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\begin{document}
\title{Math 127 Homework}
\author{Mary Radcliffe}
\date{Due 14 February 2019}
\maketitle

Complete the following problems. Fully justify each response. You need only turn in those problems marked with a (*).

\begin{enumerate}

\item Let $x$ be a real number, and write a decimal expansion of $x$ in the form $x=n.d_1d_2d_3d_4\dots$, where $n$ is an integer, and $d_1, d_2, d_3, \dots$ are digits of the decimal expansion. Let $x_1, x_2, x_3, \dots$ be a sequence as defined on page 5 of the notes, so $x_1=n$, $x_2=n.d_1$, $x_3=n.d_1d_2$, etc. Prove that, using the definition given, $x$ is the limit of the sequence.

\item (*) Use the Peano Axioms to prove that multiplication in the natural numbers is commutative.

\item Use the Peano Axioms to prove that addition in the natural numbers is associative.

\item (*) Use mathematical induction to prove that for any $n\in\N$, $4^n+6n-1$ is divisible by 9.

\item (*) Use mathematical induction to prove that for any $n\in\N$, $\displaystyle\sum_{i=1}^n \frac{1}{i^2}\leq 2-\frac{1}{n}$.

\item (*) Prove Theorem 3 in the Induction notes.

\item (*) Here is a ``proof'' by induction of something we know to be false. Explain why it is broken.
\begin{quote}
\begin{prop}For any $n\in\N$, $\displaystyle\sum_{i=0}^n 2^i = 2^{n+1}$.
\end{prop}

\begin{proof}We work by induction on $n$. Suppose that for some $n\in\N$, the result is true, so that $\displaystyle\sum_{i=0}^n 2^i = 2^{n+1}$. Consider the case of $n+1$. We have
\begin{eqnarray*}
\displaystyle\sum_{i=0}^{n+1} 2^i &=& \displaystyle\sum_{i=0}^n 2^i + 2^{n+1}\\
&=& 2^{n+1}+2^{n+1}\quad\hbox{(by the inductive hypothesis)}\\
& = & 2\cdot 2^{n+1}=2^{n+2}.
\end{eqnarray*}

Therefore, if the result is true for $n$, it is also true for $n+1$. Hence, the result holds for all $n\in\N$.
\end{proof}
\end{quote}

\item Here is a ``proof'' by induction of something we know to be false. Explain why it is broken.
\begin{quote}
\begin{prop}If you have finitely many horses, then every horse is the same color.
\end{prop}

\begin{proof}
Number the horses $h_1, h_2, h_3, \dots$. We work by induction on $n$ to show that $h_1, h_2, \dots, h_n$ are all the same color for any fixed $n$. Since there are finitely many horses, this implies that all horses are the same color.

Now, for the base case, suppose $n=1$. Then we have only one horse, and hence all the horses we have are the same color.

Now, suppose the result is true if we have $n$ horses. Consider a set $h_1, h_2, \dots, h_n, h_{n+1}$ of $n+1$ horses. Note that by the induction hypothesis, the set $h_1, h_2, \dots, h_n$ of horses are all the same color, say brown. Also by the induction hypothesis, since $h_2, \dots, h_n, h_{n+1}$ is a set of only $n$ horses, they are also all the same color. Since $h_2$ is in this set, and we know already that $h_2$ is brown, we must have that all these horses are also brown. Hence, all of $h_1, h_2, \dots, h_n, h_{n+1}$ are the same color.

Therefore, by mathematical induction, if you have finitely many horses, then every horse is the same color.
\end{proof}
\end{quote}

\end{enumerate}
\end{document}