Order of Real Numbers is Dual of Order of their Negatives/Proof 2
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Theorem
- $\forall x, y \in \R: x > y \iff \paren {-x}< \paren {-y}$
Proof
| \(\ds x\) | \(<\) | \(\ds y\) | ||||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds y - x\) | \(>\) | \(\ds 0\) | Inequality iff Difference is Positive | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \) | \(>\) | \(\ds 0\) | Real Number Axiom $\R \text A2$: Commutativity of Addition | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds -x + -\paren {-y}\) | \(>\) | \(\ds 0\) | Negative of Negative Real Number | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds -x - \paren {-y}\) | \(>\) | \(\ds 0\) | Definition of Real Subtraction | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds -y\) | \(<\) | \(\ds -x\) | Inequality iff Difference is Positive |
$\blacksquare$