\documentclass[12pt]{article} \usepackage{ntnu} \usepackage{ntnu-math} \author{Øystein Tveit} \title{MA0301 Exercise 9} \usepackage{amsthm} \usepackage{mathabx} \begin{document} \ntnuTitle{} \break{} \begin{excs} \exc{} \begin{align*} p \rightarrow (q \vee r) &\equiv \neg p \vee (q \vee r) \\ &\equiv \neg p \vee q \vee r \\ &\equiv \neg (p \vee \neg q) \vee r \\ &\equiv (p \vee \neg q) \rightarrow r \\ \end{align*} \exc{} R does not define a partial orderering, because it is not transitive. $aRb$ and $bRc$, however $\neg aRc$ \exc{} \begin{subexcs} \subexc{} \begin{gather*} xyz + xy\overline{z}+\overline{x}y \\ xy + \overline{x}y \\ y \end{gather*} \subexc{} \begin{gather*} y + \overline{x}z + x\overline{y} \\ y + \overline{x}z + x \\ y + z + x \\ x + y + z \end{gather*} \end{subexcs} \exc{} Step 1: \begin{align*} \sum^1_{n=1}\frac{1}{(2n-1)(2n+1)} &= \frac{1}{2\cdot1 + 1} \\ \frac{1}{(2\cdot1-1)(2\cdot1+1)} &= \frac{1}{3} \\ \frac{1}{(1)(3)} &= \frac{1}{3} \\ \frac{1}{3} &= \frac{1}{3} \\ \end{align*} Step 2: Assume \[ \sum^k_{n=1} \frac{1}{(2n-1)(2n+1)} = \frac{k}{2k+1} \] then \begin{align*} \sum^{k+1}_{n=1} \frac{1}{(2n-1)(2n+1)} &= \frac{k}{2k+1} \\[2ex] &= \frac{1}{(2\cdot1 - 1)(2\cdot1+1)} + \frac{1}{(2\cdot2 - 1)(2\cdot2+1)} + \ldots \\[2ex] &\qquad + \frac{1}{(2\cdot k - 1)(2\cdot k+1)} + \frac{1}{(2\cdot(k+1) - 1)(2\cdot(k+1)+1)} \\[2ex] &= \frac{k}{2k+1} + \frac{1}{(2\cdot(k+1) - 1)(2\cdot(k+1)+1)} \\[2ex] &= \frac{k}{2k+1} + \frac{1}{(2k+1)(2k+3)} \\[2ex] &= \frac{k(2k1)(2k+3) + (2k+1)}{(2k+1)^2(2k+3)} \\[2ex] &= \frac{k(2k+3) + 1}{(2k+1)(2k+3)} \\[2ex] &= \frac{2k^2+3k + 1}{(2k+1)(2k+3)} \\[2ex] &= \frac{(2k+1)(k+1)}{(2k+1)(2k+3)} \\[2ex] &= \frac{(k+1)}{(2k+3)} \\[2ex] &= \frac{k+1}{2(k+1)+1} \end{align*} \exc{} \textbf{Injective:} Suppose $a,b \in \R$ \begin{align*} f(a) &= f(b) \\ 2a-3 &= 2b-3 \\ 2a &= 2b \\ a &= b \\ \end{align*} thus \[ f(a) = f(b) \Leftrightarrow a = b \] which means that $f$ is injective \\ \textbf{Surjective:} Suppose $a \in \R$ \begin{align*} a &= 2x-3 \\ x &= \frac{a+3}{2} \\ a &\in \R \end{align*} therefore $f$ is surjective \\ \textbf{Inverse:} \begin{align*} y &= 2x-3 \\[2ex] x &= \frac{y+3}{2} \\[2ex] f^{-1}(y) &= \frac{y+3}{2} \end{align*} \exc{} \begin{gather*} (\overline{X \cap Y \cap Z}) \\ \{ x \mid x \in \overline{X \cap Y \cap Z} \} \\ \{ x \mid x \notin X \cap Y \cap Z \} \\ \{ x \mid x \notin X \wedge x \notin Y \wedge x \notin Z \} \\ \{ x \mid x \in \overline{X} \vee x \in \overline{Y} \vee x \in \overline{Z} \} \\ \{ x \mid x \in \overline{X} \cup \overline{Y} \cup \overline{Z} \} \\ \overline{X} \cup \overline{Y} \cup \overline{Z} \end{gather*} \exc{} Assuming 'dozen' is to be interpreted as 12 \begin{subexcs} \subexc{} \[ \nPr{31}{12} = 67\ 596\ 957\ 267\ 840\ 000 \] \subexc{} \[ 31^{12} = 787\ 662\ 783\ 788\ 549\ 761 \] \end{subexcs} \exc{} \begin{subexcs} \subexc{} If we imagine a row of numbers going from 1 to 40, we can rephrase the question as how many ways we can split the numbers into $5$ chunks. Imagine a chunk as inserting $4$ delimiters like this: \[ 1\ 2\ 3\ |\ 4\ 5\ 6\ 7\ |\ 8\ 9\ 10\ |\ 11\ \ldots\ 39\ |\ 40 \] In this case, we split the amount of numbers so that $x_1 = 3, x_2 = 4, x_3 = 3, x_4 = 29, x_5 = 1$ By doing $\nCr{n}{r}$ where n is the number of numbers and delimiters, and r is the number of delimiters, we will get all combinations of $x_1 + x_2 + x_3 + x_4 + x_5 = 40$A In order to make it $x_1 + x_2 + x_3 + x_4 + x_5 \leq 40$, we will add a fifth delimiter, indicating the last block of unused numbers. $x_1 + x_2 + x_3 + x_4 + x_5 < 40 \Rightarrow x_1 + x_2 + x_3 + x_4 + x_5 \leq 39$ This leaves us with $\nCr{39 + 5}{5} = 1086008$ different combinations. \subexc{} In this case, we modify the problem by adjusting the inequality of $x_i$ like the following \begin{align*} x_1 + x_2 + x_3 + x_4 + x_5 &< 40 \qquad &&x_i \geq -3 \\ y_1 - 3 + y_2 - 3 + y_3 - 3 + y_4 - 3 + y_5 - 3 &< 40 &&(y_i - 3 = x) \\ y_1 + y_2 + y_3 + y_4 + y_5 &< 40 + 5 \cdot 3 \\ y_1 + y_2 + y_3 + y_4 + y_5 &< 55 \end{align*} \[ (y_i - 3 = x) \Rightarrow (y_i - 3 \geq - 3) \Leftrightarrow (y_i \geq 0) \] With this information, we use the same way of solving as in \textbf{a)} \[\nCr{54 + 5}{5} = 5006386 \] \end{subexcs} \break{} \exc{} This is a Venn diagram of the set containing the elements that satisfy $\overline{c}_1$, $\overline{c}_2$ and $\overline{c}_3$. \includeDiagram[width=11cm, caption={$N(\overline{c}_2\overline{c}_3\overline{c}_4$)}]{diagrams/ex9_1.tex} The following diagrams show the terms of the RHS. \includeDiagram[width=11cm, caption={$N(c_1\overline{c}_2\overline{c}_3\overline{c}_4)$}]{diagrams/ex9_2.tex} \includeDiagram[width=11cm, caption={$N(\overline{c_1}\overline{c}_2\overline{c}_3\overline{c}_4)$}]{diagrams/ex9_3.tex} When we lay these two diagrams on top of each other, we can see that $N(c_1\overline{c}_2\overline{c}_3\overline{c}_4) + N(\overline{c}_1\overline{c}_2\overline{c}_3\overline{c}_4)$ is equal to $N(\overline{c}_2\overline{c}_3\overline{c}_4)$ \includeDiagram[width=11cm, caption={$N(c_1\overline{c}_2\overline{c}_3\overline{c}_4) + N(\overline{c}_1\overline{c}_2\overline{c}_3\overline{c}_4)$}]{diagrams/ex9_4.tex} \exc{} let $c_1 = 2 \mid n $ \\ $c_2 = 3 \mid n $ \\ $c_3 = 5 \mid n $ \\ $c_4 = 7 \mid n $ \begin{align*} N(\overline{c}_1\overline{c}_2\overline{c}_3c_4) &= N(c_4) - \left( N(c_1c_7) + N(c_2c_7) + N(c_3c_7) \right) \\ &\quad + \left( N(c_1c_2c_4) + N(c_1c_3c_4) + N(c_2c_3c_4) \right) \\ &\quad - N(c_1c_2c_3c_4) \\[2ex] &= \left\lfloor \frac{2000}{7} \right\rfloor - \left( \left\lfloor \frac{2000}{2 \cdot 7} \right\rfloor + \left\lfloor \frac{2000}{3 \cdot 7} \right\rfloor + \left\lfloor \frac{2000}{5 \cdot 7} \right\rfloor \right) \\[2ex] &\quad + \left( \left\lfloor \frac{2000}{2 \cdot 3 \cdot 7} \right\rfloor + \left\lfloor \frac{2000}{2 \cdot 5 \cdot 7} \right\rfloor + \left\lfloor \frac{2000}{3 \cdot 5 \cdot 7} \right\rfloor \right) \\[2ex] &\quad - \left\lfloor \frac{2000}{2 \cdot 3 \cdot 5 \cdot 7} \right\rfloor \\[2ex] &= 76 \end{align*} \end{excs} \end{document}