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$ \textbf{Question} $ (From Mendelson's Intro to Topology): Let $ P \subset \mathbb{R} $ such that (i) $ 1 \in P $, (ii) if $ a, b \in P $ then $ a + b \in P $, and (iii) for each $ x \in \mathbb{R} $, one and only one of the three statements, $ x \in P, \; x = 0, $ or $ -x \in P $ is true. Let $ Q = \{ (a, b) \mid (a, b) \in \mathbb{R} \times \mathbb{R} $ and $ a - b \in P \} $. Prove that $ Q $ is a transitive relation.

My approach for this problem is to argue that if $ (x, y) $ and $ (y, z) $ are in $ Q $, then $ (x, z) $ is in $ Q $ as well. Since $ (x, y) \in Q, \; x - y \in P $ and similarly $ y - z \in P $. Hence $ x - z = (x - y) + (y - z) \in Q $, so $ Q $ is transitive, but I didn't use any of the information about the set $ P $ with properties (i) and (iii). Am I missing something?

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    You have a small typo. It should be "$ x - z = (x - y) + (y - z) \in P$". But otherwise, yes, it is correct. Maybe you need the other properties for something else? – 2017-02-06
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    $Q$ is a strict total order thanks to (iii) (and (ii)), while the purpose of (i) should be to decide which "half" of $\mathbb{R}$ is $P$. All the positive rationals are easily seen to be in $P$. – 2017-02-06

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