Integers under Addition form Abelian Group

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The set of integers under addition $\struct {\Z, +}$ forms an abelian group.


From the definition of the integers, the algebraic structure $\struct {\Z, +}$ is an isomorphic copy of the inverse completion of $\struct {\N, +}$.

From Natural Numbers under Addition form Commutative Semigroup, $\struct {\N, +}$ is a commutative semigroup.

From Natural Number Addition is Cancellable all elements of $\struct {\N, +}$ are cancellable.

The result follows from Inverse Completion of Commutative Semigroup is Abelian Group.

Thus addition on $\Z$ is well-defined, closed, associative and commutative on $\Z$.


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.

Identity is Zero

From Construction of Inverse Completion: Identity of Quotient Structure, the identity of $\struct {\Z, +}$ is $\eqclass {c, c} {}$ for any $c \in \N$:

\(\ds \forall a, b, c \in \N: \, \) \(\ds \) \(\) \(\ds \eqclass {a, b} {} + \eqclass {c, c} {}\)
\(\ds \) \(=\) \(\ds \eqclass {a, b} {}\)
\(\ds \) \(=\) \(\ds \eqclass {c, c} {} + \eqclass {a, b} {}\)

$\eqclass {c, c} {}$ is the equivalence class of pairs of elements $\N \times \N$ whose difference is zero.

Thus the identity of $\struct {\Z, +}$ is seen to be $0$.

Note that a perfectly good representative of $\eqclass {c, c} {}$ is $\eqclass {0, 0} {}$.

This usually keeps to a minimum the complexity of any arithmetic that is needed.


Construction of Inverses

From Construction of Inverse Completion: Invertible Elements in Quotient Structure, we see that every element of $\struct {\Z, +}$ has an inverse.

We can see that:

\(\ds \forall a, b \in \N: \, \) \(\ds \eqclass {a, b} {} + \eqclass {b, a} {}\) \(=\) \(\ds \eqclass {a + b, b + a} {}\)
\(\ds \) \(=\) \(\ds \eqclass {a + b, a + b} {}\) Natural Number Addition is Commutative

The above construction is valid because $a$ and $b$ are both in $\N$ and hence cancellable.

From Construction of Inverse Completion: Identity of Quotient Structure, $\eqclass {a + b, a + b} {}$ is a member of the equivalence class which is the identity of $\struct {\Z, +}$.

Thus the inverse of $\eqclass {a, b} {}$ is $\eqclass {b, a} {}$.