It all depends how much you're allowed to assume.  Most of the time, such a proof is done without having to prove the basic results of inequalities, for example, if a > 0, a^2 > 0.  Or even more basic, if a > 0, and b > 0, ab > 0.  But this may not be the case.  Anyways, here is a proof assuming you don't have to show the basic properties of inequalities.

Let a > 0.  Then a^2 > 0.  Thus, a/a^2 > 0.  So a/a^2 = 1/a > 0.

Ricky wrote:

...Thus, a/a^2 > 0...

Are you really allowed to say that here though? It seems to me that all this does is assume 1/a² > 0, which is the same concept as assuming 1/a > 0. It looks like you are using the statement we need to prove to prove the statement, if you know what i mean... kind of like defining a word and using the word in your definition.

Well, not exactly.  You would just first have to prove that:

If a > 0 and b > 0, then a/b > 0.

Which is what I would consider a basic property of inequalities.

*shrug*

A lot of times, inequality proofs such as these are fairly trivial.

I just saw this thread for the first time while browsing around. So let me add my comments.

The way to go about proving this is through the axioms for the real numbers. True, the real numbers can be built up from the rational numbers, which can be built up from the integers, which in turn can be built up from the natural numbers by the Peano axioms. (And you can go back even further by building the Peano axioms from set-theoretic axioms, particularly the axiom of infinity.) However, in an elementary textbook on analysis, one could bypass the complications and just start by axiomatizing the real numbers as a complete ordered field. (In other words, one merely takes the set of real numbers as “given”, defined by set of axioms which together uniquely characterize it.) The way to prove that a > 0 ⇒ a⁻¹ > 0 is then through the axioms for an ordered field.

First, note that (−x)y = x(−y) = −(xy). These can be easily proved using the distributive axiom. We then use the following axioms for an ordered field: (i) If 0 ≤ x and 0 ≤ y, then 0 ≤ xy. (ii) If x ≤ y, then x+z ≤ y+z.

Assume to the contrary that 0 < a but a⁻¹ ≤ 0, i.e. 0 ≤ −a⁻¹). Then 0 ≤ a·(−a⁻¹) = −1. But 0 ≤ −1 and 0 < a ⇒ 0 ≤ (−1)a = −a ⇒ a ≤ 0. This contradicts the fact that 0 < a. Hence it can’t be true that a⁻¹ ≤ 0; ∴ a⁻¹ > 0. (Note: An ordered field is a totally ordered set and so the law of trichotomy for totally ordered sets applies: Given any x. y, exactly one of the following holds: x < y, x = y, x > y.)

Well, as you said, it depends on exactly what you can assume – so I’m making clear exactly what is assumed when dealing with real numbers; in this case, it’s the axioms for an ordered field.

Anyway, as this is in the Euler Avenue section, I thought I could be more “theoretical” in my comments than in, say, the Help Me section.