# Prove that $\lfloor x + y + z \rfloor = \lfloor x + y \rfloor + \lfloor z + \langle y + x \rangle \rfloor$

The original question is as follows:

For any real number $x$, let $\lfloor x \rfloor$ denote the largest integer less than or equal to $x$ and $\langle x \rangle = x - \lfloor x \rfloor$, that is, the fractional part of $x$. For arbitrary real numbers $x$, $y$ and $z$, only one of the following statements is correct. Which one is it?

1. $\lfloor x + y + z \rfloor = \lfloor x \rfloor + \lfloor y \rfloor + \lfloor z \rfloor$
2. $\lfloor x + y + z \rfloor = \lfloor x + y \rfloor + \lfloor z \rfloor = \lfloor x \rfloor + \lfloor y + z \rfloor = \lfloor x + z \rfloor + \lfloor y \rfloor$
3. $\langle x +y + z \rangle = y + z - \lfloor y + z \rfloor + \langle x \rangle$
4. $\lfloor x + y + z \rfloor = \lfloor x + y \rfloor + \lfloor z + \langle y + x \rangle \rfloor$

For options $(1)$, $(2)$ and $(3)$, I proved those relations to be false (actually special cases) from the definitions. Therefore, Option $(4)$ must be the correct one. In fact, it is.

However, my question is, how should I prove it as it were, if there were no other options? I tried using the definitions alone, but I got stuck. Should I consider proving it using cases? Or is there a simple approach?

• Try a simpler version $\lfloor a+b \rfloor = \lfloor a \rfloor + \lfloor b + \langle a \rangle \rfloor$ using $a= \lfloor a \rfloor + \langle a \rangle$ and $\lfloor \lfloor a \rfloor +c \rfloor = \lfloor a \rfloor +\lfloor c \rfloor$ – Henry Jan 3 '17 at 21:19

## 1 Answer

We have $\def\fl#1{\left\lfloor#1\right\rfloor}\def\fr#1{\left<#1\right>}$ $$x+y+z = \fl{x+y} + \fr{x+y} + z$$ by the very definition of $\fl\cdot$ and $\fr\cdot$, hence as $\fl{x+y}$ is an integer, we have $$\fl{x+y+z} = \fl{\fl{x+y} + \fr{x+y} + z} = \fl{x+y} + \fl{\fr{x+y} + z}$$