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\title{
    Principles of Mathematical Analysis (3rd Edition)\\
    by Walter Rudin
}
\author{
    Peter Cerno\\
    \small petercerno[at]gmail.com
}

\begin{document}
\begin{flushleft}

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% __BEGIN

\date{}
\maketitle

\begin{abstract}
These are my solutions to exercises from the book \emph{Principles of Mathematical Analysis} (3rd edition) by \emph{Walter Rudin}.
\end{abstract}

% __NOTE: Principles of Mathematical Analysis, third edition, by Walter Rudin.
% __NOTE: These are my solutions to exercises from the book Principles of Mathematical Analysis, third edition, by Walter Rudin.

% __SLIDE

\setcounter{section}{0}
\section{The Real and Complex Number Systems}

\setcounter{subsection}{5}
\subsection{Exercise}

Fix $b > 1$.

% __NOTE: Chapter 1: The Real and Complex Number Systems.
% __NOTE: Exercise 6.
% __NOTE: Let us fix a real number b greater than 1.

% __BREAK

\begin{enumerate}[label=(\alph*)]

\item
If $m, n, p, q$ are integers, $n > 0, q > 0$, and $r = \frac{m}{n} = \frac{p}{q}$, prove that
\[ (b^m)^{1/n} = (b^p)^{1/q} \]
Hence it makes sense to define $b^r = (b^m)^{1/n}$.

% __ADD::\end{enumerate}

% __NOTE: Assume that m, n, p, q are integers, n is bigger than 0, q is bigger than 0, and r is equal to m over n, which is equal to p over q.
% __NOTE: First, we would like to prove that if we raise b to the power of m, and then compute its nth root, we get the same result as if we raise b to the power of p, and then compute its quth root.
% __NOTE: This would allow us to define b to the power of the rational number r, since it does not matter how the rational number r is represented.

% __BREAK

\item
Prove that $b^{r+s} = b^r b^s$ if $r$ and $s$ are rational.

\end{enumerate}

% __NOTE: Second, we would like to prove that raising b to the power of r plus s gives us the same result as multiplying b to the power of r and b to the power of s, if both r and s are rational numbers. Note that this law holds for integers. Therefore, we would like to prove that it generalizes also to rational numbers.

% __SLIDE

\textbf{Proof}.

\vspace{0.1in}

(a) If $m, n, p, q$ are integers, $n > 0, q > 0$, and $r = \frac{m}{n} = \frac{p}{q}$, prove that $(b^m)^{1/n} = (b^p)^{1/q}$.
% __NOTE: Proof.
% __NOTE: Let us focus on the first statement, which claims that it does not matter how the rational number r is represented.

% __BREAK

\vspace{0.1in}

$\frac{m}{n} = \frac{p}{q}$ implies $mq = pn$, thus $b^{mq} = b^{pn} > 0$, i.e $(b^{mq})^{1/nq} = (b^{pn})^{1/nq}$.
% __NOTE: Suppose that we have two equivalent representations of the rational number r. In other words, let us assume that m over n is equal to p over q. In that case, m q is obviously equal to p n. Therefore, b raised to the power of m q is going to be equal to to b raised to the power p n. Let us compute the n q root of both sides.
% __BREAK
Let us show that $(b^{mq})^{1/nq} = (b^m)^{1/n}$ and $(b^{pn})^{1/nq} = (b^p)^{1/q}$.
% __NOTE: If we now show that the left hand side
% __BREAK
% __SUB::(b^{mq})^{1/nq}::\highlightb{(b^{mq})^{1/nq}}
% __NOTE: is equal to the nth root of b raised to the power of m, and that the right hand side
% __BREAK
% __SUB::(b^{pn})^{1/nq}::\highlightt{(b^{pn})^{1/nq}}
% __NOTE: is equal to the quth root of b raised to the power of p, then the first statement will follow easily.
% __BREAK

Apparently, $\left( (b^m)^{1/n} \right)^{nq} = \left[ \left( (b^m)^{1/n} \right)^n \right]^q = (b^m)^q = b^{mq}$, which implies that $(b^{mq})^{1/nq} = (b^m)^{1/n}$.
% __NOTE: To prove that the left hand side is equal to the nth root of b raised to the power of m, we simply raise both sides to the power of n q, and check that we get the same number.
% __BREAK
Similarly, $(b^{pn})^{1/nq} = (b^p)^{1/q}$.
% __NOTE: It is easy to see that the other equality can be proved in the same way.
% __BREAK
Therefore, $(b^m)^{1/n} = \highlightb{(b^{mq})^{1/nq}} = \highlightt{(b^{pn})^{1/nq}} = (b^p)^{1/q}$. $\square$
% __NOTE: Putting everything together we get that the nth root of b to the power of m is equal to the quth root of b to the power of p.
% __BREAK

\vspace{0.1in}

(b) Prove that ${b^{r+s}} = {b^r b^s}$ if $r$ and $s$ are rational.
% __NOTE: Let us now prove the second statement, which generalizes a similar statement for integers to all rational numbers.
% __BREAK

\vspace{0.1in}

Let us assume that $r = \frac{p}{n}$ and $s = \frac{q}{n}$ for some positive integer $n$.
% __NOTE: Let us represent the rational number r as p over n, and the rational number s as q over n, where p, q, n are integers and n is bigger than 0. Note that for simplicity we assume the same denominator for both r and s.
% __BREAK
According to (a): $b^{r+s} = b^{(p+q)/n} = (b^{p+q})^{1/n}$, thus $({b^{r+s}})^n = {b^{p+q}}$.
% __NOTE: According to the previously proved first statement, b to the power of r plus s is equal to the nth root of b to the power of p plus q. Let us raise both sides to the power of n. What we get is that the left hand side of equation raised to the power of n
% __BREAK
% __SUB::{b^{r+s}}::\highlightb{b^{r+s}}
% __NOTE: is equal to b raised to the power of p plus q.
% __BREAK
On the other hand, $({b^r b^s})^n = (b^r)^n (b^s)^n = \left( (b^p)^{1/n} \right)^n \left( (b^q)^{1/n} \right)^n = b^p b^q = {b^{p+q}}$.
% __NOTE: Now we only need to prove, that if we raise the right hand side of equation to the power of n
% __BREAK
% __SUB::{b^r b^s}::\highlightt{b^r b^s}
% __NOTE: we get the same number.
% __BREAK
% __SUB::{b^{p+q}}::\highlighto{b^{p+q}}
% __NOTE: But this is easy to see by simple rearrangement of the terms, leveraging the commutative law for multiplication.
% __BREAK
Together we have $(\highlightb{b^{r+s}})^n = \highlighto{b^{p+q}} = (\highlightt{b^r b^s})^n$, which implies (b). $\square$
% __NOTE: This proves the second statement for all rational numbers.

% __END

\vspace{4in}

\hrule

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