Strict Ordering on Integers is Asymmetric
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Theorem
Let $\eqclass {a, b} {}$ denote an integer, as defined by the formal definition of integers.
Then:
| \(\ds \eqclass {a, b} {}\) | \(<\) | \(\ds \eqclass {c, d} {}\) | ||||||||||||
| \(\, \ds \implies \, \) | \(\ds \eqclass {c, d} {}\) | \(\nless\) | \(\ds \eqclass {a, b} {}\) |
That is, strict ordering on the integers is asymmetric.
Proof
By the formal definition of integers, we have that $a, b, c, d, e, f$ are all natural numbers.
To eliminate confusion between integer ordering and the ordering on the natural numbers, let $a \prec b$ denote that the natural number $a$ is less than the natural number $b$.
We have:
| \(\ds \eqclass {a, b} {}\) | \(<\) | \(\ds \eqclass {c, d} {}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a + d\) | \(\prec\) | \(\ds b + c\) | Definition of Strict Ordering on Integers | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds b + c\) | \(\nprec\) | \(\ds a + d\) | Definition of Ordering on Natural Numbers | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \eqclass {c, d} {}\) | \(\nless\) | \(\ds \eqclass {a, b} {}\) | Definition of Strict Ordering on Integers |
$\blacksquare$
Sources
- 1951: Nathan Jacobson: Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts ... (previous) ... (next): Introduction $\S 5$: The system of integers