# Integer Multiplication has Zero

## Theorem

The set of integers under multiplication $\struct {\Z, \times}$ has a zero element, which is $0$.

## Proof

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

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

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

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

From the method of construction, $\eqclass {c, c} {}$, where $c$ is any element of the natural numbers $\N$, is the identity of $\struct {\Z, +}$.

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

We need to show that:

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

From Natural Numbers form Commutative Semiring, we can take it for granted that addition and multiplication are commutative on the natural numbers $\N$.

\(\ds \eqclass {a, b} {} \times \eqclass {0, 0} {}\) | \(=\) | \(\ds \eqclass {a \times 0 + b \times 0, a \times 0 + b \times 0} {}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \eqclass {0, 0} {}\) | Construction of Inverse Completion: Equivalence Class of Equal Elements | |||||||||||

\(\ds \) | \(=\) | \(\ds \eqclass {0 \times a + 0 \times b, 0 \times a + 0 \times b} {}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \eqclass {0, 0} {} \times \eqclass {a, b} {}\) |

$\blacksquare$

## Sources

- 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: Exercise $3$