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Proof.

  • For all $x\in\mathbb{Z}$, since $x-x=0\in\mathbb{Z}$, R is reflexivity.
  • For all $x,y\in\mathbb{Z}$, since if $x-y\in\mathbb{Z}$ then $y-x\in\mathbb{Z}$ ,R is symetric.
  • For all $x,y,z\in\mathbb{Z}$, since if $x-y\in\mathbb{Z}$ and $y-z\in\mathbb{Z}$ then we have $x-z\in\mathbb{Z}$, R is transitivity.

Therefore, R is equivalence relation.

Can you check my proof?

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    _Why_ is $y-x\in \Bbb Z$?. And _why_ is $x-z\in \Bbb Z$. The proof isn't over just because _you're_ convinced, it's over when it is capable of convincing others. – 2017-01-29
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    I would argue more why symmetricity and transitivity is true. Now it is just copy pasted from the definition of a general relation. – 2017-01-29
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    @Arthur You are right. How can I show? – 2017-01-29

4 Answers 4

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The relation $x \sim y$ defined on $\mathbb{Z}$ by $x-y\in\mathbb{Z}$ is the total relation: all pairs are related.

The total relation is trivially an equivalence relation.

The same relation defined on $\mathbb{Q}$ or $\mathbb{R}$ would be more interesting.

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let $x-y=m$ for any $m\in\mathbb{Z}$. Then $(y-x)=-m\in\mathbb{Z}$. This proves symmetry.

Next suppose $x-y=m$ and $y-z=n$, for any $n\in\mathbb{Z}$.

Then $x-z=(x-y)+(y-z)=m+n\in\mathbb{Z}$, which proves transitivity.

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You simply need to be clearer. You have understood the concept, no doubt.

If $x \in \mathbb Z$, then $x-x = 0 \in \mathbb Z$ is known.

If $x-y \in \mathbb Z$, then $y-x = -1(x-y) \in \mathbb Z$ is true because it's the negative of an integer, so it's an integer itself.

If $x-y,y-z \in \mathbb Z$, then $x-z = (x-y) + (y-z) \in \mathbb Z$ is true because the sum of two integers is also an integer.

Then, you are done.

The point is, when proofs are seemingly obvious, one must still not leave the trivial looking details out. That improves the quality of the answer.

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Clearly, we have $y-x=-(x-y)$, so if $n=x-y\in \Bbb Z$, then $y-x=-n\in \Bbb Z$.

As for transitivity, set $k=x-y, l=y-z, m=x-z$. We assume $k,l\in \Bbb Z$. Now, in more or less the save way as above, express $m$ in terms of $k,l$ so that we can see that $m\in \Bbb Z$ as well.