Homework Template
Author:
Caleb McWhorter
Last Updated:
5 лет назад
License:
Creative Commons CC BY 4.0
Аннотация:
A template for submitting course homework using TeX.
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
\documentclass[11pt,letterpaper]{article}
\usepackage[lmargin=1in,rmargin=1in,tmargin=1in,bmargin=1in]{geometry}
% -------------------
% Packages
% -------------------
\usepackage{
amsmath, % Math Environments
amssymb, % Extended Symbols
amsthm, % Theorem Environments
cancel, % Use Cancels
enumerate, % Enumerate Environments
graphicx, % Include Images
lastpage, % Reference Lastpage
multicol, % Use Multi-columns
multirow, % Use Multi-rows
xcolor % Use Colors
}
% -------------------
% Font
% -------------------
\usepackage[T1]{fontenc}
\usepackage{charter}
%\usepackage[T1]{fontenc}
%\usepackage{mathpazo}
%\usepackage[bitstream-charter]{mathdesign}
%\usepackage[T1]{fontenc}
% -------------------
% Tikz & PGF
% -------------------
\usepackage{tikz}
\usepackage{tikz-cd}
\usetikzlibrary{
calc,
decorations.pathmorphing,
matrix,arrows,
positioning,
shapes.geometric
}
\usepackage{pgfplots}
\pgfplotsset{compat=newest}
% -------------------
% Commands
% -------------------
% Problem Labels
\newcounter{problem}
\newcommand{\problem}{
\stepcounter{problem} %
\noindent \textbf{Problem \theproblem. }}
\newcommand{\solution}{\noindent \textbf{Solution: }}
\newcommand{\pf}{\noindent\emph{Proof. }}
% Special Characters
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
% Math Operators
\DeclareMathOperator{\im}{im}
\DeclareMathOperator{\disc}{disc}
\DeclareMathOperator{\Span}{span}
% Special Commands
\newcommand{\pspace}{\par\vspace{\baselineskip}}
\newcommand{\ds}{\displaystyle}
\newcommand{\defeq}{\stackrel{\text{def}}{=}}
\newcommand{\ov}[1]{\overline{#1}}
\newcommand{\ma}[1]{\stackrel{#1}{\longrightarrow}}
\newcommand{\twomatrix}[4]{\begin{pmatrix} #1 & #2 \\ #3 & #4 \end{pmatrix}}
% -------------------
% Header & Footer
% -------------------
\usepackage{fancyhdr}
\fancypagestyle{title}{
%Headers
\fancyhead[L]{\Large My Name}
\fancyhead[C]{\Large MAT 999}
\fancyhead[R]{\Large Homework \#}
\renewcommand{\headrulewidth}{0.2pt}
%Footers
\fancyfoot[L]{}
\fancyfoot[C]{}
\fancyfoot[R]{\thepage \,of \pageref{LastPage}}
\renewcommand{\footrulewidth}{0.0pt}
}
\fancypagestyle{pages}{
%Headers
\fancyhead[L]{}
\fancyhead[C]{}
\fancyhead[R]{}
\renewcommand{\headrulewidth}{0.0pt}
%Footers
\fancyfoot[L]{}
\fancyfoot[C]{}
\fancyfoot[R]{\thepage \,of \pageref{LastPage}}
\renewcommand{\footrulewidth}{0.0pt}
}
\headheight=18pt
\footskip=14pt
\pagestyle{pages}
% -------------------
% Content
% -------------------
\begin{document}
\thispagestyle{title}
% Problem 1
\problem Show that there exists no nontrivial unramified extensions of $\Q$. \pspace
\solution If $K/\Q$ is a nontrivial number field, then $|\disc K|>1$. But then $\disc K$ has a prime factor so that some prime ramifies in $K$. \qed \pspace
% Problem 2
\problem Complete the following:
\begin{enumerate}[(a)]
\item How does one prove a cotheorem?
\item Compute $\ds \int \cos x \;dx$.
\item How does one square $\twomatrix{a}{b}{c}{d}$?
\end{enumerate}
\solution
\begin{enumerate}[(a)]
\item Use rollaries.
\item We have
\begin{equation} \label{eq:integral}
\int \cos x \;dx= \sin x + C
\end{equation}
We can check \eqref{eq:integral}:
\[
\dfrac{d}{dx} \left( \sin x + C \right)= \cos x
\]
\item This is routine.
\end{enumerate} \qed \pspace
% Problem 3
\problem Prove that $\sqrt{2}$ is irrational. \pspace
\pf Assume that $\sqrt{2}= \dfrac{a}{b}$, where $a,b \in \Z$. Without loss of generality, we may assume $\gcd(a,b)= 1$. Then we have
\begin{align}
\sqrt{2}&= \dfrac{a}{b} \nonumber \\
\sqrt{2}^2&= \left( \dfrac{a}{b} \right)^2 \label{eq:implication1} \\
2&= \dfrac{a^2}{b^2} \nonumber \\
a^2&= 2b^2 \label{eq:implication2}
\end{align}
But then from \eqref{eq:implication2}, we know that $a^2$ is even so that $a$ is even. But then we must have
\[
2a^2= b^2
\]
so that $b^2$ is even, implying $b$ is even. But then $\gcd(a,b) \geq 2$, a contradiction. \qed \\
\end{document}