Vtyjkyuk

Prove
$∀x,y (x,yin mathbbR: lfloorx+yrfloor=lfloorxrfloor+lflooryrfloor∨lfloorx+yrfloor=lfloorxrfloor+lflooryrfloor+1)$

So, I let

$lfloorxrfloor=m ≡ m≤x<m+1$

$lflooryrfloor=n ≡ n≤y<n+1$

Now, I have that $m+n≤x+y<m+n+2$

... And I get stuck in here.

I found this proof over the the internet:

Let $[x] = m$ and $[y] = n$, then we have $$ m leq x < m + 1 quad
> textand quad n leq y < n + 1.$$ So, adding, we obtain, $$ m+n
> leq x+y < m+n+2.$$ Thus, $$ [x+y] = m+n = [x] + [y] quad textor
quad [x+y] = m+n+1 = [x] + [y] + 1.$$

Which seems to state that where I am is sufficient to conclude the theorem, however I don't see how,because it escalates too quickly.

How do you think i should follow?

Thanks in advance.

Hint:
How many $k in mathbbZ$ are there such that

$m + n leq k < m + n + 2$

and what are they? Now, what does this, along with the fact that

$m + n leq x + y < m + n + 2$

tell you about $lfloor x + y rfloor $?

Remark that if $a = lfloor x+y rfloor$, then the inequalities $$a leq x+y < a+1$$ can be rearranged as $$x+y-1 < a leq x+y$$ Do you see how combining this with $$m+n leq x+y < m+n+2$$ gives you the possible values of $a$?

Perhaps one may add that either $x+y<m+n+1$ or $m+n+1 leq x+y$, from which the result readily follows.