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A proof

  1. Nov 3, 2005 #1
    I answered this wrong on a test, but now I've come up with a different solution.
    Problem: Prove that a relation [itex]xRy\Leftrightarrow x-y\in\mathbb{Z}[/itex] defined on [itex]\mathbb{R}[/itex] is an equivalence relation.
    1.) Reflexivity: [itex]xRx,\forall x\in\mathbb{R}[/itex]
    For every [itex]x[/itex] we have [itex]x-x=0[/itex] which is an integer, so reflexivity holds.
    2.) Symmetricity: [itex]xRy\Rightarrow yRx,\forall x,y\in\mathbb{R}[/itex]
    If for all [itex]x,y\in\mathbb{R}[/itex] we have [itex](x-y)\in\mathbb{Z}[/itex], then [itex]y-x=-1\cdot(x-y)[/itex] (any integer multiplied by -1 is also an integer) and thus [itex](y-x)\in\mathbb{Z}[/itex] and the relation is symmetric.
    3.) Transitivity: [itex]xRy\wedge yRz\Rightarrow xRz,\forall x,y,z\in\mathbb{R}[/itex]
    For some [itex]x,y,z\in\mathbb{R}[/itex] we have [itex](x-y)+(y-z)=x-y+y-z=x-z[/itex] (the sum of two integers is also an integer) and thus [itex](x-z)\in\mathbb{Z}[/itex]. The relation is also transitive.
    Is this it???
    PS: sorry for the poor spelling
    - Kamataat
  2. jcsd
  3. Nov 3, 2005 #2


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    Staff Emeritus
    Science Advisor

    Yes, that is precisely what you should have done!
    (And don't you just hate it when that comes right after the test!)
  4. Nov 3, 2005 #3
    Yeah, especially since it's such an elementary thing. I'd think anyone with a (future) professional interest in maths/physics should know this stuff like the back of his/her hand! Oh well, it's happened before to me, so...


    Anyway, thanks!

    - Kamataat
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