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\begin{center} \textbf{Axioms of the integers $\mathbb{Z}$.} \end{center}

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The word ``integer'' is our undefined term (sometimes referenced as the ``primitive'' term) for this section.  The set $\mathbb{Z}$ is the set of all integers (Axiom D3 implies that $\mathbb{Z}$ has at least two elements, so I am grammatically correct in using the plural).  The set $\mathbb{Z}$  satisfies the following axioms.

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The usual rules (axioms) of logic are to be used to prove theorems from these axioms.  As needed these rules will be discussed and stated.  As a first such, following are the properties of the equality symbol.

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\noindent i.) [Reflexive]  $x=x$ for all $x$.

\noindent ii.) [Symmetric] If $x=y$ then $y=x$.

\noindent iii.) [Transitive] If $x=y$ and $y=z$ then $x=z$.

\noindent iv.) [Uniqueness of function values] If $f$ is a function and $x=y$ then $f(x) = f(y)$.

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Axioms about addition and multiplication: There exists two
operations on the integers: addition denoted by ``$+$'' and
multiplication denoted by ``$\cdot$''.  [Strictly speaking  ``$+$'' is a map from the cross product of the integers with itself into the
integers with certain properties as defined by the axioms, and
similarly for multiplication ``$\cdot$''.  Cross products of sets will be more formally defined later in the semester.] Note that $a \cdot b$ is usually written as $ab$.  It is this functional definition and properties of the $=$ symbol that yields the following (axioms) which will be needed for our proofs and which you may assume as part of our logic system.

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\qquad If $a=b$ and $c=d$ then $a+c = b+d$.

\qquad If $a=b$ and $c=d$ then $a \cdot c = b \cdot d$.

\noindent [Observation: the symbols $\wedge$ and $\vee$ also satisfy these properties.]

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\noindent \textbf{Axioms about addition.}

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 A1.  If $a \in \mathbb{Z}$ and $b \in \mathbb{Z}$ then $a+b
\in \mathbb{Z}$. [Closure.]

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 A2. If $a \in \mathbb{Z}$ and $b \in \mathbb{Z}$ then $a+b
= b+a$. [Commutativity.]

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 A3. If $a \in \mathbb{Z}$, $b \in \mathbb{Z}$ and $c \in
\mathbb{Z}$ then $a+(b+c)=(a+b)+c$. [Associativity.]

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 A4. There exists an element $0\in \mathbb{Z}$ so that if $a
\in \mathbb{Z}$ then $a+0 = a$. [Additive Identity element.]

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 A5. If $a \in \mathbb{Z}$ then there exits an element in $\mathbb{Z}$
denoted by $-a$ so that $-a + a =0$. [Additive inverse.]

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Definition:  $a-b$ means $a+(-b)$.

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\noindent \textbf{Axioms about multiplication.}

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B1.  If $a \in \mathbb{Z}$ and $b \in \mathbb{Z}$ then
$a\cdot b \in \mathbb{Z}$. [Closure.]

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B2. If $a \in \mathbb{Z}$ and $b \in \mathbb{Z}$ then
$a\cdot b = b\cdot a$. [Commutivity.]

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B3. If $a \in \mathbb{Z}$, $b \in \mathbb{Z}$ and $c \in
\mathbb{Z}$ then $a\cdot (b\cdot c)=(a\cdot b)\cdot c$.
[Associativity.]

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B4. There exists an element $1\in \mathbb{Z}$ so that if $a
\in \mathbb{Z}$ then $a\cdot 1 = a$. [Multiplicative Identity element.]

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B5. If $a \in \mathbb{Z}$, $b \in \mathbb{Z}$, $c \in
\mathbb{Z}$ with $c \ne 0$ and $ac=bc$ then $a=b$. [Cancellation rule.]

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\noindent \textbf{Axiom on the relationship between addition and multiplication.}

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C1. If $a \in \mathbb{Z}$, $b \in \mathbb{Z}$ and $c \in
\mathbb{Z}$ then $a\cdot (b + c)= a \cdot b + a \cdot c $.
[Distributive law.] (Note the assumption that the order of
operation is to perform $\cdot $ first then $+$. In other words $ab + cd$ means $(ab) + (cd)$.)

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\noindent \textbf{Axioms on order.}

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There exists an order relation ``$<$'' so that:

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D1. If each of $a$ and $b$ is an integer then exactly one of the
following is true:
\begin{eqnarray*}
a<b, & a=b, & b<a.
\end{eqnarray*}

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D2. If each of $a$, $b$ and $c$ is an integer $a<b$ and $b<c$,
then $a<c$.

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D3. $0<1$.

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D4. If each of $a$, $b$ and $c$ is an integer and $a<b$, then $a
+ c < b + c$.

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D5.  If each of $a$, $b$ and $c$ is an integer $a<b$ and $0<c$,
then $a \cdot c < b \cdot c$.

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Definition.  The integer $p$ is said to be positive if and only if $0<p$; $\mathbb{N} = \{ n \in \mathbb{Z} | 0 < n \}$.

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D6. If $n$ is an integer then exactly one of the following is true [Trichotomy Law.]:
\begin{eqnarray*}
n \in \mathbb{N}, & n = 0, & -n \in \mathbb{N}.
\end{eqnarray*}

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\noindent \textbf{Induction axiom.}

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E1. Suppose that $S$ is a subset of $\mathbb{N}$ containing $1$ such
that if $n \in S$ then $n+1 \in S$. Then $S=\mathbb{N}$.


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