# Abstract Algebra ordered integral domains

I have a question about ordered integral domains. What do I need to do to prove that for $a \in D^p$, $-a \in D^p$, or $a = 0$ and $b \in D^p$, $-b \in D^p$ or $b = 0$ then

1) $\mathrm{abs}(a) \geq a \geq -\mathrm{abs}(a)$

and

2) $\mathrm{abs}(a) + \mathrm{abs}(b) \geq \mathrm{abs}(a+b)$.

Do I need to split each equation by switching the inequality sign?

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Is $D^p$ meant to be the positive class? It's usually denoted $D_+$ or $D^+$, not $D^p$ (which might be confused with the set of $p$th powers in $D$). – Arturo Magidin Apr 20 '11 at 14:08

1) Trivial if $\rm\ a = 0\:;\:$ else multiply $\rm\ 1 \ge a/|a| \ge -1\$ by $\rm\:|a|$

2) Add 1) to the same inequality for $\rm\:b\:$

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By definition, $x\geq y$ if and only if $x-y=0$ or $x-y$ is the positive class.

By definition, $\mathrm{abs}(a) = a$ if $a$ is in the positive class or $a=0$, and $\mathrm{abs}(a)=-a$ if $a$ is not in the positive class and not equal to $0$ (that is, if $-a$ is in the positive class).

So, for example, to prove (1), if $a=0$, then the inequalities hold. If $a$ is positive, then $a+a$ is positive (because the positive class is closed under sums), so $a - (-a)$ is positive, proving that $a\geq -a=-\mathrm{abs}(a)$; proving $\mathrm{abs}(a)\geq a$ when $a$ is positive is easy. Now you can consider the case where $-a$ is positive.

For (2), if either $a$ or $b$ are $0$, the inequality is easy. If $a$ and $b$ are both positive or both negative, the inequality is again easy. If $a+b=0$ then it's easy as well. Check what happens when $a$ is positive and $-b$ is positive (there are two cases, depending on whether $a+b$ is positive, or $-(a+b)$ is positive).

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Forgive my ignorance, but if for 1) -a ∈ the positive class won't the inequality abs(a) ≥ a ≥ -abs(a) = -a ≥ a ≥ --a, which isn't true? – user9808 Apr 21 '11 at 15:25
@kevin: If $-a$ is in the positive class, then $\mathrm{abs}(a) = -a$. In addition, $-a-a$ is positive (sum of two positive elements), so $-a\gt a$. And $-(-a) = a$, so $a=-\mathrm{abs}(a)$. So the first inequality becomes $-a\geq a\geq a$, which is true in this situation. – Arturo Magidin Apr 21 '11 at 16:57
@kevin: Don't use the equal sign to connect two distinct inequalities; when you write $\mathrm{abs}(a)\geq a\geq-\mathrm{abs}(a)=-a\geq a\geq --a$, you are asserting that $-\mathrm{abs}(a)=-a$, which is not the case here. And I don't understand what makes you think that $-a\geq a\geq -(-a)$ is false when $-a$ is positive. – Arturo Magidin Apr 21 '11 at 18:58