<s>[INST] Understanding why a binary operator is associative. ( On a property of "fractional part of a sum" operation) Define the binary operation which gives the fractional part of a real number $x$ like so:
$$x +_1 y = \{x + y\}$$
where 
$$\{x\}=x-\lfloor x \rfloor$$
Now we want to show that this operator is associative on the interval $[0, 1)$.
The proof I found in my book is like this:

We make use of the fact that $\lfloor x \rfloor$ is an integer:
  \begin{align*}
x +_1 y &= \{x + y\} \\
        &= x + y - \lfloor x + y \rfloor
\end{align*}
So $x +_1 y$ is equal to $x + y$ minus some integer.
Then for $x, y, z \in  [0, 1)$:
  \begin{align*}
(x +_1 y) +_1 z &= (x + y - p) +_1 z \\
                &= (x + y - p) + z - q \\
                &= (x + y + z) - (p + q)
\end{align*}
And:
  \begin{align*}
x +_1 (y +_1 z) &= x +_1 (y + z - r) \\
                &= x + (y + z - r) - s \\
                &= (x + y + z) - (r + s)
\end{align*}
Where $p, q, r, s \in \mathbb{Z}$.
Since $(x +_1 y) +_1 z$ and $x +_1 (y +_1 z)$ both lie in the same interval,
  the integers $(p + q)$ and $(r + s)$ must be equal and hence the following holds:
\begin{equation*}
(x +_1 y) +_1 z = x +_1 (y +_1 z)
\end{equation*}

I don't quite follow the bolded deduction.


*

*They say that both "$(x +_1 y) +_1 z$ and $x +_1 (y +_1 z)$ lie in the same interval", ok, that makes sense, because of the definition of the $+_1$ operator.

*But why does this imply that $(p + q)$ and $(r + s)$ must be equal?

 [/INST] We have that $(x +_1 y) +_1 z = (x+y+z) - (p+q)$ and that $x +_1 (y +_1 z) = (x+y+z) - (r+s)$. If we take the floor of both of these expressions both of the left hand sides will be equal to zero so we get.
$$0 = \left \lfloor (x+y+z) - (p+q) \right \rfloor = \left \lfloor (x+y+z) - (r+s) \right \rfloor$$
Because $p+q$ and $r+s$ are integers, they can be taken out of the floor function to give
$$0 = \left \lfloor x+y+z \right \rfloor - (p+q) = \left \lfloor x+y+z \right \rfloor - (r+s).$$
Ignoring the zero at the start and just focusing on the second part of the equality we can subtract $\left \lfloor x+y+z \right \rfloor$. This gives us
$$ -(p+q) = -(r+s) $$
from which the result easily follows.
</s>