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The example is: For all $x,y\in\mathbb{Z}$,

$x\equiv y\Leftrightarrow x-y$ is even number.

Questions:

  • Show that this is an equivalence relation.
  • Find the classes of this.

My answer is:

  • Firstly, for all $x$ in $\mathbb{Z}$, we know that $x-x=0$ is even number.

    Secondly, for all $x,y$ in $\mathbb{Z}$, if $x-y$ is even number then also $y-x$ is even number because $y-x=-(x-y)$ is even number. Clear.

    Thirdly, for all $x,y,z$ in $\mathbb{Z}$, if $x-y$ and $y-z$ are even number then also $x-z$ is even number because $x-z=(x-y)+(y-z)$.

Therefore, this is an equivalence relation.

  • Only, there are two clases: $2\mathbb{Z}$ and $2\mathbb{Z} +1$. Simple example for this case: $0\equiv 2\Leftrightarrow 0-2$ is even number. Finally, $3\equiv 1\Leftrightarrow 3-1$ is even number.

Can you check my proof? Especially, can you check my proof-writing?

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    It looks just fine. – 2017-01-07
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    @LeeMosher Thanks. – 2017-01-07
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    You kind of pulled the two classes out of a hat (even though they're obvious here). A more systematic approach: $x\equiv y$ iff $y=x+2n$ for some $n\in\mathbb{Z}$. Hence $[x]=x+2\mathbb{Z}$. So if $x$ is even then $[x]=2\mathbb{Z}$ and if $x$ is odd then $[x]=2\mathbb{Z}+1$. – 2017-01-07

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