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Let $a$ be a real number. If $a$ is positive, then $-a$ is negative. Conversely, if $a$ is negative, then $-a$ is positive.

Having a hard time with intuition and the obvious answer getting in the way of my thought process. We are supposed to "verify" the statement above, given Axiom II -- the order axioms.

Since $a\in\mathbb{R}^{+}$, then $-(-a)$ is positive for all $a\in\mathbb{R}^{+}$.

Something about $\mathbb{R}^{+}$ is confusing me.

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  • $\begingroup$ Perhaps you should be thinking along the lines of: “$-x$” means “$x$ with its sign changed”. $\endgroup$ – Lubin Jan 17 at 2:55
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  1. By definition, if $a$ is positive then $a>0$.
  2. By the order axioms, we have that $a+(-a)>0+(-a)$
  3. This simplifies to $0>-a$ (by definition of additive inverse and additive identity)
  4. Therefore $-a<0$ and by definition $-a$ is negative.
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  • $\begingroup$ Thank you! I wasn't sure if we could use Field Axioms, but that was my own brain fart. $\endgroup$ – Ryan Jan 22 at 17:41
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According to the order axioms, if $a > b$ then $c + a > c + b$.

Thus if $a > 0$ then $-a + a > -a + 0$, i.e., $0 > -a$, i.e., $-a < 0$.

Conversely, if $a < 0$ then $-a + a < -a + 0$, i.e., $0 < -a$, i.e., $-a > 0$.

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