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\noindent \textbf{Author:} Ed Bueler
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\noindent \textbf{Date:} 1 September 2013
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\exer{6.66} The square root of 2 is irrational.
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\begin{proof}
Suppose to the contrary that $\sqrt{2}$ is rational. Then
there are integers $a$ and $b$ with no common factors such that
\begin{equation*}
\sqrt{2} = \frac{a}{b}.
\end{equation*}
Squaring this equation we find that
\begin{equation} \label{step1}
2 b^2 = a^2.
\end{equation}
Hence $2$ divides $a^2$ and therefore $a^2$ is even.
If $a$ were odd, then $a^2$ would also be odd, which it is not, so we conclude
that $a$ is even. So $a=2k$ for some integer $k$. It follows from equation
\eqref{step1} that
\begin{align*}
2 b^2 &= (2k)^2\\
&= 4 k^2.
\end{align*}
Hence
\begin{equation*}
b^2 = 2 k^2.
\end{equation*}
Arguing as before we see that $b^2$, and thus also $b$ itself, must be even.
So $2$ is a common factor of $a$ and $b$, which is a contradiction.
\end{proof}
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