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\begin{document}
\title[]{There is no fraction whose square is equal to two}
\author[]{Anne Onymous}
% Address of record for the research reported here
\address{Somewhere in ancient Greece.}
% Information for third author
%\thanks{}
\date{\today}
\begin{abstract}
We show that there is no fraction whose square is equal to two.
\end{abstract}
\maketitle
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\section{Introduction}
It has long been assumed that there is some fraction $m/n$ (where $m$
and $n$ are natural numbers), such that $(m/n)^2=2$. In this paper we
show that this is in fact not the case. We accordingly propose to call
the square root of two an {\em irrational number}.
\section{Main theorem}
Before stating our theorem we will need the following standard definition.
\begin{definition}
The {\em natural numbers} $\mathbb{N}$ are the set $\{1,2,\ldots\}$.
\end{definition}
Our main result is the following theorem.
\begin{theorem}
\label{thm:sqrt}
There are no two natural numbers $m$ and $n$ such that $(m/n)^2=2$.
\end{theorem}
\section{Proof}
We first state and prove a simple lemma.
\begin{lemma}
\label{lem:sqr}
Let $m$ be a natural number. If $m^2$ is even then it is divisible
by $4$ and $m$ is even.
\end{lemma}
\begin{proof}
We prove this lemma by showing that if $m$ is odd then $m^2$ is odd,
and that if $m$ is even then $m^2$ is divisible by $4$.
Assume first that $m$ is odd. Hence there exists a $k \in \mathbb{N}$
such that $m = 2k-1$. Hence
\begin{align*}
m^2 = (2k-1)^2 = 4k^2-4k+1 = 4(k^2-k) + 1.
\end{align*}
Since $4(k^2=k)$ is even, it follows that $m^2$ is odd.
Assume now that $m$ is even, so that it is equal to $2k$, where $k$
is again some natural number. Then $m^2=4k^2$, which is
divisible by $4$.
\end{proof}
We are now ready to prove Theorem~\ref{thm:sqrt}.
\begin{proof}[Proof of Theorem~\ref{thm:sqrt}]
Assume by contradiction that there exist $m',n' \in \mathbb{N}$ such
that $(m'/n')^2 = 2$. Let $k$ be the largest common divisor of $m'$ and
$n'$, and let $m=m'/k$ and $n=n'/k$. Then
\begin{align*}
\left(\frac{m}{n}\right)^2 &= \left(\frac{m'/k}{n'/k}\right)^2\\
&= \left(\frac{m'}{n'}\right)^2\\
&= 2.
\end{align*}
It follows that $2n^2=m^2$, and therefore $m^2$ is even. It follows
from Lemma~\ref{lem:sqr} that $m$ is even. It also follows from this
lemma that $m^2$ is divisible by $4$, and so $2n^2$ is divisible by
$4$. Hence $n^2$ is even, and, again by Lemma~\ref{lem:sqr}, $n$ is
even. We have shown that $m$ and $n$ are both divisible by
$2$. Hence $2k$ is a divisor of both $m'$ and $n'$, in contradiction
to our assumption.
\end{proof}
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