Ordering of Integers is Reversed by Negation
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Theorem
Let $x, y \in \Z$ such that $x > y$.
Then:
- $-x < -y$
Proof
From the formal definition of integers, $\eqclass {a, b} {}$ is an equivalence class of ordered pairs of natural numbers.
Let $x = \eqclass {a, b} {}$ and $y = \eqclass {c, d} {}$ for some $x, y \in \Z$.
We have:
\(\ds x\) | \(>\) | \(\ds y\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \eqclass {a, b} {}\) | \(>\) | \(\ds \eqclass {c, d} {}\) | Definition of Integer | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a + d\) | \(>\) | \(\ds b + c\) | Definition of Strict Ordering on Integers | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds b + c\) | \(<\) | \(\ds a + d\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \eqclass {b, a} {}\) | \(<\) | \(\ds \eqclass {d, c} {}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds -\eqclass {a, b} {}\) | \(<\) | \(\ds -\eqclass {c, d} {}\) | Negative of Integer | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds -x\) | \(<\) | \(\ds -y\) | Definition of Integer |
$\blacksquare$
Sources
- 1951: Nathan Jacobson: Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts ... (previous) ... (next): Introduction $\S 5$: The system of integers: Exercise $1$