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I need to verificate or falsify the relation. In this previouse post (Showing $ x-y\in\mathbb{Q}$ is an equivalence relation?) they just showed that $x\in\mathbb{R}, x-x=0\in\mathbb{Q}$

Does this mean, that I only have to proof that $x\in\mathbb{R}, x-x=0$, which is element of $\mathbb{Z}$? Because I thought that I have to proof it somehow for $(x-y)$, like $(x-y)-(x-y)=0$

a17cx42d1
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  • Well, for all $x,y$, $x\sim y$, so trivially, this is a equivalence relation, but its not very interesting. – Rushabh Mehta Nov 09 '18 at 17:11
  • $x-x=0$ is enough for proving reflexivity. You also must prove symmetry and transitivity. – drhab Nov 09 '18 at 17:16
  • @drhab ah okay, but for example $x+y$ is not reflexive, is it? Because $x+x=0$ is never true? – a17cx42d1 Nov 09 '18 at 18:31
  • $x+y$ on its own is not a relation so cannot carry the label "reflexive". The relation $x+y=0$ is not reflexive unless it is a relation on the set ${0}$. – drhab Nov 09 '18 at 20:22
  • If $x,y$ are real, your relation is a little bit more interesting. Otherwise, it just doesn't say much. – Doug M Nov 09 '18 at 21:49

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Your question and notation are not very clear. I will try to restate it in a way that makes more sense to me.

You write:

$\{(x,y)\in\{\mathbb{Z}\times\mathbb{Z}\} | x-y\in\mathbb{Z}\}$

I interpret $(x,y)\in\{\mathbb{Z}\times\mathbb{Z}\}$ to mean you are taking some $x\in\mathbb{Z}$ and relating it to some $ y\in\mathbb{Z}$.

The condition for this relation is: $x-y\in\mathbb{Z}$

In other words, for integers x and y, we have xRy ("x is related to y") precisely when x - y is an integer.

Now, you want to prove that this relation is reflexive, that is, xRx.

Since x is an integer, you need to check the condition: $x-x\in\mathbb{Z}$

Since for all integers x, it is true that: $x-x =0\in\mathbb{Z}$, then the relation is reflexive.

Cassius12
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