Find the value of $\lfloor x+y \rfloor$ where $x \in \mathbb{R}$, $y \in \mathbb{Z}$ I'm trying to understand properties of the greatest integer function and I am struggling to find the value of $\lfloor x+y \rfloor$  where $x \in \mathbb{R}$, $y \in \mathbb{Z}$, and prove that it is correct value.
I don't really know how to prove this, but I have been dividing it into different cases.
I think that it equals $\lfloor x \rfloor + y$ when $x,y$ are both positive but not sure how to prove it. Depending on if one or both $x$ and $y$ are negative, and their ultimate sum, I get different values. I am having trouble determining when exactly this happens though and then proving the results. Any help would be great, thanks!
 A: It is always $\lfloor x \rfloor + y$.  Write $x = \lfloor x \rfloor + \{ x \}$ where $0 \le \{ x \} \lt 1$.  Then you just throw away the $\{ x \}$ because the rest is the integer.  Remember that $\lfloor -2.3 \rfloor = -3$, not $-2$
A: Note, lets use $x$ as the real and $n$ as the integer.
Since $x - 1 \le  \lfloor x \rfloor \le x$, it follows that $-x \le - \lfloor x \rfloor \lt -x+1.$
Combining this inequality with $x+n - 1 \lt \lfloor x+n \rfloor \le x + n$, we obtain $n-1 \lt \lfloor x+n \rfloor - \lfloor x \rfloor \lt n+1.$ 
Hence $\lfloor x+n \rfloor - \lfloor x \rfloor = n.$  
A: You can work directly from the definition. Let $n=\lfloor x\rfloor$; then $n\le x<n+1$. Now add the integer $y$ to the inequality to get (after very minor rearrangement of the righthand side) $$n+y\le x+y<(n+y)+1\;.$$ But since $n+y$ is an integer, that means precisely that $\lfloor x+y\rfloor=n+y$, by the definition of the floor function.
A: Hint $\ $ Using the universal property of floor makes such proofs mechanical, e.g.
$$\rm n  \le \color{#0A0}{x + k}\iff n\!-\!k \le x \iff n\!-\!k\le \lfloor x\rfloor\iff n\le \color{#C00}{\lfloor x\rfloor \!+\!k} $$ 
Therefore we deduce $\rm\ \lfloor \color{#0A0}{x + k}\rfloor = \lfloor \color{#C00}{\lfloor x\rfloor \!+\!k}\rfloor = \color{}{\lfloor x\rfloor \!+\!k}$
