I am trying to prove that $a >0$ if and only if $\frac{1}{a} >0$, But I am having a lot of trouble doing so.
So I would have to prove both sides of the implication. That is $a>0 \implies \frac{1}{a} >0$ and $\frac{1}{a} >0 \implies a>0$.
I think that in order to prove this I would have to show that
$a>0 \wedge b>0 \implies ab>0$,
$a <0 \wedge b<0 \implies ab>0$,
and $a<0 \wedge b>0 \implies ab<0$
And then I would have to show that the multiplicative inverse of a positive number is positive, and the multiplicative inverse of a negative number is negative. Would this be what I need to show in my proof?
I started my proof by assuming $a,b,c \in \Bbb {R}$, with $a>0$, $a>b$, and $c>0$. So then $a-b$ and $c-0$ would both be positive, and so $(a-b)(c-0)$ would be positive, by the axiom that states if $a$ and $b$ are positive then $ab$ is positive also. So then $c(a-b)=ca-cb$ also. And so $ca>cb$. So $ca>0$.
I am very confused if this is correct so far and what to do from here. How do I proceed with this proof? Thank you for any help you can give.
Hint: Prove that, if $a>0$ and $b<0$ then $ab<0$. Prove that $1>0$. Now suppose that $a>0$ and $\frac{1}{a} <0$