# Product of Real Numbers is Positive iff Numbers have Same Sign

## Theorem

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

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

## Proof

### 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:

$x \times y = 0$

$y \ne 0$ and $x \ne 0$

$\Box$

Let $x > 0$.

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

 $\displaystyle x$ $>$ $\displaystyle 0$ $\displaystyle \implies \ \$ $\displaystyle x \times y$ $<$ $\displaystyle 0 \times 0$ Order of Real Numbers is Dual of Order Multiplied by Negative Number $\displaystyle \implies \ \$ $\displaystyle x \times y$ $<$ $\displaystyle 0$ Definition of Dual Ordering

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

$y > 0$

$\Box$

Let $x < 0$.

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

 $\displaystyle x$ $<$ $\displaystyle 0$ $\displaystyle \implies \ \$ $\displaystyle x \times y$ $<$ $\displaystyle 0 \times 0$ Real Number Axioms: $\R O2$: compatibility with multiplication $\displaystyle \implies \ \$ $\displaystyle x \times y$ $<$ $\displaystyle 0$ Definition of Dual Ordering

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

$y > 0$

$\Box$

Thus:

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

$\Box$

### Necesssary Condition

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

$x \times y > 0$

$\Box$

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

 $\displaystyle x$ $<$ $\displaystyle 0$ $\displaystyle \implies \ \$ $\displaystyle x \times y$ $>$ $\displaystyle 0 \times 0$ Order of Real Numbers is Dual of Order Multiplied by Negative Number $\displaystyle \implies \ \$ $\displaystyle x \times y$ $>$ $\displaystyle 0$ Definition of Dual Ordering

$\Box$

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

$x \times y > 0$

$\blacksquare$