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

Question

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.

Answer 1

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.

Answer 2

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., $-a > 0$.