I am typing up a sheet of course problems, and wanted to offset the question numbers from the text of my solutions slightly.
Based upon an answer to this question, I tried adjusting leftskip. The code is below:
\documentclass{article}
\usepackage{amsmath,amsthm,amssymb,amsfonts}
\begin{document}
\noindent 14.5) \\
\setlength{\leftskip}{8mm} \noindent Suppose that $f(a) = f(b)$. This says that $2a + 4 = 2b + 4$, so subtracting \noindent 4 from each side and dividing by 2 yields $a = b$. We conclude that $f$ is injective. \\\\
\noindent Suppose that $c \in E$. This says $2|c $; since $\mathbb{Z}$ distributes multiplication over addition, we have (where $c = 2d$)
\begin{equation*}
c - 4 = 2d - 2(2) = 2(d-2),
\end{equation*}
\vspace{2mm}
\noindent so $c-4$ is divisible by 2; therefore $\frac{c-4}{2} \in \mathbb{Z}$. But then $f(\frac{c-4}{2})$ clearly equals $c$; so $E \subseteq f[\mathbb{Z}]$. This means that $f$ is surjective. \\\\
\noindent We conclude that $f$ is a bijection from $\mathbb{Z}$ to $E$. \\
\setlength{\leftskip}{0mm} \noindent 9.10)\\
\setlength{\leftskip}{8mm} \noindent The first statement is false (take $a=3$, $b=8$ to see that there is an integer - namely, 6 - which is in $\mathcal{D}_{ab}$ but not in $\mathcal{D}_a \bigcup \mathcal{D}_b$); the second statement is true, since $p \in \mathcal{PD}_{ab} \implies p \in \mathcal{PD}_a \lor p \in \mathcal{PD}_b$ by Euclid's Lemma (and thus $p \in \mathcal{PD}_a \bigcup \mathcal{PD}_b$), while $p \in \mathcal{PD}_{a} \bigcup \mathcal{PD}_b \implies p|a \lor p|b$, either of which says $p|ab$, so that $p \in \mathcal{PD}_{ab}$. (The second statement follows from these implications.)\\
\setlength{\leftskip}{0mm} \noindent 6.1)\\
\setlength{\leftskip}{8mm} \noindent We show that the $n$ consecutive integers following $(n+1)! + 2$, $n \geq 2$, are composite. Let $S = \{(n+1) + k: 2 \leq k \leq n+1 \}$, and let $K = \{k : 2 \leq k \leq n+1 \}$; it is clear that $S$ is a set of $n$ consecutive integers. Taking any $k$ in $K$, we consider the (obvious) corresponding $s_k = (n+1)! + k \in S$; since $k|(n+1)!$ and $k|k$, we have $k|s_k$; the quotient $\frac{(n+1)!}{k} + 1$ is greater than 1, so each $s_k$ in $S$ thus admits a non-trivial factorization (i.e., is composite). \\
\setlength{\leftskip}{0mm} \noindent 6.2)\\
\setlength{\leftskip}{8mm} \noindent We have
\begin{eqnarray*}
n^4 + 4 &=& (n^4 + 4n^2 + 4) -4n^2 \\ &=& (n^2 + 2)^2 - (2n)^2\\
&=&(n^2 - 2n + 2)(n^2 + 2n + 2);
\end{eqnarray*}
\vspace{1mm} \noindent both factors are nontrivial for $n \geq 2$. \\
\setlength{\leftskip}{0mm} \noindent 6.5) \\
\setlength{\leftskip}{8mm}
\noindent If we had $d > 1: d|m, n$, then we would have $m = \mu d, n = \nu d$;
\end{document}
The code was fine for its intended purpose on the first page; however, on the second page, it no longer dedented the question numbers in the way I wanted. (They appeared directly above the solution text, despite my using the same commands as on the first page.)
What's going on here? How can I get my desired effect?
\leftskiptake effect at the start of the second page in the OP's example? – Ian Thompson Oct 17 '12 at 10:54leftskipdoes take effect, and everything on the second page is indented, which makes it look at first glance as though nothing is indented. – Ian Thompson Oct 17 '12 at 14:48