},+)$ to $(\mathbb{Q}-\{0\},.)$ such that $f(2)=\frac{1}{3}$, then what is the value of $f(-8)$?
By property of group homomorphism, we can write - $f(8) = f(2+2+2+2) = f(2)^4 = \frac{1}{81}$.
But we are asked to find $f(-8)$. How does the negative sign change the answer (i.e. if it changes, I am not sure)?
Thank you.

REPLY [7 votes]: Since $$f(x)f(-x)=f(x-x)=f(0)=1$$ by the homomorphism property then $f(-x)=\frac{1}{f(x)}$.

REPLY [4 votes]: This question is intentionally confusing.
The fact that it's a group homomorphism means $f(x+y)=f(x)f(y)$, since the domain is an additive group, and the codomain is a multiplicative group. Thus actually, 
$f(8)=f(2+2+2+2)=f(2)f(2)f(2)f(2)=\frac{1}{81}$.
Since $1=f(0)=f(8-8)=f(8)f(-8)$, $f(-8)=\frac{1}{f(8)}=81$.