rational numbers are real numbersRationalNumbersAreRealNumbers2006-03-13 13:47:272006-04-18 02:31:46Resultaxiomatic definition of the real numbers% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\sR}[0]{\mathbb{R}}
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\newcommand{\Z}{\mathbbmss{Z}}
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\newtheorem{cor}{Corollary}%\subsubsection*{Rational numbers are real numbers}
Let us first show that the natural numbers $0,1,2,\ldots$ are
contained in the real numbers as constructed above.
Heuristically, this should be clear. We start with $0$.
By adding $1$ repeatedly we obtain the natural numbers
$$
0, \quad 0+1, \quad (0+1)+1, \quad ((0+1)+1)+1, \ldots,
$$
To make this precise, let $\N$ be the natural numbers.
(We assume that these exist. For example, all the usual constructions
of $\R$ rely on the existence of the natural numbers.)
Then we can define a map $f\colon \N\to \R$ as
\begin{enumerate}
\item $f(0)=0$, or more precisely, $f(0_\N)=0_\R$,
\item $f(a+1)=f(a)+1$ for $a\in \N$.
\end{enumerate}
By induction on $a$ one can prove that
\begin{eqnarray*}
f(a + b) &=& f(a) + f(b), \\
f(a b) &=& f(a) f(b), \quad a,b\in \N
\end{eqnarray*}
and
\begin{eqnarray*}
f(a) &\ge & 0, \ a\in \N\ \mbox{with equality only when}\ a=0.
\end{eqnarray*}
The last claim follows since $f(a)>0$ for $a=1,2,\ldots$ (by induction),
and $f(0)=0$.
It follows that $f$ is an injection: If $a\le b$, then $f(a)=f(b)$ implies
that $f(a)=f(a)+f(b-a)$, so $a=b$.
To conclude, let us show that
$f(\N)\subset\R$ satisfies the Peano axioms with zero element $f(0)$ and
sucessor operator
\begin{eqnarray*}
S\colon f(\N) &\to& f(\N) \\
k &\mapsto & f(f^{-1}(k)+1)
\end{eqnarray*}
First, as $f$ is a bijection, $x=y$ if and only if $S(x)=S(y)$
is clear.
Second, if $S(k)=0$ for some $k=f(a)\in f(\N)$, then $a+1=0$; a contradiction.
Lastly, the axiom of induction follows since $\N$ satisfies this axiom.
We have shown that $f(\N)$ are a subset of the real numbers that
behave as the natural numbers.
From the natural numbers, the integers and rationals can
be defined as
\begin{eqnarray*}
\Z&=& \N\cup \{-z\in \R : z\in \N\}, \\
\Q&=& \left\{ \frac{a}{b} : a\in \Z, b\in \N\setminus\{0\} \right\}.
\end{eqnarray*}
Mathematically, $\Z$ and $\Q$ are subrings of $\R$ that are
ring isomorphic to the integers and rationals, respectively.
\subsubsection*{Other constructions}
The above construction follows \cite{royden}. However, there are also
other constructions. For example, in \cite{spivak}, natural numbers in $\R$
are defined as follows. First, a set $L\subseteq \R$ is \emph{inductiv{e}} if
\begin{enumerate}
\item $0\in L$,
\item if $a\in L$, then $a+1\in L$.
\end{enumerate}
Then the natural numbers are defined as real numbers that are contained in all
inductiv{e} sets.
A third approach is to explicitly exhibit the natural numbers when
constructing the real numbers. For example, in \cite{rudin},
it is shown that the rational numbers form a subfield of $\R$
using explicit Dedekind cuts.
\begin{thebibliography}{9}
\bibitem{royden} H.L. Royden,
\emph{Real analysis}, Prentice Hall, 1988.
\bibitem{spivak} M. Spivak,
\emph{Calculus}, Publish or Perish.
\bibitem{rudin} W. Rudin,
\emph{Principles of mathematical analysis},
McGraw-Hill, 1976.
\end{thebibliography}