# 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:

- $x \times y = 0$

Therefore by Proof by Contradiction:

- $y \ne 0$ 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, $x \times y > 0$.

Therefore by Proof by Contradiction:

- $y > 0$

$\Box$

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, $x \times y > 0$.

Therefore by Proof by Contradiction:

- $y > 0$

$\Box$

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 $y > 0$.

Then from Product of Strictly Positive Real Numbers is Strictly Positive:

- $x \times y > 0$

$\Box$

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 $x, y < 0$:

- $x \times y > 0$

$\blacksquare$

## Sources

- 2000: James R. Munkres:
*Topology*(2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(h)}$