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\newtheorem{lem}{Lemma}
\newtheorem{thm}{Theorem}

\title{Fraleigh 9.1.37}
\author{Jason Wojciechowski}

\begin{document}
\maketitle

\begin{lem}
  An automorphism of a field $E$ carries elements that are squares of
  elements in $E$ onto elements that are squares of elements in $E$.
\end{lem}
\begin{proof}
  Suppose $\phi$: $E \rightarrow E$ is an automorphism.  Then for
  every $a_k \in E$ such that $\phi(a_k) = a_m, \phi(a_k a_k) = a_m
  a_m$, so all $a_k a_k$'s, or all squares in $E$, go to another
  square in $E$.
\end{proof}

\begin{lem}
  An automorphism of the field $\R$ carries positive numbers onto
  positive numbers.
\end{lem}
\begin{proof}
  Since all positive $r \in \R$ have two real square roots, we can say
  that a real number is a square if and only if it is positive.  Thus,
  since squares get carried to squares in any field (Lemma 1),
  positive numbers get carried to positive numbers in the real
  numbers.
\end{proof}

\begin{lem}
  If $\phi$ is an automorphism of $\R$ and $a < b, a, b \in \R$, then
  $\phi(a) < \phi(b)$.
\end{lem}
\begin{proof}
  Suppose $b > a$.  Then $b - a > 0$, so, by Lemma 2, $\phi(b-a) > 0$,
  so $\phi(b)-\phi(a) > 0$, so $\phi(b) > \phi(a)$.
\end{proof}

\begin{thm}
  The only automorphism of $\R$ is the identity automorphism.
\end{thm}

\begin{proof}
  The image of $1$ under any automorphism must be $1$, since it is the
  multiplicative identity.  Then, since automorphisms respect the
  addition operation, every integers must be taken to itself by an
  automorphism on the reals.  Since automorphisms must also respect
  multiplication (and thus division), every rational must be taken to
  itself.  
  
  Now the only reals left in question are the irrationals.  Every
  irrational number can be squeezed infinitely closely by two
  rationals.  For example, $\pi$ is between $3$ and $4$, $3.1$ and
  $3.2$, $3.14$ and $3.15$, and so on.  By Lemma 3, an irrational
  which sits between rationals $a$ and $b$ must be taken to an
  irrational between $a$ and $b$.  Since we can squeeze every
  irrational infinitely closely with a sequence of rationals on either
  side of it, every irrational must end up being taken to itself.
  Thus, every real number must be taken to itself by an automorphism
  on the reals, so the only automorphism on the reals is the identity
  automorphism.
\end{proof}

\end{document}