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 $0$.
- $\forall x, y \in \R: x \times y > 0 \iff \paren {x, y \in \R_{>0} } \lor \paren {x, y \in \R_{<0} }$
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 $y < 0$.
| \(\ds x\) | \(>\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x \times y\) | \(<\) | \(\ds 0 \times 0\) | Real Number Ordering is Compatible with Multiplication: Negative Factor | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x \times y\) | \(<\) | \(\ds 0\) | 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 $y > 0$.
| \(\ds x\) | \(<\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x \times y\) | \(<\) | \(\ds 0 \times 0\) | Real Number Axiom $\R \text O2$: Usual Ordering is Compatible with Multiplication | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x \times y\) | \(<\) | \(\ds 0\) | Definition of Dual Ordering |
But by hypothesis, $x \times y > 0$.
Therefore by Proof by Contradiction:
- $y > 0$
$\Box$
Thus:
- $x \times y > 0 \implies \paren {x > 0 \land y > 0} \lor \paren {x < 0 \land y < 0}$
$\Box$
Necesssary Condition
Let $x > 0$ and $y > 0$.
Then from Strictly Positive Real Numbers are Closed under Multiplication:
- $x \times y > 0$
$\Box$
Let $x < 0$ and $y < 0$.
| \(\ds x\) | \(<\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x \times y\) | \(>\) | \(\ds 0 \times 0\) | Real Number Ordering is Compatible with Multiplication: Negative Factor | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x \times y\) | \(>\) | \(\ds 0\) | 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 $2 \ \text{(h)}$