witt_irrationals_FEEDBACK.tex

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\title{Irrational Numbers Project}
\author{Carson Witt}

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\begin{document}

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\mbox{\LARGE Irrationals }\hfill Carson Witt

\vspace{1cm}

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\vspace{1cm}

Grade: 100

\vspace{1cm}

\begin{tabularx}{15cm}{ |p{6cm} | p{8cm}|}
\hline
Approriate Introduction &
Very nice! I really like the use of subsections within the abstract
to help the reader with your report.
\\
\hline
Proof of the irrationality of $\sqrt{2}$ &
Excellent outline of the proof.
\\
\hline
Explanation of continued fractions &
Very nice! It's great to see you add in the additional detail about
history and background. Stuff like that really adds to the quality of
your report -- and also helps the reader to enjoy reading it!
\\
\hline
Working Python code to approximate one of $e$, $\pi$, or $\sqrt{2}$ &
Yes. Good explanations of the code as well.
\\
\hline
Use of the lstlisting environment &
Yes.
\\
\hline
Proper formatting &
Yes.
\\
\hline
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\section*{Abstract}

\noindent{\textbf{Introduction:}}

An irrational number is a number that cannot be expressed as a ratio of two
numbers, or a fraction. Commonly known irrational numbers are the ratio $\pi$ of
a circle's circumference to its diameter, Euler's number $e$, the golden ratio
$\phi$, and the square root of two. All square roots of natural numbers, other
than of perfect squares, are irrational. When expressed as decimals, irrational
numbers do not repeat or terminate. \par

\noindent{\textbf{History:}}

According to Wikipedia, "The first proof of the existence of irrational numbers
is usually attributed to a Pythagorean (possibly Hippasus of Metapontum), who
probably discovered them while identifying sides of the pentagram. The then-
current Pythagorean method would have claimed that there must be some
sufficiently small, indivisible unit that could fit evenly into one of these
lengths as well as the other." \par

\noindent{\textbf{Task For This Project:}}

In this project, I will be proving that $\sqrt{2}$ is irrational (by
contradiction), explaining continued fractions and showing the continued
fraction for $e$, $\sqrt{2}$, and $\pi$, and writing Python code that
approximates $e$, $\sqrt{2}$, and $\pi$ to a requested number of digits.

\section*{Context/Work}

\textbf{Proving that $\sqrt{2}$ is irrational by contradiction:}

\begin{enumerate}
\item Let's assume that $\sqrt{2}$ is rational, meaning it can be written as the
ratio of two integers, $a$ and $b$:
\end{enumerate}

$$
\sqrt{2} = \frac{a}{b}
$$
Where $b \neq 0$ and we assume that $a$ and $b$ have no common factors. If
common factors exist, we cancel them in the numerator and denominator.

\begin{enumerate}[resume]
\item Squaring both sides of the equation gives us:
\end{enumerate}

$$
2 = \frac{a^2}{b^2}
$$

\begin{enumerate}[resume]
\item Which implies:
\end{enumerate}

$$
a^2 = 2b^2
$$

\begin{enumerate}[resume]
\item This means that $\sqrt{a}$ must be even, since $\sqrt{a}$ is $2$
multiplied by some number. We know this to be true because the multiplication of
two even numbers will always be even.
\item This also means that $a$ itself is even because if $a$ was odd, $a*a$
would be odd as well.
\item Since $a$ is an even number, it is $2$ times another whole number.
\end{enumerate}

$$
a = 2k
$$

\begin{enumerate}[resume]
\item If we substitute $a = 2k$ into the squared original equation, we get:
\end{enumerate}

$$
2 = \frac{(2k)^2}{b^2}
$$
$$
2 = \frac{4k^2}{b^2}
$$
$$
2b^2 = 4k^2
$$
$$
b^2 = 2k^2
$$

\begin{enumerate}[resume]
\item This means that $b^2$ is even, which follows that $b$ itself is even.
\item \textbf{This is where there is a contradiction.} If $a$ and $b$ are both
even numbers, then $\frac{a}{b}$ is not in its simplest form and still has
common factors. This is a contradiction because we assumed that the equation was
rational and had no common factors from the start. \textbf{Therefore, $\sqrt{2}$
must be irrational.}
\end{enumerate}

\newpage

\noindent{\textbf{Continued Fractions:}} \par

A continued fraction is a fraction of infinite length whose denominator is a
quantity plus a fraction, which latter fraction has a similar denominator, and
so on. Continued fractions are great ways to express irrational numbers like
$e$, $\pi$, and $\sqrt{2}$. \par

\vspace{0.02cm}{\noindent{Interesting Facts:}}

\begin{itemize}
\item John Wallis first used the term "continued fraction" in his Arithmetica
Infinitorum of 1653.
\item Another word for a continued fraction is anthyphairetic ratio.
\end{itemize}

\noindent{The basic form of a continued fraction is as follows:}

$$
a_0 +\cfrac{b_1}{a_1+\cfrac{b_2}{a_2+\cfrac{b_3}{a_3+\cdots}}}
$$

where $a_n$ and $b_n$ are either rational numbers, real numbers, or complex
numbers. If $b_n = 1$ for all $n$ the expression is called a simple continued
fraction. If the expression has a finite amount of terms, it is called a finite
continued fraction. Similarly, if the expression has an infinite number of
terms, it is called an infinite continued fraction.

\vspace{0.02cm}{$e$ as a continued fraction:}

$$
e = 2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{2}{3+\cfrac{3}{4+\cdots}}}}
$$

$\pi$ as a continued fraction:

$$
\pi = {\cfrac{4}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\cfrac{4^2}{9+\cdot
s}}}}}}
$$

$\sqrt{2}$ as a continued fraction:

$$
\sqrt{2} = 1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cdots}}}}
$$

\newpage

\noindent\textbf{{Python Code For Approximating e:}} \par

Below is the code I wrote to approximate $e$:

\begin{lstlisting}[language=Python, caption=Estimator for $e$]
n = input("How many decimals of e would you like to approximate?")

sum = 0
desired_e = N(e, digits = n + 1)
term_number = 0

while sum != desired_e:
sum += 1/factorial(term_number)
term_number += 1

print "NOTE: The code will approximate to the " + str(n) + " digits you
requested, but it will show " + str(n+1) + " digits to prevent rounding errors."
print "-----------------------------------------------------------"
print "Estimated value of e: " + str(N(sum, digits = n + 1))
print "-----------------------------------------------------------"
print "Actual value of e:     " + str(desired_e)
print "-----------------------------------------------------------"
print "Difference:           " + str((N(desired_e - sum, digits = 2)))
print "-----------------------------------------------------------"
\end{lstlisting}

\vspace{0.2cm}{\noindent\textbf{{Explanation:}}} \par

The code will ask how many digits of $e$ you would like to approximate and store
it in variable $n$. The variable "sum" has an initial value of $0$. While "sum"
is not equal to the actual value of $e$ ("desired{\_}e"), the series expansion
for $e$ (shown below) will be continuously added to "sum", increasing the
"term{\_}number" ($k$) by $1$ integer each time until "sum" does equal
"desired{\_}e". When "sum" does equal "desired{\_}e", the code will exit the
While loop. The code will then print the value of $e$ estimated with the
variable "sum" with $n+1$ digits. Underneath that, the code will print the
actual value of $e$ (stored as a constant by SageMathCloud) with the variable
"desired{\_}e" with $n+1$ digits. The code will then calculate the difference
between the estimated and actual value of $e$ and print that number. NOTE: The
difference should always be zero if the code is properly functioning. \par

\vspace{0.2cm}{\noindent\textbf{{Series Expansion Used for $e$:}}} \par

$$
e =\sum ^{\infty }_{k=0}\dfrac{1}{k!}
$$

\newpage

\noindent\textbf{{Python Code For Approximating $\sqrt{2}$:}} \par

Below is the code I wrote to approximate $\sqrt{2}$:

\begin{lstlisting}[language=Python, caption=Estimator For $\sqrt{2}$]
n_2 = input("How many decimals of sqrt(2) would you like to approximate?")

sum_2 = 0
desired_2 = N(sqrt(2), digits = n_2 + 1)
term_number_2 = 0

while sum_2 != desired_2:
sum_2 += (factorial(2*(term_number_2)+1))/((2^(3*(term_number_2)+1))*(factorial(
term_number_2))^2)
term_number_2 += 1

print "NOTE: The code will approximate to the " + str(n_2) + " digits you
requested, but it will show " + str(n_2+1) + " digits to prevent rounding
errors."
print "-----------------------------------------------------------"
print "Estimated value of 2: " + str(N(sum_2, digits = n_2 + 1))
print "-----------------------------------------------------------"
print "Actual value of 2:    " + str(desired_2)
print "-----------------------------------------------------------"
print "Difference:           " + str((N(desired_2 - sum_2, digits = 2)))
print "-----------------------------------------------------------"
\end{lstlisting}

\vspace{0.2cm}{\noindent\textbf{{Explanation:}}} \par

The code for approximating $\sqrt{2}$ is essentially the same as the code for
approximating $e$. The only difference is in the variables and the series
expansions.\par

\vspace{0.2cm}{\noindent\textbf{{Series Expansion Used for $\sqrt{2}$:}}} \par

$$
\sqrt{2} =\sum ^{\infty }_{k=0}\dfrac {\left( 2k+1\right) !}{2^{3k+1}\left(
k!\right) ^{2}}
$$

\newpage

\noindent\textbf{{Python Code For Approximating $\pi$}} \par

Below is the code I wrote to approximate $\pi$:

\begin{lstlisting}[language=Python, caption=Estimator For $\pi$]
n_pi = input("How many decimals of pi would you like to approximate?")

sum_pi = 0
desired_pi = N(pi, digits = n_pi + 1)
term_number_pi = 0

while sum_pi != desired_pi:
sum_pi += (1/(16^term_number_pi))*((4/(8*(term_number_pi)+1))-(2/(8*(term_number
_pi)+4))-(1/(8*(term_number_pi)+5))-(1/(8*(term_number_pi)+6)))
term_number_pi += 1

print "NOTE: The code will approximate to the " + str(n_pi) + " digits you
requested, but it will show " + str(n_pi+1) + " digits to prevent rounding
errors."
print "-----------------------------------------------------------"
print "Estimated value of pi: " + str(N(sum_pi, digits = n_pi + 1))
print "-----------------------------------------------------------"
print "Actual value of pi:    " + str(desired_pi)
print "-----------------------------------------------------------"
print "Difference:            " + str((N(desired_pi - sum_pi, digits = 2)))
print "-----------------------------------------------------------"
\end{lstlisting}

\vspace{0.2cm}{\noindent\textbf{{Explanation:}}} \par

The code for approximating $\pi$ is essentially the same as the code for
approximating $e$ and $\sqrt{2}$. The only difference is in the variables and
the series expansions.\par

\vspace{0.2cm}{\noindent\textbf{{Series Expansion for $\pi$
(Bailey–Borwein–Plouffe Formula):}}} \par

$$
\pi =\sum ^{\infty }_{k=0}\left[ \dfrac {1}{16^{k}}\left( \dfrac
{4}{8k+1}-\dfrac {2}{8k+4}-\dfrac {1}{8k+5}-\dfrac {1}{8k+6}\right) \right]
$$

Interestingly enough, the code for approximating $\pi$ gave me the most trouble.
The series expansions that I had previously used in the code either did not
converge fast enough or somehow made the variable "sum{\_}pi" infinitely locked
in the While loop. After some research and trial and error, I discovered the
Bailey–Borwein–Plouffe Formula and decided to try it. It worked perfectly.

\newpage

\section*{Conclusion}

Overall, this has been my favorite project. The code was fairly challenging and
I initially ran into a couple of issues, but this was the first project where I
figured the code out on my own. While I went to Mr. Abell when technical issues
arose, the basic structure of the code was my own. In addition, this is the
first project that I started well in advance of the due date. I usually try to
figure the project out the day it is assigned, but then I put it off until the
last few days. This time, I had finished the code a week or two before the due
date, making it possible for me to write the report without any stress. \par

I have really enjoyed brushing up my Python skills and learning \LaTeX \space
this year, and I hope I can continue to use these tools in the future, whether
it be for school or just for fun.

\end{document}