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\begin{document}
\begin{center}
{\LARGE Week 1, Due Fri 10/4}
\end{center}


\begin{enumerate}


\item Let~$x$ be a number such that~$x^2 = 2$ and assume that $x > 0$.
Prove (using only the properties $\mathrm{P}1$ to $\mathrm{P}12$) that
\[\frac{17}{12} > x > \frac{17}{12} - \frac{1}{12^2}\]

\item Spivak, Chapter 1, Problems: $1,7, 12, 19, 25$.



\item {\bf Rational numbers are poorly approximated by other rational numbers 
\ldots other than themselves.\rm} Let $\alpha$ be a number (ultimately
a real number but for now just in the sense of Spivak, Chapter 1.)
\begin{enumerate}
\item Suppose that $\alpha = a/b$ is a rational number. Suppose that $p/q$
is another rational number
which is different from $\alpha$. Prove that
\[\left| \alpha - \frac{p}{q} \right| \ge \frac{1}{|bq|}.\]
\item Suppose that $\alpha = a/b$ is a rational number. Suppose that $p/q$
is another rational number
which is different from $\alpha$. 
 Suppose that
\[\left| \alpha - \frac{p}{q} \right| \le \frac{1}{q^2}.\]
Prove that~$|b| \ge |q|$.
\item Drop the assumption that $\alpha$ is a rational number.
Suppose there exist infinitely many rational numbers $p_n/q_n$
with $p_n$ and $q_n$  and with $|q_n|$ increasing so that
\[0 < \left| \alpha - \frac{p_n}{q_n}  \right| \le \frac{1}{q^2_n}.\]
Prove that $\alpha$ is not a rational number.
\item (*) Let~$p_n = 1,1,3,7,17 \ldots$ and~$q_n = 0,1,2,5,12, \ldots$ be sequences
defined by the initial values~$p_0=1, p_1=1$ and~$q_0=0$, $q_1=1$ and then
\[\begin{aligned}
p_{n} & = 2 p_{n-1} + p_{n-2}, \\
q_{n} & = 2 q_{n-1} + q_{n-2}. \end{aligned} 
\]
Prove that, for~$n \ge 1$,
\(\displaystyle{\left| \sqrt{2} - \frac{p_n}{q_n} \right| \le \frac{1}{q^2_n}}\), 
and deduce that~$\sqrt{2}$ is irrational.
\end{enumerate}






\end{enumerate}

\end{document}

