I'm trying to prove that for all $a\in\Bbb{R}$ if $a>0$ then $a^{-1}>0$
I'm going for a direct proof, via axioms of real numbers.
For every $a\in\Bbb{R}$, there must be a additive inverse such that $$a+(-a)=0, a\in\Bbb{R}$$
This means that $-a<0$, if $-a+a<a$ ... $a>0$
Now $-a$ must have a multiplicative inverse such that $$-a\cdot (-a)^{-1}=1$$
now $-a^{-1}<0$, it follows (...additive inverse ) that $$-a^{-1}+a^{-1}<a^{-1}$$$$0<a^{-1}$$$$a^{-1}>0$$
What are the flaws in this proof?
Before you attempt to prove something is positive, it might help if you begin with axioms which define what 'positive' means. I have an Analysis text which gives the following axioms:
There is a subset of the Real Numbers called 'positive' which satisfies the following properties: (1) For every Real Number x, either x is positive or x is $0$ or -x is positive (and only one of these conditions holds); (2) Any sum or product of finitely many positive numbers is positive.
[Note: It then defines '$a > b$' as '($a - b$) is positive']
From this we can first show that 1 is positive and -1 is not (exercise left for the student). Afterwards we assume $a$ is positive, hence NOT '$a$ is $0$', so $a^{-1}$ exists and $a \times a^{-1} = 1$ (by definition). By condition (1), either $a^{-1}$ is positive or $a^{-1}$ is $0$ or $-(a^{-1})$ is positive. If $a^{-1}$ is $0$ then by (2), $a \times a^{-1} = a \times 0 = 0$, contradicting $a \times a^{-1} = 1$; if $-(a^{-1})$ is positive, by (2) $a \times -(a^{-1}) = -(a \times a^{-1}) = -1$ is positive, which it is not. By exclusivity the only remaining possibility is $a^{-1}$ is positive. QED