Identity with the floor function

Question

I'm struggling to complete proof of the following identity:

$$ \Bigl\lfloor \frac{m+ n}{2} \Bigr\rfloor + \Bigl\lfloor \frac{m - n +1}{2} \Bigr\rfloor =m, $$ where $m$ and $n$ are both integer. By definition of floor function, $x-1 < \lfloor x \rfloor \leq x$.

Then \begin{align} \frac{m+n}{2} -1 < & \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor \leq & \frac{m+n}{2} \\ \frac{m-n + 1}{2} -1 < & \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor \leq & \frac{m-n +1}{2}. \end{align} By adding member to member, we obtain the following result: $$ m +\frac{1}{2} -2 < \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor \leq m +\frac{1}{2}. $$ Which implies

$$ -1.5 < \Bigl\lfloor \frac{m+ n}{2} \Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2} \Bigr\rfloor -m \leq .5 $$ This results in

$$ \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor \in \left\{0,1\right\}. $$

How to decide that the result is $0$?

Any help is welcome.

Answer 1

hint

Observe that $$\frac{m-n+1}{2}=\frac{m+n+1}{2}-n$$

So, you just need to prove that $$\lfloor a\rfloor +\lfloor a+\frac 12\rfloor=2a$$

where $ a=\frac{m+n}{2}$.

There are two cases

$$a\in \Bbb N \implies \lfloor a\rfloor +\lfloor a+\frac 12\rfloor=a+a$$

$$a\notin \Bbb N \implies a+\frac 12\in \Bbb N$$ $$\implies \lfloor a\rfloor +\lfloor a+\frac 12\rfloor=$$ $$a+\frac 12-1+a+\frac 12=2a$$

Done.

Answer 2

If $m+n$ is even, then $\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor = \frac{m+n}{2}$ and as $m-n=m+n -2m$ is also even, $\Bigl\lfloor \frac{m- n+1}{2}\Bigr\rfloor = \frac{m-n-1}{2}$.

So $\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor = m$

If $m+n$ is odd, then $\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor = \frac{m+n-1}{2}$ and as $m-n=m+n -2m$ is also odd, $\Bigl\lfloor \frac{m- n+1}{2}\Bigr\rfloor = \frac{m-n}{2}$.

So $\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor = m$

Answer 3

Notice that $\forall x \in \mathbb Z, \left\lfloor \frac x2 \right\rfloor = \frac x2 - \left\lfloor \frac{x \mod 2}{2}\right\rfloor$, therefore

$$\left\lfloor \frac{m+n}{2}\right\rfloor + \left\lfloor \frac{m-n+1}{2}\right\rfloor\\ =\frac{m+n}{2} - \left\lfloor \frac{m+n \mod 2}{2}\right\rfloor + \frac{m-n+1}{2}- \left\lfloor \frac{m-n+1 \mod 2}{2} \right\rfloor$$ $$= m + \frac 12 - \left\lfloor \frac{m+n \mod 2}{2}\right\rfloor - \left\lfloor \frac{m-n+1 \mod 2}{2} \right\rfloor\tag1$$

Since $m+n$ and $m-n$ have the same parity, the last two terms in $(1)$ sum to $-\frac 12$, and we are done.