How do we prove : .
Proof:
Let and suppose ,then by trichotomy law we have :
a<0 or a-0
For a<0 => ,hence contradiction ,since 1>0
For a=0,how do we cope with this case??
Given $\dfrac 1 a > 0$
suppose $a < 0$ then
$a = -|a|$
$\dfrac 1 a = \dfrac {1}{-|a|}=-\dfrac{1}{|a|}=-\left|\dfrac 1 a\right|$
by definition
$\left|\dfrac 1 a\right| > 0$
so
$-\left|\dfrac 1 a\right| < 0$
and thus
$\dfrac 1 a < 0$
but we're given that
$\dfrac 1 a > 0$
and thus we have a contradiction so it must be that
$a > 0$
The product of two positive numbers (and so the quotient of two positive numbers) gives a positive number.
The product of two negative numbers (and so the quotient of two negative numbers) gives a negative number.
The product of two numbers of different sign (and so the quotient of two numbers of different sign) gives a negative number.
Since your quantity 1/a is positive, that means the two numbers must be of the same sign. Since 1 is positive, so must be a. Q.E.D.
You don't need it. If there exist x such that then . But 0 times any number is 0 so that 1/x exist only if "0" is the only number in your set (and most field axioms include the axiom that the set of "numbers" contains more than one number). In any field, 0 has no multiplicative inverse: the is NO .