How to find whether a function is injective or surjective

Question

$g$ is a function from $\mathbb Z$ to $\mathbb Z$ such that $g(x)=\lfloor2x\rfloor$.

Can anyone explain how to find whether this function is injective or surjective

Answer 1

Most students asking this sort of question are having trouble grasping the definitions (in this case, of injective and surjective).

Also, be careful: The domain of $g$ is the integers, not the reals (at least as the problem is posed here).

Let's examine whether $g(x)$ is injective; by definition, that means let's ask "can any two different values of $x$ map to the same value of $g(x)$"?

Clearly not, since if $x_1 \neq x_2$ then $|x_1-x_2| \geq 1$ so the values of the floors of double the $x_1$ and $x_2$ must differ more than $(2-1-1)$ (actually, by at least $2$). So the function is injective.

Now see if $g(x)$ is surjective: "Can there be any value $g_0 \in \Bbb{Z}$ such that $\forall x \in \Bbb{Z} : \lfloor 2x \rfloor \neq g_0$?"

Well, consider $g_0 = 1$ (or indeed any odd integer). No floor of double an integer is odd, so the function cannot be surjective.

If the same question were asked using the domain $x \in \Bbb{R}$, and the codomain for $g(x)$ remains $\Bbb{Z}$, then the function would not be injective ($1$ and $1.1$ lead to the same $g(x) = 2$) but now it would be surjective since any integer $n$ can be expressed as $\lfloor 2(n/2)$ for $x$ equal to the real number $n/2$.