I answered this wrong on a test, but now I've come up with a different solution.(adsbygoogle = window.adsbygoogle || []).push({});

Problem: Prove that a relation [itex]xRy\Leftrightarrow x-y\in\mathbb{Z}[/itex] defined on [itex]\mathbb{R}[/itex] is an equivalence relation.

Solution:

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

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# Homework Help: A proof

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