Product of Real Numbers is Positive iff Numbers have Same Sign

Theorem
The product of two real numbers is greater than $0$ either both are greater than $0$ or both are less than $0$.


 * $\forall x, y \in \R: x \times y > 0 \iff \paren {x, y \in \R_{>0} } \lor \paren {x, y \in \R_{<0} }$

Sufficient Condition
Let $x \times y > 0$.

either $x = 0$ or $y = 0$.

Then from Real Zero is Zero Element:
 * $x \times y = 0$

Therefore by Proof by Contradiction:
 * $y \ne 0$ and $x \ne 0$

Let $x > 0$.

$y < 0$.

But, $x \times y > 0$.

Therefore by Proof by Contradiction:
 * $y > 0$

Let $x < 0$.

$y > 0$.

But, $x \times y > 0$.

Therefore by Proof by Contradiction:
 * $y > 0$

Thus:
 * $x \times y > 0 \implies \paren {x > 0 \land y > 0} \lor \paren {x < 0 \land y < 0}$

Necesssary Condition
Let $x > 0$ and $y > 0$.

Then from Strictly Positive Real Numbers are Closed under Multiplication:
 * $x \times y > 0$

Let $x < 0$ and $y < 0$.

Thus if either $x, y > 0$ or $x, y < 0$:
 * $x \times y > 0$