How many cases are there in this specific example?

$\lfloor (n+m)/2\rfloor\ = \lfloor m/2\rfloor\ + \lceil n/2 \rceil$

I thought it was 4, but I am not sure

n greater than 1 and greater than 2

m greater than 1 and greater than 2

trying to do a proof by case

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I’m going to assume that you omitted a necessary pair of parentheses, and that you really meant

$$\left\lfloor\frac{m+n}2\right\rfloor=\left\lfloor\frac{m}2\right\rfloor+\left\lceil\frac{n}2\right\rceil\;,\tag{1}$$

where $m$ and $n$ are integers. You do want four cases to investigate this.

• If $m$ and $n$ are both even, say $m=2a$ and $n=2b$, both sides of $(1)$ reduce to $a+b$.

• If $m$ is even and $n$ is odd, say $m=2a$ and $n=2b+1$, the left side of $(1)$ reduces to $a+b$, but the right side reduces to $a+b+1$.

• If $m$ is odd and $n$ is even, say $m=2a+1$ and $n=2b$, both sides of $(1)$ reduce to $a+b$.

• If $m$ and $n$ are both odd, say $m=2a+1$ and $n=2b+1$, both sides of $(1)$ reduce to $a+b+1$.

Thus, $(1)$ is true unless $m$ is even and $n$ is odd.

If $m$ and $n$ are not required to be integers, $(1)$ can fail even more easily. For instance, if $m=n=\frac12$, the lefthand side of $(1)$ is $0$, while the righthand side is $0+1=1$.

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