# Multiplication of Cross-Relation Equivalence Classes on Natural Numbers is Well-Defined

## Theorem

Let $\struct {\N, +}$ be the semigroup of natural numbers under addition.

Let $\struct {\N \times \N, \oplus}$ be the (external) direct product of $\struct {\N, +}$ with itself, where $\oplus$ is the operation on $\N \times \N$ induced by $+$ on $\N$.

Let $\boxtimes$ be the cross-relation defined on $\N \times \N$ by:

$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$

Let $\eqclass {x, y} {}$ denote the equivalence class of $\tuple {x, y}$ under $\boxtimes$.

Let $\otimes$ be the binary operation defined on these equivalence classes as:

$\forall \eqclass {a, b} {}, \eqclass {c, d} {} \in \N \times \N: \eqclass {a, b} {} \otimes \eqclass {c, d} {} = \eqclass {\tuple {a \cdot c} + \tuple {b \cdot d}, \tuple {a \cdot d} + \tuple {b \cdot c} } {}$

where $a \cdot c$ denotes natural number multiplication between $a$ and $c$.

The operation $\otimes$ on these equivalence classes is well-defined, in the sense that:

 $\ds \eqclass {a_1, b_1} {}$ $=$ $\ds \eqclass {a_2, b_2} {}$ $\ds \eqclass {c_1, d_1} {}$ $=$ $\ds \eqclass {c_2, d_2} {}$ $\ds \leadsto \ \$ $\ds \eqclass {a_1, b_1} {} \otimes \eqclass {c_1, d_1} {}$ $=$ $\ds \eqclass {a_2, b_2} {} \otimes \eqclass {c_2, d_2} {}$

## Proof

Let $\eqclass {a_1, b_1} {}, \eqclass {a_2, b_2} {}, \eqclass {c_1, d_1} {}, \eqclass {c_2, d_2} {}$ be $\boxtimes$-equivalence classes such that $\eqclass {a_1, b_1} {} = \eqclass {a_2, b_2} {}$ and $\eqclass {c_1, d_1} {} = \eqclass {c_2, d_2} {}$.

Then:

 $\ds \eqclass {a_1, b_1} {}$ $=$ $\ds \eqclass {a_2, b_2} {}$ Definition of Operation Induced by Direct Product $\, \ds \land \,$ $\ds \eqclass {c_1, d_1} {}$ $=$ $\ds \eqclass {c_2, d_2} {}$ Definition of Operation Induced by Direct Product $\ds \leadstoandfrom \ \$ $\ds a_1 + b_2$ $=$ $\ds a_2 + b_1$ Definition of Cross-Relation $\, \ds \land \,$ $\ds c_1 + d_2$ $=$ $\ds c_2 + d_1$ Definition of Cross-Relation

Both $+$ and $\times$ are commutative and associative on $\N$. Thus:

 $\ds \paren {a_1 \cdot c_1 + b_1 \cdot d_1 + a_2 \cdot d_2 + b_2 \cdot c_2}$ $+$ $\ds \paren {a_2 \cdot c_1 + b_2 \cdot c_1 + a_2 \cdot d_1 + b_2 \cdot d_1}$ $\, \ds = \,$ $\ds \paren {a_1 + b_2} \cdot c_1 + \paren {b_1 + a_2} \cdot d_1$ $+$ $\ds a_2 \cdot \paren {c_1 + d_2} + b_2 \cdot \paren {d_1 + c_2}$ $\, \ds = \,$ $\ds \paren {b_1 + a_2} \cdot c_1 + \paren {a_1 + b_2} \cdot d_1$ $+$ $\ds a_2 \cdot \paren {d_1 + c_2} + b_2 \cdot \paren {c_1 + d_2}$ as $a_1 + b_2 = b_1 + a_2, c_1 + d_2 = d_1 + c_2$ $\, \ds = \,$ $\ds b_1 \cdot c_1 + a_2 \cdot c_1 + a_1 \cdot b_2 + a_1 \cdot d_1$ $+$ $\ds a_2 \cdot d_1 + a_2 \cdot c_2 + b_2 \cdot c_1 + b_2 \cdot d_2$ $\, \ds = \,$ $\ds \paren {a_1 \cdot d_1 + b_1 \cdot c_1 + a_2 \cdot c_2 + b_2 \cdot d_2}$ $+$ $\ds \paren {a_2 \cdot c_1 + b_2 \cdot c_1 + a_2 \cdot d_1 + b_2 \cdot d_1}$

So we have $a_1 \cdot c_1 + b_1 \cdot d_1 + a_2 \cdot d_2 + b_2 \cdot c_2 = a_1 \cdot d_1 + b_1 \cdot c_1 + a_2 \cdot c_2 + b_2 \cdot d_2$ and so, by the definition of $\boxtimes$, we have:

$\eqclass {a_1 \cdot c_1 + b_1 \cdot d_1, a_1 \cdot d_1 + b_1 \cdot c_1} {} = \eqclass {a_2 \cdot c_2 + b_2 \cdot d_2, a_2 \cdot d_2 + b_2 \cdot c_2} {}$

So, by the definition of integer multiplication, this leads to:

$\eqclass {a_1, b_1} {} \otimes \eqclass {c_1, d_1} {} = \eqclass {a_2, b_2} {} \otimes \eqclass {c_2, d_2} {}$

Thus integer multiplication has been shown to be well-defined.

$\blacksquare$