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#+TITLE: Assignment Three
#+AUTHOR: Logan Hunt
#+STARTUP: entitiespretty fold inlineimages
#+LATEX_HEADER: \notindent \notag \usepackage{ dsfont } \usepackage{amsmath}
#+LATEX: \setlength\parindent{0pt}
#+OPTIONS: toc:nil
* Section 2.2
** Question One
*** b
| \oplus | 0 | 1 | 2 | 3 |
| 0 | 0 | 1 | 2 | 3 |
| 1 | 1 | 2 | 3 | 0 |
| 2 | 2 | 3 | 0 | 1 |
| 3 | 3 | 0 | 1 | 2 |
| \odot | 0 | 1 | 2 | 3 |
| 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 2 | 3 |
| 2 | 0 | 2 | 0 | 2 |
| 3 | 0 | 3 | 2 | 1 |
*** c
| \oplus | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| 0 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | 1 | 2 | 3 | 4 | 5 | 6 | 0 |
| 2 | 2 | 3 | 4 | 5 | 6 | 0 | 1 |
| 3 | 3 | 4 | 5 | 6 | 0 | 1 | 2 |
| 4 | 4 | 5 | 6 | 0 | 1 | 2 | 3 |
| 5 | 5 | 6 | 0 | 1 | 2 | 3 | 4 |
| 6 | 6 | 0 | 1 | 2 | 3 | 4 | 5 |
| \odot | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| 2 | 0 | 2 | 4 | 6 | 1 | 3 | 5 |
| 3 | 0 | 3 | 6 | 2 | 5 | 1 | 4 |
| 4 | 0 | 4 | 1 | 5 | 2 | 6 | 3 |
| 5 | 0 | 5 | 3 | 1 | 6 | 4 | 2 |
| 6 | 0 | 6 | 5 | 4 | 3 | 2 | 1 |
** Question Three
$x = [1], [3], [5], [7]$
** Question Five
$x = [1], [2], [4], [5]$
** Question Eight
$x = [1], [2], [6], [7]$
** Question Eleven
*** b
$x = [0], [1], [2], [3]$
** Question Fifteen
*** c
From the Binomial Theorem,
\begin{align*}
(a + b)^5 &= a^5 + 5a^4 b + 10a^3 b^2 + 10a^2 b^3 + 5ab^4 + b^5
\end{align*}
Then,
\begin{equation*}
(a + b)^5 &\equiv_5 (a^5 + 5a^4 b + 10a^3 b^2 + 10a^2 b^3 + 5ab^4 + b^5) \\
&\equiv_5 (a^5 + b^5) \\
\end{equation*}
since each of the terms $5a^4 b, 10a^3 b^2, 10a^2 b^3, 5ab^4$ are zero as $5 \equiv_5 0$ and $10 \equiv_5 0$.
* Section 2.3
** Question One
*** b
$[1], [3], [5], [7]$ since $[7 * 7] = [49] = [1]$, $[5 * 5] = [25] = [1]$, $[3 * 3] = [9] = [1]$, and
$[1 * 1] = [1]$.
** Question Two
*** b
$[2], [4], [6]$ since $[2 * 4] = [8] = [0]$, $[4 * 4] = [16] = [0]$, and $[6 * 4] = [24] = [0]$.
** Question Eight
*** a
1. $2x = 1$
2. $2x = 3$
3. $2x = 5$
*** b
Yes, each one is equivalent to 0 when $x = 6$.
** Question Nine
*** a
By definition, there exists $b$, the inverse of $a$, such that $ab = 1$.
Assume that $a$ is a zero divisor, then there exists $c \neq 0$
such that $ac = 0$. Then, $(ab)c = 0b \Rightarrow (1)(c) = 0$ implies that $c = 0$, which is a contradiction.
*** b
By definition, there exists $b$, with $b \neq 0$ such that $ab = 0$.
Assume that $a$ is a unit, then there exists $c$ such that $ac = 1$.
Then, $(b)ac = 1b \Rightarrow 0c = b$ implies that $b = 0$, which is a contradiction.
** Question Eleven
By definition of $a$ being a unit, there exists $ay = 1$ with $y$ being an inverse of $a$.
By multiplying our target $(y)ax = (y)b \Rightarrow x = yb$.
To prove this is unique, assume that $k$ and $l$ are solutions of $ax = b$. Then, $ak = b$ and $al = b$. Since $a$ is a unit, by using our
previous strategy, $(y)ak = (y)b$ and $(y)al = (y)b$, so $k = yb$ and $l = yb$ and thus $k = l$.
* Chapter 13
** A2
As $p | c \Rightarrow c = pk$, and $p | c \Rightarrow c = ql$ then $pk = ql$ and thus by Theorem 1.5 since $p$ and $q$ are prime $p | l$
or $p | q$ and $q | p$ or $q | k$, but since $p$ and $q$ are prime, then it must be that only $p | l$ and $q | k$.
Since $p | l$ then $l = mp$ and $c = ql \Rightarrow c = qmp$ and $qp$ is a factor of $c$.
** A3 (a)
GO = 0715
715^3 (mod 2773) = 107
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