Proving That $f: \mathbb{Z} \to \mathbb{Z}$, where $f(x) = x + 10$, Is Surjective

Question

I am trying to prove that $f: \mathbb{Z} \to \mathbb{Z}$, where $f(x) = x + 10$, is surjective. I have included my reasoning and would appreciate it if people could check whether it is correct.

My Proof

Let $b \in \mathbb{Z}$.

We want to show that, for any $b \in \mathbb{Z}$, there exists some $x \in \mathbb{Z}$ such that $b = f(x) = x + 10$.

$\therefore x = b - 10$ We constructed an integer $x$.

$\therefore f(x) = f(b - 10)$

$= (b - 10) + 10$ (By the hypothesis.)

$= b$

$= x + 10 \ \ \ Q.E.D. \ \ \ $ We have now demonstrated that $b = f(x) = x + 10$.

I would greatly appreciate it if people could please take the time to review my proof and the reasoning I have included in it.

Answer 1

Your proof is fine.

Here with minute differences in style:

Need to show that $f$ is surjective.

Let $b \in {Z}.$

Need to find an $a \in \mathbb{Z}$ such that

$f(a)=b$.

Choose $a= b-10 $, then

$f(a)=(b-10)+10 = b.$

Done.