Proving a floor function is surjective

Question

I have a floor function $f: \Bbb R → \Bbb Z$,

$$ f(x)=\lfloor x-7 \rfloor $$

That I am trying to prove is surjective or onto. I know by definition that the floor function's domain is the set of reals and the range is the set of integers. I also know how to prove a function is surjective, but in this case I feel like I have hit a wall. I don't know how to proceed from here, any help?

Answer 1

To prove that a function is surjective, you need to prove that for each $y$ in the codomain there exists some $x$ (which will depend on the value of $y$ and will likely be represented as some function of $y$) such that $f(x)=y$.

Given some integer $y$, can we find some real number $x$ such that $\lfloor x - 7\rfloor = y$?

What is $\lfloor (y+7)-7\rfloor$ for $y$ an integer?

Answer 2

Of course it is, just map every integer $x$ and you will get whole $\mathbb{Z}$. For every integer $x$ we have: $$f(x) = x-7$$ \begin{eqnarray} % \nonumber to remove numbering (before each equation) &\vdots &\\ 4 &\longmapsto & -3 \\ 5 &\longmapsto & -2\\ 6 &\longmapsto & -1\\ 7 &\longmapsto & 0\\ 8 &\longmapsto & 1\\ 9 &\longmapsto & 2 \\ &\vdots & \end{eqnarray}

Answer 3

All you need is to find, for given $n\in\Bbb Z$ some real number $x$ with $\lfloor x-7\rfloor =n$, i.e., with $n\le x-7<n+1$.