$\mathbb R^*$ is $\mathbb R/ \{ 0\}$ under the multiplication operation.
$f: \mathbb R^{*} \rightarrow \mathbb Z_{2}$ such that $f(r) = 0$ if $r>0$, and $f(r) = 1$ if $r<0$. I think that $f$ is a homomorphism because:
For $r>0$:
$f(a \cdot b) = 0 = [f(a) + f(b)]_2 = [0+0]_2 = 0$ and $f(-a \cdot -b) = 0 = [f(-a) + f(-b)]_2 = [1+1]_2 = 0$
For $r<0$
$f(-a \cdot b) = 1 = [f(-a)+f(b)]_2 = [1+0]_2 = 1$
Also, the kernel of $f$ be all $n \in R^*$ such that $n>0$. Is my logic correct?