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 $0$.


 * $\forall x, y \in \R: x \times y > 0 \iff \left({x, y \in \R_{>0}}\right) \lor \left({x, y \in \R_{<0}}\right)$

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

Aiming for a contradiction, suppose 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 $x > 0$.

Aiming for a contradiction, suppose $y < 0$.

But by hypothesis, $x \times y > 0$.

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

Let $x < 0$.

Aiming for a contradiction, suppose $y > 0$.

But by hypothesis, $x \times y > 0$.

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

Thus:
 * $x \times y > 0 \implies \left({x > 0 \land y > 0}\right) \lor \left({x < 0 \land y < 0}\right)$

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

Then from Product of Strictly Positive Real Numbers is Strictly Positive:
 * $x \times y > 0$

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

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