Prove $\forall x,y \in \mathbb{R} :\lfloor{x+y}\rfloor=\lfloor{x}\rfloor+\lfloor{y}\rfloor∨\lfloor{x+y}\rfloor=\lfloor{x}\rfloor+\lfloor{y}\rfloor+1$

Question

Prove $∀x,y\ (x,y\in \mathbb{R}: \lfloor{x+y}\rfloor=\lfloor{x}\rfloor+\lfloor{y}\rfloor∨\lfloor{x+y}\rfloor=\lfloor{x}\rfloor+\lfloor{y}\rfloor+1)$

So, I let

$\lfloor{x}\rfloor=m ≡ m≤x<m+1$

$\lfloor{y}\rfloor=n ≡ n≤y<n+1$

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 $$ m \leq x < m + 1 \quad > \text{and} \quad n \leq y < n + 1.$$ So, adding, we obtain, $$ m+n > \leq x+y < m+n+2.$$ Thus, $$ [x+y] = m+n = [x] + [y] \quad \text{or} \quad [x+y] = m+n+1 = [x] + [y] + 1.$$

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.

Answer 1

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 $a$?

Answer 2

Hint: How many $k \in \mathbb{Z}$ 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 $?

Answer 3

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