# Prove $xRy$ if $\exists \space k \in \mathbb{Z}(y+x=2k)$ is an equivalence relation

Given the binary relation $xRy$ defined in $\mathbb{Z}$ such that $\exists \space k \in \mathbb{Z}(y+x=2k)$.

To prove that the relation $R$ is an equivalence relation, I must prove that $R$ is symmetric, reflexive and transitive.

-Symmetric

$\forall x,y\space(x+y=2k \wedge y+x=2k)$. So one have two equations that are equal to $2k$, I can write $x+y=y+x$. That will result in $0=0$, that is true in $\mathbb{Z}$.

-Reflexive

Let be $x \in \mathbb{Z}$. Then $\space \forall x \space (x+x=2k)$. So I write $2x=2k \Leftrightarrow x=k$. If $k$ is a integer then $x$ will be too.

-Transitive

$\forall x,y,z \space (x+y=2k \wedge y+z=2k \Rightarrow x+z=2k)$. Before all, one can see that $2k$ will allways be an integer pair number. Then I can sum $x+y=2k$ with $y+z=2k$ and get $x+2y+z=4k$. After, one can write $x+z=2(2k-y)$, that is an integer pair number too.

Are my arguments correct?

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I’m guessing that integer pair number is an attempt to translate número inteiro par or something like that; if so, the English is even integer. – Brian M. Scott Dec 19 '12 at 11:33

You have the right idea, but a lot of the details need more work or more careful expression, especially in the argument for transitivity.

Transitivity: Suppose that $x\,R\,y$ and $y\,R\,z$. Then there are integers $k$ and $\ell$ such that $x+y=2k$ and $y+z=2\ell$. We want to show that $x+z$ is even. Adding the equations $x+y=2k$ and $y+z=2\ell$ to get $x+2y+z=2k+2\ell$, doesn’t quite give us $x+z$, but a simple rearrangement does: $$x+z=2k+2\ell-2y=2(k+\ell-y)\;,$$ where $k+\ell-y$ is certainly an integer. Thus, $x+y$ is indeed even, $x\,R\,z$, and $R$ is transitive.

Note that you need to use different letters for $k$ and $\ell$, since you’ve no reason to think that $x+y=y+z$.

Symmetry: Suppose that $x\,R\,y$. Then there is an integer $k$ such that $x+y=2k$. But then $y+x=x+y=2k$, so by definition $y\,R\,x$, and $R$ is symmetric.

Reflexivity: Let $x\in\Bbb Z$. Then $x+x=2x$, where $x\in\Bbb Z$, so by definition $x\,R\,x$, and $R$ is reflexive.

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Your proof is wrong because you assume what you are trying to prove and use it to prove other things instead of prove what you are trying to!

Reflexive: For any $x\in\mathbb Z$, $x+x=2x$ where $x\in\mathbb Z$.

Symmetric: For any $x,y\in\mathbb Z$, if $y+x=2k$ for some $k\in\mathbb Z$, then $x+y=2k$ where $k\in\mathbb Z$.

Transitive: For any $x,y,z\in\mathbb Z$, if $y+x=2k$ and $z+y=2l$ for some $k,l\in Z$, then $z+x=2(k+l-y)$ where $k+l-y\in\mathbb Z$.

Note that this equivalence relation just partitions the integers into two classes: one contains all the even integers and the other all the odd integers.

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The relation is symmetric. However, it appears that you only proved that if $x + y = y + x = 2k$, then $0 = 0$. This is not the statement that $R$ is symmetric. A possible proof of symmetry is the following:

$x R y$ if and only if there exists a $k$ such that $x + y = 2k$ if and only if there exists a $k$ such that $y + x = 2k$ if and only if $y R x$.

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