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\author{Lizzy Hunt}
\date{\today}
\title{Assignment Five}
\hypersetup{
 pdfauthor={Lizzy Hunt},
 pdftitle={Assignment Five},
 pdfkeywords={},
 pdfsubject={},
 pdfcreator={Emacs 28.2 (Org mode 9.6.1)}, 
 pdflang={English}}
\begin{document}

\maketitle
\setlength\parindent{0pt}

\section{Section 3.2}
\label{sec:org258154b}
\subsection{Question One}
\label{sec:orgf02bb85}
\subsubsection{a}
\label{sec:orgccb1c0c}
\(a^2 - ab + ba - b^2\)
\subsubsection{b}
\label{sec:org9164b85}
\(a^3 + ba^2 + 2a^2b + 2ab^2 + ab^2 + b^3\)
\subsubsection{c}
\label{sec:orgdaba485}
(a) could be \(a^2 - b^2\)

(b) could be \(a^3 + 3a^2b + 3ab^2 + b^3\)

\subsection{Question Three}
\label{sec:org4d02a16}
\subsubsection{a}
\label{sec:orgb743fca}
\(\begin{smallmatrix}
0 & 0 \\
0 & 0 \\
\end{smallmatrix}\), \(\begin{smallmatrix}
1 & 0 \\
0 & 1 \\
\end{smallmatrix}\), \(\begin{smallmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{smallmatrix}\), \(\begin{smallmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{smallmatrix}\)
\subsubsection{b}
\label{sec:org8300d7e}
\begin{verbatim}
>>> set(filter(lambda y: all([y**2 % 12 == y for x in range(12)]), range(12)))
{0, 1, 4, 9}
\end{verbatim}

\subsection{Question Seven}
\label{sec:org9152891}
S is closed under multiplication:
\(i \in S, j \in S \Rightarrow i \cdot j = n1_R \cdot m1_R = (nm)1_R\).

S is closed under subtraction:
\(i \in S, j \in S \Rightarrow i - j = n1_R - m1_R = (n-m)1_R\)

\subsection{Question Eight}
\label{sec:org3ed31b0}
Considering \(n,m \in T\) then \(n = xb, m = yb\) with \(x,y \in R\):

\begin{enumerate}
\item T is closed under multiplication: \(n \cdot m = xb \cdot yb = (x \cdot y)b\), which follows the rule.
\item T is closed under subtraction: \(n - m = xb - yb = (x - y)b\), which also follows the rule.
\end{enumerate}

We also know \(T\) is not empty since it must have at least 0\textsubscript{R}.

\subsection{Question Ten}
\label{sec:orgc4efb36}
\subsubsection{a}
\label{sec:orgbc2ab9d}
\(\bar{R} = {(0, 0), (1, 0), (2, 0)}\)

\(\bar{S} = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4)}\)

\subsubsection{b}
\label{sec:org564675e}
Considering \(n, m \in \bar{R}\), then \(n = (x, 0_S), n = (y, 0_S)\) with \(x,y \in R\).

\begin{enumerate}
\item \(\bar{R}\) is closed under multiplication: \(n \cdot m = (x, 0_S) \cdot (y, 0_S) = (x \cdot y, 0_S)\) and \(x \cdot y \in R\), so thus \(n \cdot m \in R \times S\).
\item \(\bar{R}\) is closed under subtraction: \(n - m = (x, 0_S) - (y, 0_S) = (x - y, 0_S)\) and \(x - y \in R\), so thus \(n - m \in R \times S\).
\end{enumerate}

We also know that \(\bar{R}_0 = (0_R, 0_S) \in R \times S\) so \(\bar{R}\) is not empty.

\subsubsection{c}
\label{sec:org5905763}
Considering \(n, m \in \bar{S}\), then \(n = (0_R, x), n = (0_R, y)\) with \(x,y \in S\).

\begin{enumerate}
\item \(\bar{S}\) is closed under multiplication: \(n \cdot m = (0_R, x) \cdot (0_R, y) = (x \cdot y, 0_S)\) and \(x \cdot y \in S\), so thus \(n \cdot m \in R \times S\).
\item \(\bar{S}\) is closed under subtraction: \(n - m = (0_R, x) - (0_R, y) = (0_R, x - y)\) and \(x - y \in R\), so thus \(n - m \in R \times S\).
\end{enumerate}

We also know that \(\bar{S}_0 = (0_R, 0_S) \in R \times S\) so \(\bar{S}\) is not empty.

\subsection{Question Thirteen}
\label{sec:orgfc1192c}
\subsubsection{a}
\label{sec:orgdc9f5c9}
Considering \(n, m \in S \cap T\), then \(n \in S\) and \(n \in T\), \(m \in S\) and \(m \in T\), therefore:

\begin{enumerate}
\item \(S \cap T\) is closed under multiplication as \(m \cdot n\) must also be in \(S \cap T\)
\item \(S \cap T\) is closed under addition as \(m + n\) must also be in \(S \cap T\)
\end{enumerate}

Since \(S\) and \(T\) are both subrings of \(R\), \(0_R \in S \cap T\).

\subsubsection{b}
\label{sec:orgeb2e7d6}
No, consider \(S\) being the integer multiples of 8 and \(T\) being the integer multiples of 3 being subrings of \(\mathds{Z}\) (proof of these being subrings is in
Question Six of Section 3.1 in Assignment Four), then \(n \in S\) with \(n = 8\) and \(m \in T\) with \(m = 3\) then \(n + m = 11 \notin S \cup T\).

\subsection{Question Fifteen - TODO}
\label{sec:org675f8a8}

\subsection{Question Twenty-One}
\label{sec:org0c713ff}
\subsubsection{a}
\label{sec:org6158430}
From \(ab = ac \Rightarrow ab - ac = 0_R \Rightarrow a(b - c) = 0_R\), we know \(b-c\) is equivalent to \(0_R\) since we're given that \(a\) is a non-zero element.

\(b-c = 0_R \Rightarrow b = c\)

\subsubsection{b}
\label{sec:org4391edc}
From \(ba = ca \Rightarrow ba - ca = 0_R \Rightarrow (b - c)(a) = 0_R\) we come to the same conclusion as (a)

\(b-c = 0_R \Rightarrow b = c\)

\section{Section 3.3}
\label{sec:orgff5f7f5}
\subsection{Question One}
\label{sec:org08aff62}
In the true nature of being a computer science student, automate something for 2 hours that you could've done in 20 minutes!
\begin{verbatim}
cartesian_prod = lambda zn, zm: list([(i, j) for i in range(zn) for j in range(zm)])
mult_tup = lambda a, b, zn, zm: ((a[0] * b[0]) % zn, (a[1] * b[1]) % zm)
add_tup = lambda a, b, zn, zm: ((a[0] + b[0]) % zn, (a[1] + b[1]) % zm)
empty_table = lambda col, row, symbol: [[symbol] + [str(x) for x in col]] + [[str(x)] + [0 for i in range(len(col))] for x in row]

def make_cartesian_congruence_table(zn, zm, op, mapping, symbol):
  cp = cartesian_prod(zn, zm)
  mapped_cp = [cp[mapping[i]] for i in range(len(cp))]
  table = empty_table(mapped_cp, mapped_cp, symbol)
  for i in range(len(cp)):
    for j in range(len(cp)):
      table[i+1][j+1] = str(op(mapped_cp[i], mapped_cp[j], zn, zm))
  return table

def make_normal_table(n, op, symbol):
  table = empty_table(list(range(n)), list(range(n)), symbol)
  for i in range(n):
    for j in range(n):
      table[i+1][j+1] = str(op(i, j) % n)
  return table
\end{verbatim}

\subsubsection{Z\textsubscript{2} \texttimes{} Z\textsubscript{3} with bijection}
\label{sec:orgcac27c3}
\begin{verbatim}
bijection = [0, 4, 2, 3, 1, 5]
\end{verbatim}

\begin{enumerate}
\item Multiplication Tables
\label{sec:org1f4aa9f}
\begin{verbatim}
make_cartesian_congruence_table(2, 3, mult_tup, bijection, '\odot')
\end{verbatim}

\begin{center}
\begin{tabular}{lllllll}
\(\odot\) & (0, 0) & (1, 1) & (0, 2) & (1, 0) & (0, 1) & (1, 2)\\[0pt]
(0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0)\\[0pt]
(1, 1) & (0, 0) & (1, 1) & (0, 2) & (1, 0) & (0, 1) & (1, 2)\\[0pt]
(0, 2) & (0, 0) & (0, 2) & (0, 1) & (0, 0) & (0, 2) & (0, 1)\\[0pt]
(1, 0) & (0, 0) & (1, 0) & (0, 0) & (1, 0) & (0, 0) & (1, 0)\\[0pt]
(0, 1) & (0, 0) & (0, 1) & (0, 2) & (0, 0) & (0, 1) & (0, 2)\\[0pt]
(1, 2) & (0, 0) & (1, 2) & (0, 1) & (1, 0) & (0, 2) & (1, 1)\\[0pt]
\end{tabular}
\end{center}

\item Addition Tables
\label{sec:org807c598}
\begin{verbatim}
make_cartesian_congruence_table(2, 3, add_tup, bijection, '\oplus')
\end{verbatim}

\begin{center}
\begin{tabular}{lllllll}
\(\oplus\) & (0, 0) & (1, 1) & (0, 2) & (1, 0) & (0, 1) & (1, 2)\\[0pt]
(0, 0) & (0, 0) & (1, 1) & (0, 2) & (1, 0) & (0, 1) & (1, 2)\\[0pt]
(1, 1) & (1, 1) & (0, 2) & (1, 0) & (0, 1) & (1, 2) & (0, 0)\\[0pt]
(0, 2) & (0, 2) & (1, 0) & (0, 1) & (1, 2) & (0, 0) & (1, 1)\\[0pt]
(1, 0) & (1, 0) & (0, 1) & (1, 2) & (0, 0) & (1, 1) & (0, 2)\\[0pt]
(0, 1) & (0, 1) & (1, 2) & (0, 0) & (1, 1) & (0, 2) & (1, 0)\\[0pt]
(1, 2) & (1, 2) & (0, 0) & (1, 1) & (0, 2) & (1, 0) & (0, 1)\\[0pt]
\end{tabular}
\end{center}
\end{enumerate}

\subsubsection{Z\textsubscript{6}}
\label{sec:orgbe8adf1}
\begin{enumerate}
\item Multiplication Tables
\label{sec:orga98b12b}
\begin{verbatim}
make_normal_table(6, lambda a, b: a * b, '\odot')
\end{verbatim}

\begin{center}
\begin{tabular}{rrrrrrr}
\(\odot\) & 0 & 1 & 2 & 3 & 4 & 5\\[0pt]
0 & 0 & 0 & 0 & 0 & 0 & 0\\[0pt]
1 & 0 & 1 & 2 & 3 & 4 & 5\\[0pt]
2 & 0 & 2 & 4 & 0 & 2 & 4\\[0pt]
3 & 0 & 3 & 0 & 3 & 0 & 3\\[0pt]
4 & 0 & 4 & 2 & 0 & 4 & 2\\[0pt]
5 & 0 & 5 & 4 & 3 & 2 & 1\\[0pt]
\end{tabular}
\end{center}

\item Addition Tables
\label{sec:orgfcf7bde}
\begin{verbatim}
make_normal_table(6, lambda a, b: a + b, '\oplus')
\end{verbatim}

\begin{center}
\begin{tabular}{rrrrrrr}
\(\oplus\) & 0 & 1 & 2 & 3 & 4 & 5\\[0pt]
0 & 0 & 1 & 2 & 3 & 4 & 5\\[0pt]
1 & 1 & 2 & 3 & 4 & 5 & 0\\[0pt]
2 & 2 & 3 & 4 & 5 & 0 & 1\\[0pt]
3 & 3 & 4 & 5 & 0 & 1 & 2\\[0pt]
4 & 4 & 5 & 0 & 1 & 2 & 3\\[0pt]
5 & 5 & 0 & 1 & 2 & 3 & 4\\[0pt]
\end{tabular}
\end{center}
\end{enumerate}

\subsection{Question Two}
\label{sec:org9540aed}
\begin{verbatim}
bijection = [0, 3, 2, 1]
make_cartesian_congruence_table(2, 2, mult_tup, bijection, '\odot')
\end{verbatim}

\begin{center}
\begin{tabular}{lllll}
\(\odot\) & (0, 0) & (1, 1) & (1, 0) & (0, 1)\\[0pt]
(0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0)\\[0pt]
(1, 1) & (0, 0) & (1, 1) & (1, 0) & (0, 1)\\[0pt]
(1, 0) & (0, 0) & (1, 0) & (1, 0) & (0, 0)\\[0pt]
(0, 1) & (0, 0) & (0, 1) & (0, 0) & (0, 1)\\[0pt]
\end{tabular}
\end{center}

\begin{verbatim}
bijection = [0, 3, 2, 1]
make_cartesian_congruence_table(2, 2, add_tup, bijection, '\oplus')
\end{verbatim}

\begin{center}
\begin{tabular}{lllll}
\(\oplus\) & (0, 0) & (1, 1) & (1, 0) & (0, 1)\\[0pt]
(0, 0) & (0, 0) & (1, 1) & (1, 0) & (0, 1)\\[0pt]
(1, 1) & (1, 1) & (0, 0) & (0, 1) & (1, 0)\\[0pt]
(1, 0) & (1, 0) & (0, 1) & (0, 0) & (1, 1)\\[0pt]
(0, 1) & (0, 1) & (1, 0) & (1, 1) & (0, 0)\\[0pt]
\end{tabular}
\end{center}

\subsection{Question Three}
\label{sec:org421e770}
\begin{enumerate}
\item \(f\) is injective, since \(f(a) = f(b) \Rightarrow (a,a) = (b,b) \Rightarrow a = b\)
\item \(f\) is surjective since every range element in \(R^*\), \((a,a)\) is mapped to \(a \in R\) by definition
\item \(f(a) + f(b) = (a, a) + (b, b) = (a + b, a + b) = f(a + b)\) and \(f(a)f(b) = (a, a) \cdot (b, b) = (ab, ab) = f(ab)\)
\end{enumerate}

\subsection{Question Four}
\label{sec:org1f8479a}
\(f(1)f(3) = (2)(6) \equiv_{10} 2\) but \(f(3) = 6\) which doesn't hold the properties of homomorphism.

\subsection{Question Five}
\label{sec:orgeb656c3}
Consider the given function:
\begin{enumerate}
\item \(f\) is injective, since \(f(a) = f(b) \Rightarrow\) \(\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}\) = \(\begin{smallmatrix} 0 & 0 \\ 0 & b \end{smallmatrix}\) \(\Rightarrow a = b\)
\item \(f\) is surjective, since every range element \(\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}\) has an element in \(\mathds{R}\), \(a\), such that \(f(a) =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}\) by definition
\item \(f(a) + f(b) =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}\) + \(\begin{smallmatrix} 0 & 0 \\ 0 & b \end{smallmatrix}\) = \(\begin{smallmatrix} 0 & 0 \\ 0 & (a + b) \end{smallmatrix}\) = \(f(a + b)\),
and \(f(a) \cdot f(b) =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix} 0 & 0 \\ 0 & b \end{smallmatrix}\) = \(\begin{smallmatrix} 0 & 0 \\ 0 & (a \cdot b) \end{smallmatrix}\) = \(f(a \cdot b)\),
\end{enumerate}

\subsection{Question Nine - TODO}
\label{sec:org2f29e8e}

\subsection{Question Eleven}
\label{sec:org3e455ed}
\subsubsection{b}
\label{sec:org24734c1}
\begin{verbatim}
>>> f = lambda x: 3 * x
>>> list(filter(lambda x: f(x * x) != (f(x) * f(x)), range(0, 20, 2)))
[2, 4, 6, 8, 10, 12, 14, 16, 18]
\end{verbatim}
\subsubsection{d}
\label{sec:orgc2ba394}
\begin{verbatim}
>>> k = lambda x: 0 if x == 0 else (x ** -1)
>>> list(filter(lambda x: k(x * x) != (k(x) * k(x)), range(0, 20)))
[5, 10, 13, 19]
\end{verbatim}
\subsection{Question Twelve}
\label{sec:org95ce95e}
\subsubsection{c}
\label{sec:org77d2e96}
Not a homomorphism. Just from reducing \(f(x + x)\) we find \(f(x + x) \neq f(x) + f(x)\):

\begin{equation*}
f(x + x) = \dfrac{1}{(x + x)^2 + 1} = \dfrac{1}{4x^2 + 1}
\end{equation*}

\begin{equation*}
f(x) + f(x) = \dfrac{1}{x^2 + 1} + \dfrac{1}{x^2 + 1} = \dfrac{2}{x^2 + 1}
\end{equation*}

Thus \(f(x + x) \neq f(x) + f(x)\).
\subsubsection{d}
\label{sec:org1c2b201}
\begin{verbatim}
>>> import numpy as np
>>> h = lambda x: [[-x, 0], [x, 0]]
>>> list(filter(lambda x: not np.array_equiv(h(x * x), np.dot(h(x), h(x))), range(0, 20))
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
\end{verbatim}
\subsection{Question Thirteen}
\label{sec:org57e2c18}
\subsubsection{a}
\label{sec:org7515001}
If \(r \in R\) then \((r, 0_S) \in R \times S\), and \(f((r, 0_S)) = r\), so every range element has a domain element mapping to it.

For some \(a = (r, m) \in R \times S\) and \(b = (t, v) \in R \times S\) then \(f((r, m)) + f((t, v)) = r + t = f((r + t, v + m))\), and
\(f((r, m)) \cdot f((t, v)) = r \cdot t = f((r \cdot t, v \cdot m))\).

\subsubsection{b}
\label{sec:orgfb2a3ab}
If \(s \in S\) then \((0_R, s) \in R \times S\), and \(f((0_R, s)) = s\), so every range element has a domain element mapping to it.

For some \(a = (r, m) \in R \times S\) and \(b = (t, v) \in R \times S\) then \(f((r, m)) + f((t, v)) = m + v = f((r + t, m + v))\), and
\(f((r, m)) \cdot f((t, v)) = m \cdot v = f((r \cdot t, m \cdot v))\).

\subsection{Question Fifteen}
\label{sec:org4046d12}
Consider \(f: Z_4 \rightarrow Z_4 \ni f(x) = 0\), then 2 is a zero divisor, but 0 is not by definition.

\subsection{Question Twenty One}
\label{sec:org48d8cc7}
\subsubsection{Lemma}
\label{sec:org3598e64}
We assume a multiplicative identity \(x\) exists in \(\mathds{Z}^{*}\):

\(b \odot x = b \Rightarrow b + x - bx = b \Rightarrow x - bx = 0 \Rightarrow x(1-b) = 0\) and \(1-b \neq 0 \forall b \in \mathds{Z}^{*}\) so \(x = 0\).

which is verified by:

\(0 \odot b = 0 + b - 0 \cdot b = b\) and \(b \odot 0 = b + 0 - b \cdot 0 = b\).

\subsubsection{Proof}
\label{sec:orgee74248}

Consider \(f : \mathds{Z}^{} \rightarrow \mathds{Z}^{*}\) to be the isomorphism we so desire, then by Theorem 3.10,
f(1 \(\in\) \mathds{Z}) = 0 \(\in\) \mathds{Z}\textsuperscript{*}.

Therefore, f(2) = f(1 + 1) = f(1) \(\oplus\) f(1) = -1, f(3) = f(2 + 1) = -1 \(\oplus\) f(1) = -2.

\(f : \mathds{Z}^{} \rightarrow \mathds{Z}^{*} \ni f(x) = 1 - x\) seems to fit the bill nicely.

\begin{enumerate}
\item \(f\) is a bijection since we have an inverse \(x = 1 - f(x)\)
\item \(f(a \oplus b) = 1 - (a \oplus b) = 1 - (a + b - 1) = (1 - a) + (1 - b) = f(a) + f(b)\) and
\(f(a \odot b) = 1 - (a \odot b) = 1 - (a + b - ab) = (1-a) \cdot (1-b) = f(a) \cdot f(b)\)
\end{enumerate}

\subsection{Question Twenty Four}
\label{sec:org07a0b7a}
\subsubsection{a}
\label{sec:org897ff21}
\begin{enumerate}
\item By the usual coordinate addition we know that \(a, b \in R \Rightarrow a + b \in R\) and is associative, commutative, and the additive identity is \((0, 0)\).
\item \(R\) is closed under multiplication: \(a, b \in R \Rightarrow a = (c, d) \wedge b = (e, f) \Rightarrow a \cdot b = (ce, de)\) and since \(c, d, e, f \in \mathds{R}\) then \((ce, de) \in \mathds{R} \times \mathds{R}\) by definition.
\item Multiplication is associative: \(a, b, c \in R \Rightarrow a = (d, e) \wedge b = (f, g) \wedge c = (h, i) \Rightarrow (a \cdot b) \cdot c = (df, ef) \cdot (h, i) = (dfh, efh)\) and \(a \cdot (b \cdot c) = (d, e) \cdot (fh, gh) = (dfh, efh)\)
\item Multiplication is distributive: \(a, b, c \in R \Rightarrow a = (d, e) \wedge b = (f, g) \wedge c = (h, i) \Rightarrow a(b + c) = (e, f) \cdot (f + h, g + i) = (ef + eh, ff + fh) = (ef, ff) + (eh, fh) = a \cdot b + a \cdot c\)
and \(a, b, c \in R \Rightarrow a = (d, e) \wedge b = (f, g) \wedge c = (h, i) \Rightarrow (a + b) \cdot c = (d + f, e + g) \cdot (h, i) = (dh + fh, eh + gh) = (dh, eh) + (fh, gh) = a \cdot c + b \cdot c\)
\end{enumerate}
\subsubsection{b}
\label{sec:org92e2744}
Consider the function \(f: R \rightarrow M(\mathds{R})\) is an isomorphism, such that \(f((a, b)) =\) \(\begin{smallmatrix} a & 0 \\ b & 0 \end{smallmatrix}\), then:
\begin{enumerate}
\item \(f\) is surjective since every element in the range has a domain element \(\begin{smallmatrix} a & 0 \\ b & 0 \end{smallmatrix} = y \ni f((a, b)) = y\)
\item \(f\) is injective since \(f((c, d) = x) = f((e, f) = y) \Rightarrow\) \(\begin{smallmatrix} c & 0 \\ d & 0 \end{smallmatrix}\) = \(\begin{smallmatrix} e & 0 \\ f & 0 \end{smallmatrix}\) \(\Rightarrow c = e \wedge d = f \Rightarrow x = y\)
\item \(f((c, d) = x) + f((e, f) = y) \Rightarrow\) \(\begin{smallmatrix} c & 0 \\ d & 0 \end{smallmatrix}\) + \(\begin{smallmatrix} e & 0 \\ f & 0 \end{smallmatrix}\) = \(\begin{smallmatrix} c + e & 0 \\ d + f & 0 \end{smallmatrix}\)
\(= f(x + y)\), and \(f((c, d) = x) \cdot f((e, f) = y) \Rightarrow\) \(\begin{smallmatrix} c & 0 \\ d & 0 \end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix} e & 0 \\ f & 0 \end{smallmatrix}\) = \(\begin{smallmatrix} ce & 0 \\ de & 0 \end{smallmatrix}\)
\(= f(x \cdot y)\)
\end{enumerate}
\end{document}