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+#+TITLE: Assignment Four
+#+AUTHOR: Lizzy Hunt
+#+STARTUP: entitiespretty fold inlineimages
+#+LATEX_HEADER: \notindent \notag \usepackage{ dsfont } \usepackage{amsmath}
+#+LATEX: \setlength\parindent{0pt}
+#+OPTIONS: toc:nil
+
+* Section 3.1
+** Question One
+*** a
+$a \in R, b \in R \Rightarrow a + b \in R$
+*** b
+$a \in R \Rightarow x \in R \ni a + x = 0_R$
+** Question Three
+1. All operations are closed since only the elements in ${0, e, a, b}$ appear in the tables.
+2. From the second row and columns in the multiplication table we see that $e$ is the multiplicative identity.
+3. From the first row and columns in the addition table we see that $0$ is the zero element.
+4. In this field, each element is its own additive inverse.
+5. There is commutativity as the transpose of each table is identical to the original (symmetry along the diagonals).
+** Question Six
+*** a
+Since our addition and multiplication operators are the same in $\mathds{Z}$, we have
+associativity, commutativity, and distributivity.
+
+Sums of multiples of 3 are also multiples of 3: $3n + 3m = 3(n + m)$, so this set is closed under addition.
+
+Products of multiples of 3 are also multiples of 3: $(3n)(3m) = (3)(3nm)$, so this set is closed under multiplication.
+
+The additive inverse exists in the set for every element: $3n + x = 0 \Rightarrow x = 3 \cdot (-n)$.
+
+$0$ is the zero element and is a multiple of 3 since $3 \cdot 0 = 0$.
+
+Therefore ${x : x = 3n \ni n \in \mathds{Z}}$ is a subring of $\mathds{Z}$
+
+*** b
+Since our addition and multiplication operators are the same in $\mathds{Z}$, we have
+associativity, commutativity, and distributivity.
+
+Sums of multiples of $k$ are also multiples of $k$: $kn + km = k(n + m)$, so this set is closed under addition.
+
+Products of multiples of $k$ are also multiples of $k$: $(kn)(km) = (k)(knm)$, so this set is closed under multiplication.
+
+The additive inverse exists exists in the set for every element: $kn + x = 0 \Rightarrow k = k \cdot (-n)$.
+
+$0$ is the zero element and is a multiple of k since $k \cdot 0 = 0$.
+
+Therefore ${x : x = kn \ni n \in \mathds{Z}}$ is a subring of $\mathds{Z}$
+
+** Question Nine
+*** a
+${(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5)}$
+*** b
+Since our addition and multiplication operators are the same as in $R$, then we have
+associativity, commutativity, and distributivity.
+
+Sums of elements in $R*$ are also in $R*$ since any element $(r,r) \in R*, (j, j) \in R* = (r + j, r + j) \in R*$.
+
+Products of elements in $R*$ are also in $R*$ since any element $(r, r) \in R*, (j, j) \in R* = (rj, rj) \in R*$.
+
+The additive inverse exists in the set for every element: $(r, r) + x = 0 \Rightarrow x = (-r, -r) \in R*$.
+
+$(0, 0) \in R*$ is the zero element.
+
+Therefore ${(r, r) : (r, r) \in R*}$ is a subring of $R*$.
+
+** Question Ten
+Consider $a \in S \ni a = (10, -10)$ and $b \in S \ni b = (10, -10)$, then $ab \notin S$ since
+$ab = (100, 100)$ which does not follow the rule that $100 + 100 = 0$.
+** Question Eleven
+*** a
+Addition is closed since
+$\begin{smallmatrix}
+a & a \\
+b & b 
+\end{smallmatrix}$ + $\begin{smallmatrix}
+c & c \\
+d & d 
+\end{smallmatrix}$ = $\begin{smallmatrix}
+(a + c) & (a + c) \\
+(b + d) & (b + d)
+\end{smallmatrix}$ which of the form of the given rule.
+
+It is also closed under multiplication since
+$\begin{smallmatrix}
+a & a \\
+b & b 
+\end{smallmatrix}$ $\cdot$ $\begin{smallmatrix}
+c & c \\
+d & d 
+\end{smallmatrix}$ = $\begin{smallmatrix}
+(ac + ad) & (ac + ad) \\
+(bc + bd) & (bc + bd)
+\end{smallmatrix}$ which is also of the form of the given rule.
+
+The zero element is the zero matrix $\begin{smallmatrix}
+0 & 0 \\
+0 & 0
+\end{smallmatrix}$, trivially.
+
+Associativity and commutivity (of addition) come from these operations existing for 2x2 matrices in $M(\mathds{R})$.
+
+*** b
+$\begin{smallmatrix}
+a & a \\
+b & b 
+\end{smallmatrix}$ $\cdot$ $\begin{smallmatrix}
+1 & 1 \\
+0 & 0 
+\end{smallmatrix}$ = $\begin{smallmatrix}
+(a(1) + a(0)) & (a(1) + a(0)) \\
+(b(1) + b(0)) & (b(1) + b(0))
+\end{smallmatrix}$
+which is equivalent to $\begin{smallmatrix}
+a & a \\
+b & b 
+\end{smallmatrix}$
+
+*** c
+$\begin{smallmatrix}
+1 & 1 \\
+0 & 0 
+\end{smallmatrix}$ $\cdot$ $\begin{smallmatrix}
+1 & 1 \\
+2 & 2
+\end{smallmatrix}$ = $\begin{smallmatrix}
+((1)(1) + (1)(2)) & ((1)(1) + (1)(2)) \\
+((0)(1) + (0)(2)) & ((0)(1) + (0)(2))
+\end{smallmatrix}$ = $\begin{smallmatrix}
+3 & 3 \\
+0 & 0
+\end{smallmatrix}$.
+** Question Twelve
+$\mathds{Z}[i]$ is closed under addition: $(a + bi) + (c + di) = (a + c) + (b + d)i$ and since
+$\mathds{Z}$ is a ring itself, $(a + c) + (b + d)i$ is also in $\mathds{Z}[i]$.
+
+$\mathds{Z}[i]$ is closed under multiplication: $(a + bi) \cdot (c + di) = ac + adi + cbi + bdi^2 = (ac - bd) + (ad + bd)i$
+by similar logic.
+
+The additive inverse always exists in the set: $(a + bi) + x = 0 \Rightarrow x = -a - bi$.
+
+Finally, the zero element is trivially $0 + 0i$ since $(a + bi) + (0 + 0i) = a + bi$.
+
+** Question Fourteen
+$S$ is closed under addition. For example given some $f, g \in S$ then $(f + g)(x) = h$ and $h$ will still satisfy the rule that
+$h(2) = 0$ since $(f + g)(2) = f(2) + g(2) = 0 + 0$, and addition was already closed in the domain given that question 8 is a ring
+itself.
+
+Similarly, $S$ is closed under multiplication. $(f \cdot g)(x) = h$ and h will still satisfy the rule as $h(2) = f(2) \cdot g(2) = 0 \cdot 0$.
+
+The zero element is $f \ni f(x) = 0$.
+
+Finally, the additive inverse $g$ exists in the set for each $f$ since $f + g = 0$ implies that $g$ is just $-f$ (the rule still
+stands here).
+
+** Question Fifteen
+*** b
+| \odot      | (0, 0) | (0, 1) | (0, 2) | (1, 0) | (1, 1) | (1, 2) |
+| (0, 0) | (0, 0) | (0, 0) | (0, 0) | (0, 0) | (0, 0) | (0, 0) |
+| (0, 1) | (0, 0) | (0, 1) | (0, 2) | (0, 0) | (0, 1) | (0, 2) |
+| (0, 2) | (0, 0) | (0, 2) | (0, 1) | (0, 0) | (0, 2) | (0, 1) |
+| (1, 0) | (0, 0) | (0, 0) | (0, 0) | (1, 0) | (1, 0) | (1, 0) |
+| (1, 1) | (0, 0) | (0, 1) | (0, 2) | (1, 0) | (1, 1) | (1, 2) |
+| (1, 2) | (0, 0) | (0, 2) | (0, 1) | (1, 0) | (1, 2) | (1, 1) |
+
+| \oplus     | (0, 0) | (0, 1) | (0, 2) | (1, 0) | (1, 1) | (1, 2) |
+| (0, 0) | (0, 0) | (0, 1) | (0, 2) | (1, 0) | (1, 1) | (1, 2) |
+| (0, 1) | (0, 1) | (0, 2) | (0, 0) | (1, 1) | (1, 2) | (1, 0) |
+| (0, 2) | (0, 2) | (0, 0) | (0, 1) | (1, 2) | (1, 0) | (1, 1) |
+| (1, 0) | (1, 0) | (1, 1) | (1, 2) | (2, 0) | (2, 1) | (2, 2) |
+| (1, 1) | (1, 1) | (1, 2) | (1, 0) | (2, 1) | (2, 2) | (2, 0) |
+| (1, 2) | (1, 2) | (1, 0) | (1, 1) | (2, 2) | (2, 0) | (2, 1) |
+
+** Question Eighteen
+No, by definition, as the distributive axiom is violated for this to be a ring.
+
+For example, $1(1 + 1) = (1)(1) + (1)(1) = 2$ but $1(1 + 1) = 1$.