Order relation in $\mathbb{Q}$

Question

I'm with trouble with this question: verify that the relation given by $$xRy \iff \exists n \in \mathbb{Q} \text{ such that } x = y + n$$ is an order relation in $\mathbb{Q}$

My doubt is in the part of antisymmetry, because how can I get from the equations

$x = y + n$ and

$y = x + m$

That x = y? I don't see this happening, so I think this is not an order relation. Is this right?

Answer 1

Suppose $P:xRy$ and $Q:yRx$. By the order axioms you must have $$ P\wedge Q\implies \{x=y\}$$ by negation $$ x\neq y\implies \bar{P}\vee\bar{Q}$$ Meaning that if $x$ is different than $y$, one of them at least must not be in a relation with the other.

But for instance, if you take $x=1$, and $y=-1$, you can find $n=2\in \mathbb Q$ such that $1=-1+2$. Same applies the other way, you can find $n=-2\in\mathbb Q$. Which is a contradiction as at least one of $\bar{P}$ and $\bar{Q}$ must be verified.

So the relation is not an order relation