Prove that a set with two properties has to be the positive real set

[Perrault]

Homework Statement

Let S ⊆ R be such that
(i) a, b ∈ S ⇒ ab, a + b ∈ S
(ii) for all x ∈ R exactly one of the following holds
x ∈ S, x = 0, −x ∈ S.

Show that S = {x ∈ R ; x > 0} (the set of positive numbers P)
2. Relevant theorems

(T1) a² > 0 ∀ a ∈ R. (So a²∈P)

(T2) All positive real numbers have a square root.

The Attempt at a Solution

This has been tantalizing me for hours. I want to prove that S⊆P and that P⊆S so that S=P. Trying to prove that S⊆P, I assume a∈S. Then a²∈S, by property (i), and a²∈P (by theorem T1). If we assume that a∈S is negative then we should be able to arrive at a contradiction, but I can't get to it! I understand that a and a² would be of different sign, but that doesn't seem to help me much.

I did show that if 1 is in S then it is in P and vice-versa. From there I am able to obtain the same thing for all positive integers (by property (i), if 1∈S then 1+1∈S so 2+1∈S, ...)

Anyone got any clue?

Thank you so much!

 

[haruspex]

If a < 0, consider sqrt(-a).