Strict Ordering on Integers is Asymmetric

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Let $\eqclass {a, b} {}$ denote an integer, as defined by the formal definition of integers.


\(\ds \eqclass {a, b} {}\) \(<\) \(\ds \eqclass {c, d} {}\)
\(\, \ds \implies \, \) \(\ds \eqclass {c, d} {}\) \(\nless\) \(\ds \eqclass {a, b} {}\)

That is, strict ordering on the integers is asymmetric.


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