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Math Help - Well defined binary operation

  1. #1
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    Well defined binary operation

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

    How do I do this?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Well defined binary operation

    Quote Originally Posted by dwsmith View Post
    Prove 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 \mathbb{Z}^2\to\mathbb{Z} from where it's obvious since the maps (x,y,z)\to x+y-z and (x,y)\to \lfloor x+y\rfloor map into the integers
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Well defined binary operation

    Quote Originally Posted by dwsmith View Post
    Prove 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 \mathbb{R} in this case, then we have to prove that for all x,y\in\mathbb{R} , x*y=x+y-\left\lfloor x+y\right\rfloor exists, is unique and belongs to \mathbb{R} .

    Edited: For example, Drexel28 supposed the set of definition is \mathbb{Z} .
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  4. #4
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    Re: Well defined binary operation

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

    I am trying to show associativity.

    (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:

    x+y+z-\left\lfloor x+y+z-\left\lfloor y+z\right\rfloor\right\rfloor
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Re: Well defined binary operation

    It can be useful to express x*y=\begin{Bmatrix} x+y-1 & \mbox{ if }& x+y\geq 1\\x+y & \mbox{if}& x+y<1\end{matrix}
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  6. #6
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    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.
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Well defined binary operation

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