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

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.

• Perhaps you should be thinking along the lines of: “$-x$” means “$x$ with its sign changed”. – Lubin Jan 17 at 2:55

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.
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., $$-a > 0$$.