# Total Ordering and relation

Consider the relation $<$ on $\mathbb{Q}$ defined by: $(m, n) < (j, k) \iff jn-mk \in \mathbb{N}.$

Where $m, j \in \mathbb{N}$ and $n, k\in\mathbb{Z}$

I want to show that $<$ is a total order.

I have showed $\forall_x\in\mathbb{Q}$ that $<$ is transitive and $x\nless x$. That is equivalent to anti-symmetry. I have also showed that the relation is well-defined.

So the last thing I need to show is trichotomy: ($x<y$ or $y<x$ or $y=x$).

So we can satisfy the relation or not. If $jn-mk \in \mathbb{N}$ then we are done. Also, note that if this is the case than $mk-jk\notin\mathbb{N}$

Next, check if $mk-jn \in \mathbb{N}$ if so we are done. Similar to the above if this is the case than $jn-mk \notin \mathbb{N}$ Finally, the last case is that $jn=mk$. Note, in this case the relation cannot be satisfied. This is all three cases and they are mutually disjoint.

1. Is my reasoning to show trichotomy correct?

EDIT I am skipping the typing required to show the details here, and am mainly interested in feedback on my trichotomy reasoning.

All help is greatly appreciated!

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You need to begin by saying exactly what you mean by $\Bbb Q$. –  Brian M. Scott Feb 14 '13 at 3:54
Writing rational numbers as ordered pairs is a bit odd. One thing you need to do is prove it is well-defined - a rational number can be represented by more than one pair of integers. Also, your definition does not work unless in the pairs $(m,n)$ and $(j,k)$, you have the $n$ and $k$ positive –  Thomas Andrews Feb 14 '13 at 4:05
Sorry, I left out some details. I have already shown that the relation is well-defined and I just made an edit to specify what sets the $m,n,j,k$ belong to. –  CodeKingPlusPlus Feb 14 '13 at 4:26
If your $\Bbb Q$ consists of ordered pairs as you’ve written them in your question, then $<$ is not a total order. If your $\Bbb Q$ consists of equivalence classes of ordered pairs, as I suspect it does, then your notation is wrong, and you need to rewrite everything in terms of equivalence classes. –  Brian M. Scott Feb 14 '13 at 4:33
I'll assume that there's a typo and you meant to write $mk-jn\in\mathbb N$ instead of $mk-jk\in\mathbb N$.
Apart from that, your reasoning is OK (though perhaps missing some details) up to the point where you've concluded that $x\lt y$ and $y\lt x$ correspond to $jn-mk\in\mathbb N$ and $mk-jn\in\mathbb N$, respectively, that these are mutually exclusive and that neither of them obtaining leaves only the possibility $mk=jn$. However, "Note, in this case the relation cannot be satisfied." doesn't prove anything; it's true by construction, and it's not what you need to show – what you need to show is that neither of those two cases obtains if and only if $x=y$. So you need to argue that $mk=jn$ is equivalent to $x=y$. But that equation is by definition the condition for $(m,n)$ and $(j,k)$ to be equivalent, so assuming that you had implicitly introduced $x$ and $y$ as having representatives $(m,n)$ and $(j,k)$, respectively (which is one of the missing details), you're done.