Informally, the integers are the complete set of whole numbers, including negative numbers such as $-1, -2, -3, \cdots$, zero, and the positive whole numbers $1, 2, 3, \cdots$. However, since we have not defined a negative number yet, this definition must be formalized a little.
The set of natural numbers $\naturals$ includes the additive identity element $0$. For any $n \in \naturals$, $n + 0 = n$.
The additive inverse of any elment $n$ is a number $m$ such that $n + m = 0$. We will represent this by $-n$. The additive inverse of the natural numbers is not in $\naturals$. To include the additive inverse, we must define a new type of number, the Integers.
The commonly used term “subtracting a number $n$“ is actually, more formally, “ adding the additive inverse of $n$”
Formally, the integers are a set comprising zero ($0$), the natural numbers ($1, 2, 3, \cdots$), and their additive inverses ($-1, -2, -3, \cdots$). The set of integers is commonly denoted by the symbols $\integers$.
The natural numbers are a subset of the integers. $\naturals \subset \integers$.
We define the positive integers as the integers $1, 2, 3, \cdots$. We will refer to the positive integers by $\integers_+$
To be very formal, the set of integers comprises a zero element $0$, a set of elements that have a one-to-one correspondence with $\naturals$, and their additive inverses.
From this perspective, it is not immediately apparent which of the integers are positive. We could, for instance call $-1, -2, \cdots$ the positive integers and $1, 2, \cdots$ their additive integers and everything would remain consistent. All this means is that the elements $-1, -2, -3, ..$ of the set $\integers$ are mapped to the naturals.
As a result, and to remain consistent with the natural numbers, we end up having to define the positive integers by diktat.
Arithmetic operations over the integers have most of the properties of the natural numbers.
As in the case of natural numbers, the addition $+$ operation is defined over integers. The following properties hold.
Integer operations differ from operations over naturals in at least one major category, namely exponentiation. $\naturals$ are closed under exponentiation: for any $n,m \in \naturals$, $n^m \in \naturals$, where $n^m$ is defined as $n \times n \times \cdots \times n (m {\rm times})$. However, this is not true for integers because of the presence of negative numbers: $n^m$ need not be in $\integers$ even if $n,m \in \integers$.
$\integers$ forms an algebraic structure called a ring. A ring is a set equipped with two operations $+$ (addition) and $\times$ (multiplication) such that
It is easy to verify that $\integers$ satisfies all the above conditions.
Note that rings do not include multiplicative inverses. The inclusion of multiplicative inverses converts the ring to a field. Multiplicative inverses are not defined over $\integers$, which too conforms to the definition of rings.
As in the case of naturals, we can define an ordering relation $\lt$ defined as: $n \lt m$ implies that $m - n$ is positive.
Key Fact
Note that this depend on the definition of positive numbers given earlier.
Equivalently, we can say that $n \lt m$ implies that there exists an $l \in \naturals_+$ such that $n + l = m$.
The fact that all positive numbers are greater than $0$ is actually an outcome of the above relation.
For any positive number $n$, $n - 0 = n$, and $n \in \integers_+$. Therefore $0 \lt n$.
$\integers$ is a totally ordered set: the order relationship has the property of trichotomy. For any $n,m \in \integers$, $n \lt m$ or $m \lt n$ or $n = m$.
However the integers are not well ordered since there is no least element to the integers. A least element of a set is an element $l$ such that $l \leq n$ for all elements $n$ in the set. Such an element doesn't exist for the integers. No matter how far we go to the negative end of the set, there is always another element beyond it.
Prove that