Is this function bijective, surjective and injective?

$\lfloor\cdot\rfloor: Q \rightarrow\mathbb Z$ with $\lfloor x\rfloor :=$ floor of $x$.

I know a function is injective by using $f(x_1)=f(x_2) \Rightarrow x_1=x_2$ and a function is surjective if each element of the codomain, $y\in Y$, is the image of some element in the domain $x\in X$, and bijective if the function is both injective and surjective.

I don't know what floor of $x$ is.