TITLE: Homework advice-- Ordered fields and the rationals
QUESTION [1 upvotes]: This is a homework problem, so do not give answers. Does anyone have any advice about how to approach this problem? I am not too sure how or where to start. Here is the problem:
Show that the rational numbers $\mathbb{Q}$ form an ordered field which satisfies the Archimedean property.  What property of real numbers $\mathbb{R}$ is not satisfied by $\mathbb{Q}$? 
The definition for an ordered field: An ordered field is a field $\mathbb{F}$ which is also an ordered set, such that
$(i)\hspace{10mm}  x+y < x + z  $  if $x,y,z \in \mathbb{F}$ and $y<z$
$ (ii) \hspace{10mm}xy >0$ if $x,y \in \mathbb{F}, x > 0, y > 0$
For the first part do I start by assuming $\mathbb{Q}$ is a field and an ordered set, then show that it satisfies the two axioms in the definition for an ordered field? Or am I supposed to show the set $\mathbb{Q}$ is ordered field by showing it satisfies the axioms for an ordered set, then show it satisfies the axioms of a field, then show it also satisfies the axioms in the definition for an ordered field? What I mean by show is let $x=m/n$ where $x\in\mathbb{Q}$ and $n,m \in \mathbb{Z},$ and $n \neq 0$, and  similarly for $y = k/h$ and $z = p/q$. What are your thoughts and what is your advice or suggestion? I still don't even know how I would approach the second part about the Archimedean Property. Here was an attempt at a proof (I suppose) that was completely random. I started with assuming $\mathbb{Q}$ is an ordered field. Then I have for $$ x + y < x + z \implies (m/n) + (k/h) < (m/n)+(p/q) \implies (m/n)-(m/n) + (k/h) < (m/n)-(m/n) + (p/q) \implies(k/h) < (p/q) \implies y < x$$ and it also implies $qk < hp$. I don't know what this would mean as far as the progression of a proof, but the inequality appears to be like the Archimedean Property, but it isn't necessarily so.

REPLY [1 votes]: Show that the rational numbers $\mathbb{Q}$ form an ordered field...

For the first part do I start by assuming $\mathbb{Q}$ is a field and an ordered set, then show that it satisfies the two axioms in the definition for an ordered field?

No, that isn't enough. Until you prove that $\mathbb Q$ is a field, it's conceivable that it isn't!
Consider a different example. The integers $\mathbb Z$ are an integral domain -- almost a field -- and an ordered set, and $\mathbb Z$ even obeys the axioms (i) and (ii). So if you weren't careful to check every single field axiom, you might conclude that $\mathbb Z$ is an ordered field, which is wrong!

Or am I supposed to show the set $\mathbb{Q}$ is ordered field by showing it satisfies the axioms for an ordered set, then show it satisfies the axioms of a field, then show it also satisfies the axioms in the definition for an ordered field?

Yes.


... which satisfies the Archimedean property.


You need to show that for all $p/q\in\mathbb Q$, there exists an integer $n$ such that $n>p/q$. The $n$ that you find should depend on $p$ and/or $q$ somehow. There are many possible choices.
The inequality that you displayed doesn't have much to do with the Archimedean property, as far as I can tell.
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