Prove $-(a/b) = (-a)/b$ Prove $-(a/b) = (-a)/b$ where b is not zero. Using the field Axioms and ordered properties of Real numbers.
So I'm doing some supplementary exercise on my own and really got stuck in this.
Any help and insights is deeply appreciated.
 A: The definition of $-(a/b)$ is an element which you add to $a/b$ to get zero.  Now $$(-a/b) + a/b = (-a)b^{-1} + ab^{-1} = [(-a) +a]b^{-1}$$ Can you finish?
A: We have
$(-a) \cdot b^{-1} + a \cdot b^{-1} = [(-a) + a] \cdot b^{-1} = 0 \cdot b^{-1} = 0$
Then, by uniqueness, $(-a) \cdot b^{-1} = -(a \cdot b^{-1})$
A: $(-a)/b + a/b = (-a)*1/b + a*1/b$.  That is simply a matter of notation shortcut.  $x/y$ was defined to be $x*1/y$.
$(-a)*1/b + a*1/b = (-a + a)1/b $.  This is the distributive property.
$(-a + a)*1/b = 0*1/b $.  This is the definition of an additive inverse.
$0*1/b = 0$.  This is another exercise: prove $0*x = 0$ for all $x $.  That should be a proposition you've proven or seen proven.[1]
So $(-a)/b + a/b = 0$.  Thus $(-a)/b$ is the additive inverse of $a/b  $ which is, by field axioms unique and defined to be $-(a/b)$.
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[1]
$0*x = (0+0)*x = 0*x + 0*x $
Therefore.
$0=(0*x) - (0*x) = (0*x) + (0*x) - (0*x) = 0*x $.
So $0*x = 0$
A: $$-\frac{a}{b}=-1 \frac{a}{b}=\frac{-1 \cdot a}{b}=\frac{-a}{b}$$
