Integer Multiplication Identity is One

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Theorem

The identity of integer multiplication is $1$:

$\exists 1 \in \Z: \forall a \in \Z: a \times 1 = a = 1 \times a$


Proof

Let us define $\eqclass {\tuple {a, b} } \boxtimes$ as in the formal definition of integers.

That is, $\eqclass {\tuple {a, b} } \boxtimes$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxtimes$.

$\boxtimes$ is the congruence relation defined on $\N \times \N$ by $\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$.


In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxtimes$, as suggested.


From the method of construction, $\eqclass {c + 1, c} {}$, where $c$ is any element of the natural numbers $\N$, is the isomorphic copy of $1 \in \N$.

To ease the algebra, we will take $\eqclass {1, 0} {}$ as a canonical instance of this equivalence class.


Thus it is to be shown that:

$\forall a, b \in \N: \eqclass {a, b} {} \times \eqclass {1, 0} {} = \eqclass {a, b} {} = \eqclass {1, 0} {} \times \eqclass {a, b} {}$


From Natural Numbers form Commutative Semiring, we take it for granted that:

addition and multiplication are commutative and associative on the natural numbers $\N$
natural number multiplication is distributive over natural number addition.


So:

\(\ds \eqclass {a, b} {} \times \eqclass {1, 0} {}\) \(=\) \(\ds \eqclass {1 \times a + 0 \times b, 0 \times a + 1 \times b} {}\)
\(\ds \) \(=\) \(\ds \eqclass {a, b} {}\) from Construction of Inverse Completion: Members of Equivalence Classes


So:

$\eqclass {a, b} {} \times \eqclass {1, 0} {} = \eqclass {a, b} {}$


The identity $\eqclass {a, b} {} = \eqclass {1, 0} {} \times \eqclass {a, b} {}$ is demonstrated similarly.

$\blacksquare$


Sources

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