Ordering on Integers is Transitive

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

$\blacksquare$