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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
up vote 1 down vote accepted

I'll assume that there's a typo and you meant to write $mk-jn\in\mathbb N$ instead of $mk-jk\in\mathbb N$.

First, note that you've misstated the trichotomy condition – it's not that at least one of those relations holds, but that exactly one of them holds.

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.

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