Order of Real Numbers is Dual of Order Multiplied by Negative Number

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Theorem

$\forall x, y, z \in \R: x > y, z < 0 \implies x \times z < y \times z$


Proof

Let $z < 0$.

Then from Real Number is Greater than Zero iff its Negative is Less than Zero:

$-z > 0$

and so:

\(\displaystyle x\) \(>\) \(\displaystyle y\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x \times \paren {-z}\) \(>\) \(\displaystyle y \times \paren {-z}\) Real Number Axioms: $\R O2$: compatibility with multiplication
\(\displaystyle \leadsto \ \ \) \(\displaystyle -\paren {x \times z}\) \(>\) \(\displaystyle -\paren {y \times z}\) Multiplication by Negative Real Number
\(\displaystyle \leadsto \ \ \) \(\displaystyle x \times z\) \(<\) \(\displaystyle y \times z\) Order of Real Numbers is Dual of Order of their Negatives

$\blacksquare$


Sources