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

  1. Oct 8, 2013 #1
    1. The problem statement, all variables and given/known data

    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.

    3. 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!
     
  2. jcsd
  3. Oct 8, 2013 #2

    haruspex

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