Product of Real Numbers is Positive iff Numbers have Same Sign

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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