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Positivity Axioms
Let $\mathscr{P}$ be the set of positive real numbers.
(i) For $a,b \in \mathscr{P}$, $a+b \in \mathscr{P}$ and $ab \in \mathscr{P}$
(ii) For $a$ a real number only one of the three is true: $a \in \mathscr{P}$, $-a \in \mathscr{P}$, or $a=0$.

I want to show that for $a and $c, $a+c < b+d$.

Since $a, then $b-a \in \mathscr{P}$ and $d-c \in \mathscr{P}$.
Then $(b-a)+(d-c) \in \mathscr{P}$.

Ideally I would want to say: $(b+d)-(a+c) \in \mathscr{P}$. So $a+c.
Is this proof correct?

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    @peoplepower I think Fitzpatrick rearranged Royden to have the positivity axioms come first. It's odd since we are talking about elements in $\mathbb{R}$ - so I should be able to use my knowledge of fields.2012-10-18

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