Ordering on Integers is Transitive
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Theorem
Let $\eqclass {a, b} {}$ denote an integer, as defined by the formal definition of integers.
Then:
\(\ds \eqclass {a, b} {}\) | \(\le\) | \(\ds \eqclass {c, d} {}\) | ||||||||||||
\(\, \ds \land \, \) | \(\ds \eqclass {c, d} {}\) | \(\le\) | \(\ds \eqclass {e, f} {}\) | |||||||||||
\(\ds \implies \ \ \) | \(\ds \eqclass {a, b} {}\) | \(\le\) | \(\ds \eqclass {e, f} {}\) |
That is, ordering on the integers is transitive.
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 $a \preccurlyeq b$ denote that the natural number $a$ is less than or equal to the natural number $b$.
We have:
\(\ds \eqclass {a, b} {}\) | \(\le\) | \(\ds \eqclass {c, d} {}\) | ||||||||||||
\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds a + d\) | \(\preccurlyeq\) | \(\ds b + c\) | Definition of Ordering on Integers | |||||||||
\(\ds \eqclass {c, d} {}\) | \(<\) | \(\ds \eqclass {e, f} {}\) | ||||||||||||
\(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds c + f\) | \(\preccurlyeq\) | \(\ds d + e\) | Definition of Ordering on Integers | |||||||||
\(\ds \leadsto \ \ \) | \(\ds a + d + f\) | \(\preccurlyeq\) | \(\ds b + c + f\) | adding $f$ to both sides of $(1)$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a + d + f\) | \(\preccurlyeq\) | \(\ds b + d + e\) | from $(2)$: $b + \paren {c + f} \preccurlyeq b + \paren {d + e}$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a + f\) | \(\preccurlyeq\) | \(\ds b + e\) | subtracting $d$ from both sides | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \eqclass {a, b} {}\) | \(\le\) | \(\ds \eqclass {e, f} {}\) | Definition of Ordering on Integers |
$\blacksquare$