matematik ödevlerini latex ile yapmak

Matematik odevlerini kompozisyon yazar gibi latexde yazmaktir... Ilk baslarda zor gelir insana, belli bir sure sonra alisilir, hatta elle yapmaktan daha hizli gelir... Belli bir sure sonra kafanin icinde latex ile dusunmeye basladigini fark edersin...

Cok kraldir, hem hocalarin gozunde temiz bir odev verdigin icin yer edinirsin, hem odevleri guzelce arsivlersin... Problem oldu mu degistirmek daha kolaydir mesela... Alisirsan derste latexle bile not alabilirsin...

zaen bu cagda cvsini veya tezini latexsiz yazana adam gozuyle bakmiyorlar... Simdiden matematik ile alistirma yapmak ve deneyim elde etmek, ileride cok buyuk katki salayuacaktir bunyeye...

Edit: Yogun istek uzerine

Windows: http://miktex.org/
Unix: http://www.ling.upenn.edu/advice/latex/starting.html
Mac: http://www.tug.org/mactex/2011/

Misal bir odev suna benziyor, templatimi kullanmak isteyen kullanabilir...

\documentclass[a4paper,12pt]{article}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{mathtools}
\usepackage[T1]{fontenc}
\usepackage{titling}

\setlength{\droptitle}{-10em}

\title {Tutorial}
\author {Bard of Hell}

\begin{document}
\maketitle

1. I want to show that $P \Rightarrow Q$ is logically equivalent to $(\neg Q) \Rightarrow (\neg P)$.
\begin {proof}
By the truth tables below,

\begin{minipage}{2in}
\begin{tabular}{c c c}
\hline\hline
P & Q & $P \Rightarrow Q$\\ [0.5ex]
\hline
T & T & T \\
T & F & F \\
F & T & T \\
F & F & T \\ [1ex]
\hline
\end{tabular}
\end{minipage}
\begin{minipage}{2in}
\begin{tabular}{c c c c c}
\hline\hline
$P$ & $Q$ & $\neg Q$ & $\neg P$ & $(\neg Q) \Rightarrow (\neg P)$\\ [0.5ex]
\hline
T & T & F & F & T \\
T & F & T & F & F \\
F & T & F & T & T \\
F & F & T & T & T \\ [1ex]
\hline
\end{tabular}
\end{minipage}

We can see that: $P \Rightarrow Q = (\neg Q) \Rightarrow (\neg P)$
\end {proof}

\vspace{2 pc}

2. De Morgan\rq{}s Laws:

$(i)$: $\neg (P \land Q) = (\neg P) \lor (\neg Q)$

\begin {proof}
By the truth tables below,

\begin{minipage}{2in}
\begin{tabular}{c c c c}
\hline\hline
$P$ & $Q$ & $P \land Q$ & $\neg (P \land Q)$ \\ [0.5ex]
\hline
T & T & T & F \\
T & F & F & T \\
F & T & F & T \\
F & F & F & T \\ [1ex]
\hline
\end{tabular}
\end{minipage}
\begin{minipage}{2in}
\begin{tabular}{c c c c c}
\hline\hline
$P$ & $Q$ & $\neg P$ & $\neg Q$ & $(\neg P) \lor (\neg Q)$ \\ [0.5ex]
\hline
T & T & F & F & F \\
T & F & F & T & T \\
F & T & T & F & T\\
F & F & T & T & T\\ [1ex]
\hline
\end{tabular}
\end{minipage}

Thus we can say $\neg (P \land Q) = (\neg P) \lor (\neg Q)$
\end {proof}

$(ii)$: $\neg (P \lor Q) = (\neg P) \land (\neg Q)$

\begin {proof}
By the truth tables below,

\begin{minipage}{2in}
\begin{tabular}{c c c c}
\hline\hline
$P$ & $Q$ & $P \lor Q$ & $\neg (P \lor Q)$ \\ [0.5ex]
\hline
T & T & T & F \\
T & F & T & F \\
F & T & T & F \\
F & F & F & T \\ [1ex]
\hline
\end{tabular}
\end{minipage}
\begin{minipage}{2in}
\begin{tabular}{c c c c c}
\hline\hline
$P$ & $Q$ & $\neg P$ & $\neg Q$ & $(\neg P) \land (\neg Q)$ \\ [0.5ex]
\hline
T & T & F & F & F \\
T & F & F & T & F \\
F & T & T & F & F\\
F & F & T & T & T\\ [1ex]
\hline
\end{tabular}
\end{minipage}

Thus we can say $\neg (P \lor Q) = (\neg P) \land (\neg Q)$
\end {proof}

\vspace{2 pc}

3. Let $a\nmid b = \neg(a \mid b)$.This means that \lq\lq{}it is not true that $a$ divides $b$\rq\rq{}. Or to put it in other words \lq\lq{}a does not divide b\rq\rq{}. Mathematically speaking, this means $\neg(\exists n \in \mathbb{Z}, an=b)$ , this is equivalent to $\forall n \in \mathbb{Z}, an \neq b$.

I want to show that if $a,b,c \in \mathbb{Z}$ and $a \neq 0$. If $a \nmid b$, then $a \nmid b$ and $a \nmid c$.
\begin {proof}
I will prove this relation by contrapositive proof. Suppose $a \mid b$ or $a \mid c$. This means that $\exists k_1,k_2 \in \mathbb{Z}$ such that $a k_1 = b$ or $a k_2 =c$.

Case 1: $a k_1 = b$. Then $a k_1 = b $. If I multiply both sides by $c$ we get, $a k_1 c = b c$. Since $c \in \mathbb{Z}$ therefore $k_1 c \in \mathbb{Z}$ then $a \mid b c$.

Case 2: $a k_2 = c$. Then $a k_2 = c$. If I multiply both sides by $b$ we get, $a k_2 b = c b$. Since $b \in \mathbb{Z}$ therefore $k_2 b \in \mathbb{Z}$ then $a \mid b c$.

Thus, we can conclude that if $a,b,c \in \mathbb{Z}$ and $a \neq 0$. If $a \nmid b$, then $a \nmid b$ and $a \nmid c$.
\end {proof}

1.5.9) Let $R$ be a commutative ring. Let $P$ be the set with following properties:

($i$) If $a \in R$, then one and only one of the followings hold $a \in P, a=0, -a \in P$

($ii$) If $a,b \in P$, then $a+b \in P$ and $ab \in P$

Now define $a,b \in R$ $a<b$ if $b-a \in P$

Now I want to show that this relation satisfies $(O1) - (O4)$.

$(O1)$ So, let $b,a \in R$. By closure on $R$, $b-a \in R$ then by $i$ either $b-a \in P$, $a-b \in P$ or $a-b =0$.

Case 1: $b-a \in P$ then by the definition of $<$; $a<b$.

Case 2: $a-b \in P$ then by the definition of $<$; $b<a$.

Case 3: $a-b =0$, if I add $b$ to both sides I get $a =b$

$(O2)$ Let $a,b,c \in R$ and let $b-a \in P$ and $c - b \in P$. This means $a<b$ and $b<c$. By the $ii$ ; $(b-a) + (c-b) \in P$. By the additive commutativity and additive inverse properties of $R$ we can say $c-a \in P$. Thus $a<c$.

$(O3)$ Let $a,b,c \in R$. Let $b-a \in P$. This means $a<b$. By the additive identity in R $b-a = b-a + 0$. By the additive inverse in $R$, $b-a = b-a + c-c$. By the commutative property in addition in $R$, $b-a = b+c -a -c$. And lastly, by the distributive property of R, $b-a = (b+c)-(a+c)$. Then $(b+c)-(a+c) \in P$, thus $a+c < b+c$.

$(O4)$ Let $a,b,c \in R$. Let $b-a \in P$, then $a<b$. Also suppose $c - 0 \in P$, by the additive identity in addition on $R$; $c \in P$ which transcribes as $0<c$. By the $ii$ property $c(b-a) \in P$. By the distributive property $cb - ca \in P$. This means $ca < cb$.

Similarly let $ P = \{ a \in R | a > 0 \}$. And $a,b \in R$

$(i)$ Case 1: $a=0$. There are\rq{}t any conditions for $a \in A$ thus $a$ is free to be $0$.

Case 2: $a>0$. Then $a \in P$

Case 3: $a<0$. By the additive inverse property in $R$, I can add $-a$ to both sides, thus $a-a=0<-a$. Then
$-a \in P$.

$(ii)$. $a>0, b>0$. If I add $b$ to $a>0$, then $a+b>b$. Since $b>0$, $b+a > b >0$ thus $b+a >0$. Then
$b+a \in P$.

Lets multiply $a>0$ by $b$, $ab>0$. Then $ab \in P$.

\vspace{2 pc}

5. $(iii)$ I want to show that no matter how we define $>$, the order relations don\rq{}t work!

\begin{proof}
Suppose Not! Assume that all of the order relations hold for mod($n$). Now assume $\bar 1 > \bar 0$. And let $\overline{n-1} > 0$. By the $(O3)$, $\bar 1 +\overline{n-1} > \overline{n-1}$. Thus $\bar n > \overline{n-1}$. But by definition, $\bar n = \bar 0$ thus $\bar 0 > \overline{n-1}$ thus by $(O1)$ this is a contradiction. $\Rightarrow \Leftarrow$
\end{proof}

\vspace{2 pc}

$(v)$ I want to show that $\mathbb{Z}_ p$ is a field for prime $p$.
\begin{proof}
First of, I have shown in the previous homework that the line between being a field or not is $(M5)$, the multiplicative inverse property. So let me restate my statement; if $\forall j \in \mathbb{Z}_n, \exists k \in \mathbb{Z}_n$ such that $k j =mn +1$, for $m \in \mathbb{Z}$, then $n$ is prime.

Suppose not! This means that if $\forall j \in \mathbb{Z}_n, \exists k \in \mathbb{Z}_n$ such that $k j =mn +1$, for $m \in \mathbb{Z}$, then $n$ is not prime, therefore I can write $n = x p$ such that $x, p \in \mathbb{Z}$ and $p$ is prime.

We know that both $x , p < n$. Thus lets investigate the case $j=p$, then $k p = m x p + 1$. Subtract $kp+1$ from both sides, we get $ -1 =mxp -kp$. Lets multiply both sides by $-1$ and use the distributive property to claim $p (k - mx) = 1$. If I divide both sides by $p$, we get $(k - mx) = \frac{1}{p}$. We know that $k,m,x \in \mathbb{Z}$ and by the closure in $\mathbb{Z}$ any binary operation should give a value in $\mathbb{Z}$ yet $1/p \notin \mathbb{Z}$ $\Rightarrow \Leftarrow$
\end{proof}

\vspace{2 pc}

1.7.25 I want to show that for $A$ and $B$ sets and $f:A \rightarrow B$ function, and for $A_1,A_2 \subseteq A$, if $f(A_1) \cap f(A_2) \subseteq f(A_1 \cap A_2)$ is not always correct. Here is counter-example:

Let $f: A \rightarrow B$ where $A = B = \mathbb{Z}$ and let $f$ be defined as;

\begin{displaymath}
f(x) = \left\{
\begin{array}{lr}
3, & x=2 \\
x, & otherwise
\end{array}
\right.
\end{displaymath}

Now let $A_1 = \{2n+1 | n \in \mathbb{N}\}, A_2 = \{2n | n \in \mathbb{N}\}$. Since $2n+1 \neq 2n$ for any natural number, $A_1 \cap A_2 = \emptyset$. Therefore $f(A_1 \cap A_2) = \emptyset$.

Now $3 \in A_1$ since $3 = 2 \cdot 1 + 1, 1\in \mathbb{N}$. Similarly $3 \in f(A_2)$ since $2 \in A_2$. Thus $3 \in (A_1)$ and $3 \in f(A_2)$, therefore $3 \in (A_1) \cap f(A_2) \neq \emptyset$. Thus $f(A_1) \cap f(A_2) \nsubseteq f(A_1 \cap A_2)$.

\vspace{2 pc}

1.7.26 First of to prove this statement, I need to prove that the $f(A_1 \cap A_2) \subseteq f(A_1) \cap f(A_2)$ is correct.
\begin {proof}
Let $y \in f(A_1 \cap A_2)$. This means $\exists x \in A_1 \cap A_2$ such that $y = f(x)$. $x \in A_1$ and $x \in A_2$. therefore $y = f(x) \in f(A_1), y = f(x) \in f(A_2)$. Thus $y \in f(A_1 ) \cap f(A_2)$.
\end {proof}

Now I want to show that $f(A_1) \cap f(A_2) \subseteq f(A_1 \cap A_2)$ is true iff $f$ is injective.
\begin {proof}
($\rightarrow$) Let $f(A_1) \cap f(A_2) \subseteq f(A_1 \cap A_2)$ be true. This means that $y \in f(A_1) \cap f(A_2)$,
Let $x_1 \in A_1, x_2 \in A_2$ such that $y = f(x_1), y = f(x_2)$. We also know that $y \in f(A_1 \cap A_2)$. Therefore, $\exists x \in A_1 \cap A_2, y =f(x)$. Then $y = f(x) = f(x_1) = f(x_2)$, therefore $x = x_1 = x_2 \in A_1 \cap A_2$. Thus, if $f(x_1)=f(x_2)$ then $x_1 = x_2$, thus $f$ is injective!

($\leftarrow$) Let $y \in f(A_1) \cap f(A_2)$, and let $f$ be injective. So, $y \in f(A_1)$ and $y \in f(A_2)$, let $x_1 \in A_1, x_2 \in A_2$ such that $y = f(x_1)$ and $y= f(x_2)$. Thus $y = f(x_1) = f(x_2)$. Since $f$ is injective, $x_1 = x_2$. Let $x = x_1 = x_2$ and so $x \in A_1$ and $x \in A_2$, thus $x \in A_1 \cap A_2$, therefore $y = f(x) \in f(A_1 \cap A_2)$. Thus if $f$ is injective then $f(A_1) \cap f(A_2) \subseteq f(A_1 \cap A_2)$.

Therefore $f(A_1) \cap f(A_2) = f(A_1 \cap A_2)$ iff $f$ is injective.
\end {proof}

\vspace{2pc}

8. I want to show that $\sqrt{2} \notin \mathbb{Q}$.

\begin {proof}
Suppose not. Let $a,b \in \mathbb{Z}$ such that $\frac{a}{b} = \sqrt{2}$ where $a,b$ are prime to each other. Now, lets multiply both sides by $b$ so $a = b \sqrt{2}$. Now, lets square both sides of the equation, thus $a^2 =2 b^2$.

\vspace{1pc}

$Claim:$ An odd numbers square cannot be even.
\begin {proof}
Let $ k =2n+1$, where $n \in \mathbb{N}$, thus $k^2 = (2n+1)(2n+1) = 2n(2n+1) + 1 (2n+1) = 4n^2 +2n + 2n +1 = 4n^2 + 4n +1 = 2(2n^2+2n) + 1$, let $j = 2n^2 + 2n$ thus $k^2 = 2j +1$ which is odd.
\end {proof}

Thus, $a$ has to be even thus I can represent it as $a = 2k, k \in \mathbb{N}$. Then $a^2 = 4 k^2$, thus $a^2 = 4 k^2 = 2 b^2$. If I divide both sides by $2$, I get $2k^2 = b^2$ thus $b$ has to be even as well! But $a$ and $b$ were supposed to be prime to each other thus I have a contradiction! $\Rightarrow \Leftarrow $

Therefore $\sqrt{2} \notin \mathbb{Q}$.
\end {proof}

\end{document}

yazılan ödevlerin işi bitince bir kenarda biriktirilir. yeterince ağırlığa ulaşan latex'ler bir hurdacı ile anlaşılarak, geri dönüşüm yapılıp, prezervatif üretilmesi için verilir.