# Thread: Well defined binary operation

1. ## Well defined binary operation

Prove $\displaystyle x+y-\left\lfloor x+y\right\rfloor$ is a well defined binary operation.

How do I do this?

2. ## Re: Well defined binary operation

Originally Posted by dwsmith
Prove $\displaystyle x+y-\left\lfloor x+y\right\rfloor$ is a well defined binary operation.

How do I do this?
Well-it's obviously well-defined, and presumably from context you mean $\displaystyle \mathbb{Z}^2\to\mathbb{Z}$ from where it's obvious since the maps $\displaystyle (x,y,z)\to x+y-z$ and $\displaystyle (x,y)\to \lfloor x+y\rfloor$ map into the integers

3. ## Re: Well defined binary operation

Originally Posted by dwsmith
Prove $\displaystyle x+y-\left\lfloor x+y\right\rfloor$ is a well defined binary operation. How do I do this?
We need to know the set of definition. Let's suppose the set is $\displaystyle \mathbb{R}$ in this case, then we have to prove that for all $\displaystyle x,y\in\mathbb{R}$ , $\displaystyle x*y=x+y-\left\lfloor x+y\right\rfloor$ exists, is unique and belongs to $\displaystyle \mathbb{R}$ .

Edited: For example, Drexel28 supposed the set of definition is $\displaystyle \mathbb{Z}$ .

4. ## Re: Well defined binary operation

The set is $\displaystyle G=\{x\in\mathbb{R}: \ x\in [0,1)\}$

I am trying to show associativity.

$\displaystyle (x*y)*z=x+y-\left\lfloor x+y\right\rfloor +z-\left\lfloor x+y-\left\lfloor x+y\right\rfloor + z\right\rfloor$

I can't see how to manipulated that into:

$\displaystyle x+y+z-\left\lfloor x+y+z-\left\lfloor y+z\right\rfloor\right\rfloor$

5. ## Re: Well defined binary operation

It can be useful to express $\displaystyle x*y=\begin{Bmatrix} x+y-1 & \mbox{ if }& x+y\geq 1\\x+y & \mbox{if}& x+y<1\end{matrix}$

6. ## Re: Well defined binary operation

I am working on proving this an Abelian group. I found the identity to be 1. I am having trouble finding the inverse element.

7. ## Re: Well defined binary operation

The identity is $\displaystyle 0$ because $\displaystyle 0*x=x*0=x+0-\lfloor x+0\rfloor=x-0=x$ for all $\displaystyle x\in[0,1)$ .