# Strict Ordering on Integers is Transitive

## Theorem

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

Then:

 $\ds \eqclass {a, b} {}$ $<$ $\ds \eqclass {c, d} {}$ $\, \ds \land \,$ $\ds \eqclass {c, d} {}$ $<$ $\ds \eqclass {e, f} {}$ $\ds \implies \ \$ $\ds \eqclass {a, b} {}$ $<$ $\ds \eqclass {e, f} {}$

That is, strict 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 $a \prec b$ denote that the natural number $a$ is less than the natural number $b$.

We have:

 $\ds \eqclass {a, b} {}$ $<$ $\ds \eqclass {c, d} {}$ $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds a + d$ $\prec$ $\ds b + c$ Definition of Strict Ordering on Integers $\ds \eqclass {c, d} {}$ $<$ $\ds \eqclass {e, f} {}$ $\text {(2)}: \quad$ $\ds \leadsto \ \$ $\ds c + f$ $\prec$ $\ds d + e$ Definition of Strict Ordering on Integers $\ds \leadsto \ \$ $\ds a + d + f$ $\prec$ $\ds b + c + f$ adding $f$ to both sides of $(1)$ $\ds \leadsto \ \$ $\ds a + d + f$ $\prec$ $\ds b + d + e$ from $(2)$: $b + \paren {c + f} \prec b + \paren {d + e}$ $\ds \leadsto \ \$ $\ds a + f$ $\prec$ $\ds b + e$ subtracting $d$ from both sides $\ds \leadsto \ \$ $\ds \eqclass {a, b} {}$ $<$ $\ds \eqclass {e, f} {}$ Definition of Strict Ordering on Integers

$\blacksquare$