The relation A defined on $\mathbb{R}$ by $ \exists k \in \mathbb{Z} , xAy \iff x - y = 2k$ is an equivalence relation.

Question

The relation A defined on $\mathbb{R}$ by $ xAy \iff \exists k \in \mathbb{Z}, ~x - y = 2k$ is an equivalence relation.

Proof. We must show that xAy is reflexive, symmetry and transitive.

Since A is a relation on $\mathbb{R}$, so that $x, y \in \mathbb{Z}$. Then xAx is relation iff $x-x = 2k$. Hence A is reflexive Since k = 0 and $k \in \mathbb{Z}$.

Since xAy iff $\exists k \in \mathbb{Z}, x - y = 2k$, then we have yAx iff $\exists k \in \mathbb{Z}, y-x =2k$. If there exists $k \in \mathbb{Z}$, such that $x - y = 2k$, then there must exists $k \in \mathbb{Z}$, such that $y-x = 2k$. Hence A is symmerty.

Let $a \in R$, we have yAa as $y - a = 2k$, also we have xAy as $x - y = 2k$. We take the union, then we have $(xAy) \cup (yAa) ~iff~ \exists k \in \mathbb{Z}, (x-y+y-a) = 2k.$ Therefore, xAa iff $\exists k \in \mathbb{Z}, x -a = 2k$. Hence A is transitive.

Since A is reflexive, symmetry and transitive. Hence A is an equivalence relation on $\mathbb{R}$. $\blacksquare$

Does my proof correct? Do I need to worried about the domain of relation is $\mathbb{R}$ while we want to show the existence is on $\mathbb{Z}$ ?

Answer 1

Do I need to worried about the domain of relation is $\Bbb R$ while we want to show the existence is on $\Bbb Z$.

No.   It is of no concern.

Any ordered pair of real numbers will be A-related if their difference is an even integer. $$\forall (x,y)\in \Bbb R^2: \quad (x,y)\in A \iff [\exists k\in \Bbb Z: 2k=x-y]$$

$2.314$ and $12.314$ are real numbers whose difference is an even integer, $10$, so they are A-related.

$2$ and $\pi$ are real numbers whose difference is not an even integer, so they are not A-related.

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Does my proof correct?

It... needs polishing, but you have the right approach.   Here one way I'd write it.

• Reflexive:   Because $0$ is an even integer, then any real number is A-related to itself.

• Symmetric:   Because the negative of an even integer is also an even integer, then if any ordered pair of real numbers are A-related, then the reverse-ordered pair is also A-related. vis: $$x-y=2k ~\to~ y-x=2(-k)$$

• Transitive:   Because the sum of two even integers is also an even integer, then if any two ordered pairs of real numbers of the form $(x,y)$ and $(y,z)$ are both A-related, then so too is $(x,z)$. vis: $$[~x-y=2k_1~\wedge~ y-z=2k_2~]~\to~ x-z=2(k_1+k_2)$$

It covers the points you were trying to raise in clear natural language, with symbols to clarify.  

Answer 2

I'll prove the statement, following your proof as closely as I comfortably can while still maintaining proper English.

Proof. We show that $A$ is reflexive, symmetric and transitive.

For any $x$, $x-x=0=2\cdot 0$, hence $xAx$ and $A$ is reflexive.

Suppose $xAy$. Then $x-y=2k$ for some $k\in\mathbb{Z}$, and $y-x=2(-k)$, and $-k\in\mathbb{Z}$ so $yAx$. Hence $A$ is symmetric.

Suppose we have $xAy$ and $yAa$. Then $x-y=2k_1$ and $y-a=2k_2$ for some $k_1,k_2\in\mathbb{Z}$. Then $x-a=(x-y)+(y-a)=2k_1+2k_2=2(k_1+k_2)$, and $k_1+k_2\in\mathbb{Z}$, so $xAa$; hence $A$ is transitive.

Since $A$ is reflexive, symmetric and transitive, $A$ is an equivalence relation on $\mathbb{R}$. ■