Q:

Floor Function Homomorphism and Isomorphism

Is the following a homomorphism and an isomorphism?
$f\colon (\mathbb{R},+)\to (\mathbb{R}, +), f(x)=\lfloor x \rfloor$. 
I don't think it is a homomorphism because if I define $\phi$ as above. $\phi (ab) \neq \phi(a)\phi(b)$ With the above function. For example if I took values such as $a = 2.5, b = 2.5$ then $\phi(ab) = 6$ while $\phi(a)\phi(b)=4$ since these two do not equal, this is not a homomorphism. Is this correct?

A:

Group only has an addition and has not a multiplication. The group operation(addition) here is addition. 
Two show $f$ is not a homeomorphism consider $a=\dfrac{1}{2}$ and $b=\dfrac{3}{2}$.
Then $f(a)=f\left (\dfrac{1}{2}\right)=\lfloor \dfrac{1}{2}\rfloor =0$ and $f(b)=f\left (\dfrac{3}{2}\right)=\lfloor \dfrac{3}{2}\rfloor =1$ and $f(a+b)=f\left (\dfrac{1}{2}+\dfrac{3}{2}\right)=f(2)=\lfloor 2\rfloor =2$.
Therefore $f(a+b)\ne f(a)+f(b)$ and hence not a homeomorphism.

