Topic: Off-topic /
Brag post. I made this in LaTeX.

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\usepackage{amssymb,amsmath, amsthm, amsfonts, cancel, enumerate, array }

\usepackage[usenames,dvipsnames]{color, colortbl}

\usepackage{graphicx}

\usepackage{fancyhdr}

\usepackage{hyperref}

\usepackage{tikz}

\usetikzlibrary{shapes,backgrounds}

%spacing

\parindent=0in

\def\vs{\vspace{1\baselineskip}}

\def\svs{\vspace{.5\baselineskip}}

%Some shortcuts

\newcommand{\done}{\hfill $\boxtimes$}

\newcommand{\itg}{\mathbb{Z}}

\newcommand{\rat}{\mathbb{Q}}

\newcommand{\nat}{\mathbb{N}}

\newcommand{\comp}{\mathbb{C}}

\newcommand{\real}{\mathbb{R}}

\newcommand{\parti}{\mathcal{P}}

\renewcommand{\iff}{\longleftrightarrow}

\newcommand{\bprove}{\textbf{Prove: }}

\newcommand{\bproof}{\textbf{Proof: }}

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\newcommand{\onto}{\xrightarrow{\text{onto}}}

%coloring

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\newcolumntype{a}{>{\columncolor{LightCyan}}c}

\thispagestyle{fancy}

\lhead{Homework \#1 \\ Math 283}

%\chead{Math 283}

\rhead{Lucas Mann \\ \today}

\begin{document}

lukemann@fakeemail.edu \\

\hrule

\vs %this is a user defined command to insert a line space and is equivalent to \vspace{1\baselineskip}

{\bf Section 1.1 \#16}

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Write $\{6a+2b : a,b \in \itg \}$ by listing its elements between braces.

\vs

{\bf Solution:}

\begin{proof}

Any positive or negative multiple of 6 or of 2 is possible due to $a$ and $b$ being in the set of all integers. Since the multiples of 6 are a subset of the multiple of 2 and since $6a + 2b = 0$ when $a = 0$ and $b = 0$, it logically follows that the elements of the set are all possible even integers. Thus, the following set represents all elements of the above set.

$$\{…, -8, -6, -4, -2, 0, 2, 4, 6, 8, …\}.$$

\end{proof}

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\hrule

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{\bf Section 1.1 \#26} \\

Write $\{…, \frac{1}{27}, \frac{1}{9}, \frac{1}{3}, 1, 3, 9, 27, …\}$ in set-builder notation. \\

{\bf Solution:}

\begin{proof}

The shown elements of the set can be rewritten as $\{…, 3^{-3}, 3^{-2}, 3^{-1}, 3^0, 3^1, 3^2, 3^3, …\}$. \\

Thus, the following set-builder notation can be used to express the set:

$$ \{3^x : x\in \nat \}.$$

\end{proof}

\newpage

{\bf Section 1.2 \#2}

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Suppose $A = \{\pi, e, 0\}$ and $B = \{0, 1\}$.

\begin{description}

\item{(a)} Find $A \times B$. \\

This is the set consisting of all ordered pairs with first component an element in A and second component an element in B.

Thus:

$$A \times B = \{(\pi, 0), (\pi, 1), (e, 0), (e, 1), (0, 0), (0, 1)\}.$$ \done

\item{(b)} Find $B \times A$.

This is the set consisting of all ordered pairs with first component an element in B and second component an element in A.

Thus:

$$B \times A = \{(0, \pi), (0, e), (0, 0), (1, \pi), (1, e), (1, 0)\}.$$ \done

\item{(d)} Find $B \times B$.

This is the set consisting of all ordered pairs with first component an element in B and second component also an element in B.

Thus:

$$B \times B = \{(0, 0), (0, 1), (1, 0), (1, 1)\}.$$ \done

\end{description}

\vs

\hrule

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{\bf Section 1.3 \#14}

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Decide if $\real^2 \subseteq \real^3$.

\begin{proof}

$(0, 0)$ is an element within $\real^2.$ Since $\real^3$ is comprised only of ordered triplets and $(0, 0)$ is not an ordered triplet, it follows that $\real^2 \not \subseteq \real^3$.

\end{proof}

\newpage

{\bf Section 1.4 \#2}

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Find $\mathcal{P}(\{1, 2, 3, 4\})$.

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Since the cardinality of $\{1, 2, 3, 4\}$ is 4, the cardinality for $\mathcal{P}(\{1, 2, 3, 4\})$ is $2^4$.

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Using the definition of a power series, we find:

\begin{multline}

\mathcal{P}(\{1, 2, 3, 4\}) = \{ \{\}, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}, \{1, 4\}, \{2, 3\}, \{2, 4\}, \{3, 4\}, \\

\{1, 2, 3\}, \{1, 2, 4\}, \{1, 3, 4\}, \{2, 3, 4\}, \{1, 2, 3, 4\}\}. \notag

\end{multline}

\done

\vs

\hrule

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{\bf Section 1.4 \#8}

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Find $\mathcal{P}(\{1,2\} \times \{3\})$.

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\begin{description}

\item{(1)} $\{1,2\} \times \{3\} = \{(1,3),(2,3)\}.$

\vs

\item{(2)} $\{(1,2),(1,3)\}$ has cardinality = 2, so $\mathcal{P}(\{(1,3),(2,3)\})$ has $2^2$ elements.

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\item{(3)} By definition, $\mathcal{P}(\{(1,3),(2,3)\}) = \{ \{\}, \{(1,2)\}, \{(1,3)\}, \{(1,2),(1,3)\}\}$.

\end{description}

\done

\newpage

{\bf Section 1.5 \#4}

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Suppose $A = \{b, c, d\}$ and $B = \{a,b\}$.

\begin{description}

\item{(d)} Find $(A\cap B) \times A$.

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The only element in both $A$ and $B$ is $b$, so $A\cap B = \{b\}$.

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$\{b\} \times \{b,c,d\} = \{(b,b), (b,c), (b,d)\}$ by definition.

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Thus, $(A\cap B) \times A = \{(b,b), (b,c), (b,d)\}$.

\vs

\done

\item{(e)} Find $(A \times B) \cap B$.

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$A \times B = \{(b,a),(b,b),(c,a),(c,b),(d,a),(d,b)\}$ by definition.

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$B$ has none of these elements, as none of the elements of $B$ are ordered pairs.

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Thus, $(A \times B) \cap B = \{\}$.

\vs

\done

\item{(f)} Find $\mathcal{P}(A)\cap \mathcal{P}(B)$.

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$\mathcal{P}(A) = \{ \{\}, \{b\}, \{c\}, \{d\}, \{b,c\}, \{b,d\}, \{c,d\}, \{b,c,d\} \}$ by definition. \\

$\mathcal{P}(B) = \{ \{\}, \{a\}, \{b\}, \{a,b\} \}$ by definition.

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The only elements in both $\mathcal{P}(A)$ and $\mathcal{P}(B)$ are $\{\}$ and $\{b\}$.

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Thus $\mathcal{P}(A)\cap \mathcal{P}(B) = \{ \{\}, \{b\} \}$.

\vs

\done

\newpage

\item{(g)} Find $\mathcal{P}(A) – \mathcal{P}(B)$.

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$\mathcal{P}(A) = \{ \{\}, \{b\}, \{c\}, \{d\}, \{b,c\}, \{b,d\}, \{c,d\}, \{b,c,d\} \}$ by definition. \\

$\mathcal{P}(B) = \{ \{\}, \{a\}, \{b\}, \{a,b\} \}$ by definition.

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The elements in $\mathcal{P}(A)$ that are not in $\mathcal{P}(B)$ are given below.

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$$\mathcal{P}(A) – \mathcal{P}(B) = \{ \{c\}, \{d\}, \{b,c\}, \{b,d\}, \{c,d\}, \{b,c,d\} \}.$$

\vs

\done

\item{(h)} Find $\mathcal{P}(A\cap B)$.

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The only element in both $A$ and $B$ is $b$, so $A\cap B = \{b\}$.

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$\mathcal{P}(A\cap B) = \mathcal{P}(\{b\}) = \{ \{\}, \{b\}\}$.

\done

\end{description}

\newpage

{\bf Section 1.6 \#2}

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Let $A = \{0,2,4,6,8\}$, $B = \{1,3,5,7\}$, and the universal set $U = \{0,1,2,3,4,5,6,7,8\}$.

\vs

\begin{description}

\item{(g)} Find $\overline{A} \cap \overline{B}$.

\vs

$\overline{A} = U – A = \{1,3,5,7\}$ by definition.

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$\overline{B} = U – B = \{0,2,4,6,8\}$ by definition.

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$\overline{A}$ and $\overline{B}$ have no elements in common, thus $\overline{A} \cap \overline{B} = \{\}$.

\done

\vs

\item{(h)} Find $\overline{A \cap B}$.

\vs

$A$ and $B$ have no elements in common, thus $A \cap B = \{\}$.

$$\overline{A\cap B} = U – (A\cap B) = U – \{\} = U.$$

\done

\vs

\item{(i)} Find $\overline{A} \times B$.

\vs

$\overline{A} = U – A = \{1,3,5,7\}$ by definition.

\begin{multline}

\{1,3,5,7\} \times \{1,3,5,7\} = \{(1,1),(1,3),(1,5),(1,7),(3,1),(3,3),(3,5),(3,7),\\

(5,1),(5,3),(5,5),(5,7),(7,1),(7,3),(7,5),(7,7)\}. \notag

\end{multline}

\done

\end{description}

\newpage

{\bf Section 1.7 \#6}

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Does $A\cap (B\cup C) = (A\cap B)\cup (A\cap C)$?

\def \setA{ (0,0) circle (1cm) }

\def \setB{ (1.5,0) circle (1cm) }

\def \setC{ (60:1.5) circle (1cm) }

\def \setU{ (-2, -1.5) rectangle (3.5, 2.75) }

\begin{center}

\begin{tikzpicture}

\draw \setU node[below left]{$U$};

\begin{scope}

\fill[ForestGreen] \setB;

\fill[ForestGreen] \setC;

\end{scope}

\begin{scope}

\fill[cyan] \setA;

\end{scope}

\begin{scope}

\clip \setC;

\fill[Violet] \setA;

\end{scope}

\begin{scope}

\clip \setB;

\fill[Violet] \setA;

\end{scope}

\draw \setA node[left] {$A$};

\draw \setB node[right] {$B$};

\draw \setC node {$C$};

\end{tikzpicture}

\end{center}

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The above Venn diagram shows $B \cup C$ shaded in green. $A$ is shown shaded in cyan. The overlapping region of the two, $A\cap (B\cup C)$ is shown in violet.

\def \setA{ (0,0) circle (1cm) }

\def \setB{ (1.5,0) circle (1cm) }

\def \setC{ (60:1.5) circle (1cm) }

\def \setU{ (-2, -1.5) rectangle (3.5, 2.75) }

\begin{center}

\begin{tikzpicture}

\draw \setU node[below left]{$U$};

\begin{scope}

\clip \setB;

\fill[BurntOrange] \setA;

\end{scope}

\begin{scope}

\clip \setC;

\fill[Apricot] \setA;

\end{scope}

\begin{scope}

\clip \setB;

\clip \setC;

\fill[BrickRed] \setA;

\end{scope}

\draw \setA node[left] {$A$};

\draw \setB node[right] {$B$};

\draw \setC node {$C$};

\end{tikzpicture}

\end{center}

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Here, $A\cap B$ is shown shaded in orange. $A\cap C$ is shaded in light apricot. The deep red region satisfies both $A\cap C$ and $A\cap B$. Given the initial condition $(A\cap B)\cup (A\cap C)$, any highlighted region in this diagram satisfies the condition.

\vs

As visual inspection shows the respective regions of $A\cap (B\cup C)$ and $(A\cap B)\cup (A\cap C)$ as equivalent, it can be concluded that $A\cap (B\cup C) = (A\cap B)\cup (A\cap C)$.

\vs

\done

\newpage

{\bf Section 1.8 \#6}

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Find $\bigcup_{i\in \nat} [0,i+1]$.

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$$\bigcup_{i\in \nat} [0,i+1] = [0,2]\cup[0,3]\cup[0,4]\cup[0,5]\cup…$$

One can conclude via inspection that every set in this index has first component 0 and second component a natural number. Since this is the union of the sets, one may further conclude that since the regions included approach $[0, \inf )$, the region may be modeled as the region $0, \inf )$.

\done

\end{document}