Exercise 4.1.4 (Floor and ceil)

Proof. Suppose that the real numbers $x$ and $y$ satisfy

For the sake of contradiction, assume that $x\notin \mathbb{Z}$ and $y\notin \mathbb{Z}$. Then $\lfloor x\rfloor <x,\lfloor y\rfloor <y$, thus

Moreover, by equation (1),

thus, by equation (3),

Multiplying by $-1$, we obtain

thus

The comparison of (2) and (4) gives

or equivalently

By Theorem 4.1(7) , $-\lfloor -x\rfloor$ is the least integer $n\ge x$, which we can write $\lceil x\rceil =-\lfloor -x\rfloor$ (“ceil” function, in opposition to “floor” function). Here $x\notin \mathbb{Z}$, so $\lfloor x\rfloor <x<\lfloor x\rfloor +1$, thus $-\lfloor x\rfloor -1<-x<-\lfloor x\rfloor$, so $\lfloor -x\rfloor =-\lfloor x\rfloor -1$, and similar equations for $y$. This gives

Then equation (5) shows that

After simplification, we obtain $0=1$. This contradiction shows that $x\in \mathbb{Z}$ or $y\in \mathbb{Z}$.

Given that $\lfloor x+y\rfloor =\lfloor x\rfloor +\lfloor y\rfloor$ and $\lfloor -x-y\rfloor =\lfloor -x\rfloor +\lfloor -y\rfloor$, then $x$ or $y$ is an integer. □