$\quad$The following assertion is somewhat less obvious: If $a\lt 0$ and $b\lt 0,$ then $ab\gt0$. The only difficulty presented by the proof is unraveling of definitions. The symbol $a\lt 0$ means, by definition, $0\gt a$, which means $0-a=-a$ in in $P$. Similarly $-b$ is in $P$, and consequently, by $\text{P12}$, $(-a)(-b)=ab$ is in $P$. Thus $ab\gt 0$.
Hi, I'm trying to write this proof in my own words with all complete justifications and have "hit" a wall. Tell me where is my error
Proof :
$a\lt 0$ by definition means $0\gt a$.
We'll now show(prove.) that $0\gt a$.
By one of our definitions : $a>b$ if $a-b$ is in $P$. ($P$ by the way is a collection of positive integers.)
So : $0-a=-a$
By property 10 (trichotomy law.), for every number a, one and only one of the following holds : a=0, a is in P, -a is in P
This is where I hit the wall. In a intuitive way, I would surely go with "-a is in P" but I said to myself what if a=0 ?? what if I was really doing 0-0=0 ??? I can't rely on 0>a because what if turned out to be false ?? This is exactly the inequality I'm trying to prove ! Any suggestions people ?
This above quote comes from Calculus by Spivak !
Thank you