# Recent posts by LukeMann on Kongregate

Originally posted by Gabidou99:

Someone already make a thread with the same gif not too long ago.

Don’t give a fuck; #YOLO.

Originally posted by VforVendetta:

Frozen was alright, though they sung every 10 minutes it seemed.
The real issue is how popular Let it Go got. My sister, random neighborhood kids, they’re all singing and listening to it. D; Not to mention its popularity on the radio.

Yeah, it’s a musical.

Discuss how crows just recently discovered a property of physics that humans have been using for thousands of years!

(Source)

Originally posted by oOTrentOo:

Actually, science says that this is the only thing America has done right.

I don’t know, I think America did pretty well in creating this fine specimen.

Originally posted by denamo:

Even though the city i currently study in borders Ukraine, and it’s the place i was born, and i have not been aware of this at all. Yet i don’t feel like this is touching me. All that happened here is an actual anti-government pamphlet from one of the parties hanging on a wall of some building, which tried to discredit Putin as some USA government agent, so i’m not sure if these people know what they are doing at all. Perhaps an actual military occupation would make them feel better?

>Posted Jan 25, 2014

What the fuck, man. Why did you give Putin the idea to invade Ukraine?

There were problems with the following fields:

Please provide body for first post of topic

>flew for office

lel

Did he ever give out the presents?

Originally posted by DMinor:

Am I the only one who can’t connect to Kongregate’s TC room?

Nope.

ew ew ew what. Why do these forums look terrible now?

Originally posted by oOTrentOo:

DONT look at any girls
DONT touch any girls
DONT talk to any girls
DONT pee your pants

DO bring extra animal crackers in case you drop some
DO bring 2 capri suns because 1 never quenches your thirst
ALWAYS be the first in line for recess to get that awesome fully pumped ball
ALWAYS be the first in the lunch line so you can get the best school lunch

Are you me?

Originally posted by Aldir:
Originally posted by Fishstickz13:

Don’t even go and go straight to work and learn a trade through hands on experience and apprenticeship and save yourself 40K

Bahahaa

and I suppose at the end of your apprenticeship you’re granted a Bachelor’s and Master’s?

Enjoy your poverty.

I’m sure you’ll get some real world experience some day; don’t worry.

Epigenetics is my bitch.

Originally posted by HappyAlcoholic:
Originally posted by Strawuni:

what girls think is “sexy”? nah bro, just be yourself, the right one will show up eventually.

That’s terrible advice.
The only way Rolby’s gonna get a girlfriend, is if he doesn’t be himslef.

Right? “Be yourself” is only good advice for people who are naturally charismatic, attractive, or rich.

Originally posted by RollerCROWster:

You all look like humans

because all humans look the same lol

discuss how CROWS are the only species with individual looks :>>>>

http://www.nwf.org/news-and-magazines/national-wildlife/birds/archives/2013/crows-recognizing-faces.aspx

Originally posted by CowFriend:

It’s also a complete mindfuck that you can get the same number without converting the units into consideration.

Welcome to the glorious wonder that is the metric system.

Topic: Off-topic / Good times in TC

Originally posted by DMinor:
Originally posted by Tacket:

For clarity.
I have not attended any of these TC sessions that I constantly get PM’d about.
More likely than not, it is just somebody that I don’t know using my likeness and tag.

He’s much better than you in every way. I think you should stay away from TC forever.

+1

Christmas is canceled.

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\usepackage[usenames,dvipsnames]{color, colortbl}
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\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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\thispagestyle{fancy}
\lhead{Homework \#1 \\ Math 283}
%\chead{Math 283}
\rhead{Lucas Mann \\ \today}

\begin{document}

\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}

\vs
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}

\vs
\hrule
\vs

{\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}
\vs

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
\vs

{\bf Section 1.3 \#14}
\vs

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}
\vs

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

\vs

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

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
\vs

{\bf Section 1.4 \#8}
\vs

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

\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.
\vs

\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}
\vs

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

\begin{description}

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

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

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

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

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

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

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

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

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

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

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

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

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

\newpage

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

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

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

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

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

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

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

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

\newpage

{\bf Section 1.6 \#2}
\vs

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.
\vs

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

$\overline{A}$ 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}
\vs

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}
\vs

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}
\vs

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}
\vs

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

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

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}

>The finger is literally a penis

I can’t all of this.

Aldir post nudes.

Originally posted by Haruhion:

# 2013 + 1No presents yet

Salvador please

Oh man.

Hue

lol

Get in.

It turned out nicely this year!