Integer Addition Identity is Zero

From ProofWiki
Jump to navigation Jump to search

Theorem

The identity of integer addition is $0$:

$\exists 0 \in \Z: \forall a \in \Z: a + 0 = a = 0 + a$


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, $\eqclass {c, c} {}$, where $c$ is any element of the natural numbers $\N$, is the isomorphic copy of $0 \in \N$.


So, we need to show that:

$\forall a, b, c \in \N: \eqclass {a, b} {} + \eqclass {c, c} {} = \eqclass {a, b} {} = \eqclass {c, c} {} + \eqclass {a, b} {}$


Thus:

\(\displaystyle \eqclass {a, b} {} + \eqclass {c, c} {}\) \(=\) \(\displaystyle \left[\!\left[{a + c, b + c}\right]\!\right]\)
\(\displaystyle \) \(=\) \(\displaystyle \eqclass {a, b} {}\) Construction of Inverse Completion: Members of Equivalence Classes


So:

$\eqclass {a, b} {} + \eqclass {c, c} {} = \eqclass {a, b} {}$


The identity $\eqclass {a, b} {} = \eqclass {c, c} {} + \eqclass {a, b} {}$ is demonstrated similarly.

$\blacksquare$


Sources