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Section 3.1 The integers

We have defined the set \(\NN_0\) of natural numbers

\begin{equation*} 0, 1, 2, 3, 4, 5, 6, 7, 8, \ldots\text{,} \end{equation*}

but these don't have all the numbers we want. We'd like to be able to solve the equation \(m+1=0\text{,}\) but of course there is no \(m \in\NN_0\) such that \(m+1=0\text{.}\)

To fix this, we'll need to add some new numbers – the negative numbers. Together, these will define the integers.

How should these be defined? We want numbers that are differences \(m-n\) of natural numbers \(m,n \in \NN_0\text{,}\) but we have to accept the fact that there are situations in which \(m_1 \neq m_2\) and \(n_1 \neq n_2\text{,}\) but nevertheless \(m_1-n_1 = m_2-n_2\text{.}\)

So we could, if we wanted, start with the set \(\NN_0 \times \NN_0\) consisting of ordered pairs of natural numbers \((m,n)\text{,}\) which we think of as representing the difference \(m-n\text{,}\) but we have to cope with this ambiguity.

Definition 3.1.1. The integers.

Define an equivalence relation \(\sim\) on the set \(\NN_0\times \NN_0\) as follows: for any pairs \((m_1,n_1),(m_2,n_2)\in\NN_0\times\NN_0\text{,}\) we declare that \((m_1,n_1)\sim(m_2,n_2)\) if and only if \(m_1+n_2=m_2+n_1\text{.}\)

An integer is an equivalence class of pairs of natural numbers under this equivalence relation.

If \(m,n\in\NN_0\text{,}\) then let us write \([m,n]\) for the equivalence class of the pair \((m,n)\in\NN_0\times\NN_0\text{.}\)

We write \(\ZZ\) for the set of integers.

The way this definition works might be a little unfamiliar, so let's contemplate some examples.

Let \(m \in \NN_0\text{.}\) We will write \(+m\) for the equivalence class \([m,0]\text{.}\) Note that since

\begin{equation*} (m,0)\sim(m+1,1)\sim(m+2,2)\sim\cdots\text{,} \end{equation*}

it follows that

\begin{equation*} +m = [m,0] = [m+1,1] = [m+2,2] = \cdots \end{equation*}

For example, \(+7 = [42,49]\text{.}\)

We thus obtain an injective map \(+\colon \NN_0 \to \ZZ\) by the rule \(+m = [m,0]\text{.}\) The image of this map is the set of nonnegative integers.

Let \(m \in \NN_0\text{.}\) We will write \(-m\) for the equivalence class \([0,m]\text{.}\) Note that since

\begin{equation*} (0,m)\sim(1,m+1)\sim(2,m+2)\sim\cdots\text{,} \end{equation*}

it follows that

\begin{equation*} -m = [0,m] = [1,m+1] = [2,m+2] = \cdots \end{equation*}

For example, \(-23 = [19,42]\text{.}\)

We obtain another injective map \(-\colon \NN_0 \to \ZZ\) by the rule \(-m = [0,m]\text{.}\) The image of this map is the set of nonpositive integers.

Note that

\begin{equation*} \ZZ = \{+n : n\in\NN_0\} \cup \{-n : n\in \NN_0\}\text{,} \end{equation*}

and

\begin{equation*} \{+n : n \in \NN_0\} \cap \{-n : n \in \NN_0\} = \{0\}\text{.} \end{equation*}

The set \(\ZZ\) is what's called a ring. We can add, subtract, and multply integers. To define these, we start with addition and multiplication on the natural numbers.

Definition 3.1.4.

Let \(m,n\in\NN_0\text{.}\)

We define the sum \(m+n\in\NN_0\) recursively as follows. If \(n=0\text{,}\) we define \(m+0 = m\text{.}\) If \(n\) is the successor of a natural number \(k\in\NN_0\text{,}\) then we define \(m+n\) as the successor of \(m+k\text{.}\)

We define the product \(m\times n\in\NN_0\) recursively as well. If \(n=0\text{,}\) then we define \(m\times 0=0\text{.}\) If \(n\) is the successor of a natural number \(k \in \NN_0\text{,}\) then we define \(m\times n = (m\times k) + m\text{.}\)

We also write \(mn\) for the product \(m\times n\text{.}\)

Using this sort of definition and inductive arguments, we can confirm that addition and multiplication have all the usual properties you know and love: they are both associative and commutative; \(0\) is the additive identity; \(1\) is the multiplicative identity; multiplication is distributive over addition.

Definition 3.1.5.

Let \([m,n],[p,q]\in\ZZ\text{.}\)

We define the sum

\begin{equation*} [m,n]+[p,q] = [m+p,n+q] \end{equation*}

We define the difference \([m,n]-[p,q]=[m,n]+[q,p]\text{.}\)

We define the product

\begin{equation*} [m,n]\times[p,q]=[mp+nq,mq+np] \end{equation*}

Because \(\ZZ\) is defined as a set of equivalence classes, it is necessary to confirm that these operations are well-defined; i.e., that they do not depend upon the the choice of representative from the equivalence class. If \([m_1,n_1],[m_2,n_2],[p,q]\in\ZZ\text{,}\) and if \([m_1,n_1]=[m_2,n_2]\text{,}\) then \(m_1+n_2=m_2+n_1\text{.}\) In that case, we have \(m_1+p+n_2+q = m_2+q+n_1+p\text{,}\) so

\begin{equation*} [m_1,n_1]+[p,q] = [m_1+p,n_1+q]=[m_2+p,n_2+q] = [m_2,n_2]+[p,q]\text{.} \end{equation*}

Furthermore, \(m_1p+n_1q+m_2q+n_2p = (m_1+n_2)p+(m_2+n_1)q = (m_1+n_2)q+(m_2+n_1)p = m_1q+n_1p+m_2q+n_2p\text{,}\) so

\begin{equation*} [m_1p+n_1q,m_1q+n_1p]=[m_2p+n_2q,m_2q+n_2p]\text{.} \end{equation*}

Let \(m,n \in \NN_0\text{.}\) We can observe that \((-m)(-n) = mn\text{:}\)

\begin{equation*} [0,m]\times[0,n]=[mn,0]\text{.} \end{equation*}

We obtain another injective map \(-\colon \NN_0 \to \ZZ\) by the rule \(-m = [0,m]\text{.}\) The image of this map is the set of nonpositive integers.