Given the axiom about real numbers that:
- $a>b$ and $c>0 \implies ac>bc$
and the following facts about real numbers:
- $0\cdot a = 0$
- $-(-a)=a$
- $a(-b)=(-a)b=-(ab)$
- $a>0\iff -a<0$
- $a>b$ and $c<0\implies ac<bc$
I want to show that $xy>0\iff (x<0$ and $y<0$) or ($x>0$ and $y>0$). I'm hoping folks can check my approach.
For the $xy>0 \implies$ '$x$ and $y$ have the same sign' direction, I proceeded by proving the contrapositive: $$\text{if } (x\ge0 \text{ or } y\geq0)\text{ and }(x\leq0\text{ or }y\leq0), \text{ then } xy\leq0$$ The part of the solution I'm concerned with is the following:
This reduces to proving that if one or both of $x$ and $y$ are zero, then $xy=0$ or if w.l.o.g. $y<0<x$, then $xy<0$.
Clearly by fact $(2)$ we have $xy=0 $ if one or both of $x$ and $y$ are zero. If $y<0<x$ (we could have chosen $x<0<y)$, then letting $a=x$, $b=0$, and $c=y$ in fact $(6)$, we have $xy<0\cdot0=0$ so $xy<0$. So in any case, we have $xy\leq0$.
For the other direction, if $(x<0$ and $y<0$) or ($x>0$ and $y>0$) then $xy>0$, first assume $x<0$ and $y<0$. Then $-x>0$ and $-y>0$ by $(5)$. Then $(-x)(-y)=(-(-x))y=xy>0\cdot0=0$; where the first equality follows from $(4)$, the second equality from $(3)$, and the inequality from letting $a=-x$, $b=0$, and $c=-y$ in $(1)$. Lastly, if both $x$ and $y$ are positive, then by $(1)$ again letting $a=x$, $b=0$, and $c=y$ we have $xy>0 \cdot0=0$. Thus we have in either case, $xy>0$.
Hence $xy>0 \iff x$ and $y$ have the same sign.