# Inverses for Integer Addition

## Theorem

Each element $x$ of the set of integers $\Z$ has an inverse element $-x$ under the operation of integer addition:

- $\forall x \in \Z: \exists -x \in \Z: x + \paren {-x} = 0 = \paren {-x} + x$

## 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, the element $\eqclass {a, a + x} {}$ has an inverse $\eqclass {a + x, a} {}$ where $a$ and $x$ are elements of the natural numbers $\N$.

Thus:

\(\displaystyle \eqclass {a, a + x} {} + \eqclass {a + x, a} {}\) | \(=\) | \(\displaystyle \eqclass {a + a + x, a + x + a} {}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \eqclass {a, a} {}\) | Construction of Inverse Completion: Members of Equivalence Classes | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \eqclass {a + x + a , a + a + x} {}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \eqclass {a + x, a} {} + \eqclass {a, a + x} {}\) |

So $\eqclass {a, a + x} {}$ has the inverse $\eqclass {a + x, a} {}$.

$\blacksquare$

## Sources

- 1951: Nathan Jacobson:
*Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts*... (previous) ... (next): Introduction $\S 5$: The system of integers - 1964: W.E. Deskins:
*Abstract Algebra*... (previous) ... (next): $\S 2.5$: Theorem $2.25$ - 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: $\mathbf Z. \, 4$ - 2008: Paul Halmos and Steven Givant:
*Introduction to Boolean Algebras*... (previous) ... (next): $\S 1$