# Removing space after section heading after using baselineskip

Basically after setting a large \baselineskip there is now a huge space under my section heading: source:

\documentclass[14pt,extrafontsizes]{memoir}
\usepackage[margin=10pt,paperwidth=496pt, paperheight=588pt]{geometry}
\usepackage[fleqn]{amsmath}
\usepackage{multicol,array,collcell,stix}
\usepackage[shortlabels]{enumitem}
\usepackage[dvipsnames]{xcolor}
\everymath{\displaystyle}

\begin{document}
{\setlength{\baselineskip}{36pt}
\section*{Converting repeating decimals to fractions}
\begin{enumerate}[\color{MidnightBlue} 1.]
\item Express $0.\dot{3}$ as a fraction. \\
Let $x=0.\dot{3}$ \\
then $10x=3.\dot{3}$ \\
$10x-x=3.\dot{3}-0.\dot{3}$ \\
$9x=3$ \\
$9x=\frac{3}{9}$
\item Express $0.\dot{1}\dot{2}$ as a fraction. \\
Let $x=0.\dot{1}\dot{2}$ \\
then $100x=12.\dot{1}\dot{2}$ \\
$100x-10x=12.\dot{1}\dot{2}-0.\dot{1}\dot{2}$ \\
$90x=12$ \\
$x=\frac{12}{90}$ \\
$x=\frac{2}{15}$
\end{enumerate}}
\end{document}


output:

Since you're using enumitem, you can just insert a vertical skip as part of the before key. So, for example, try

\begin{enumerate}[.., before={\vspace*{-.5\baselineskip}}]


The following uses setspace as well to set the enumerate using \doublespacing rather than using forced linebreaks \\:

\RequirePackage{setspace}
\documentclass[14pt,extrafontsizes]{memoir}
\usepackage[margin=10pt,paperwidth=496pt, paperheight=588pt]{geometry}
\usepackage{stix}
\usepackage[shortlabels]{enumitem}
\usepackage[dvipsnames]{xcolor}
\everymath{\displaystyle}
\begin{document}

\show\doublespacing
\section*{Converting repeating decimals to fractions}
\begin{enumerate}[\color{MidnightBlue} 1., before={\doublespacing\vspace*{-.5\baselineskip}}]
\item Express $0.\dot{3}$ as a fraction.

Let $x=0.\dot{3}$

then $10x=3.\dot{3}$

$10x-x=3.\dot{3}-0.\dot{3}$

$9x=3$

$9x=\frac{3}{9}$

\item Express $0.\dot{1}\dot{2}$ as a fraction.

Let $x=0.\dot{1}\dot{2}$

then $100x=12.\dot{1}\dot{2}$

$100x-10x=12.\dot{1}\dot{2}-0.\dot{1}\dot{2}$

$90x=12$

$x=\frac{12}{90}$

$x=\frac{2}{15}$
\end{enumerate}
\end{document}

• hey werner, you are really helpful, is there any chance i could add you on skype or something, because it rather time consuming for me to post a whole thread every time i have a niggling problem. if not no worries, and thanks for your answer yet again :) – user1448093 May 27 '14 at 17:17
• @user1448093: I am very rarely on Skype. However, with your current (and growing) reputation, you can get quiet quick responses in chat. The group there are very friendly and helpful. I understand the labour of writing up a post only to get a seemingly quick and helpful answer... but that's the StackExchange model, unless you like to jump in the chat room occasionally. – Werner May 27 '14 at 17:29

You can change \setlength{\baselineskip} to a more sensible value. Or, if you have good reasons to have this value for this particular enumerate environment, as it happens that \baselineskip just before the \section* is also modified, you can:

1. insert \setlength{…} just after \begin{enumerate}

2. add just after the \section* command a \vspace{-\baselineskip}, say. Actually you should rather use a setstretch{3} and, to restore the exact vertical spacing between the section title and the following text, you should use \vspace{-2\baselineskip}, but the result doesn't look very good, in my opinion. Since I suggest this (I used the book class, as memoir is not installed on my system):

\documentclass[14pt,extrafontsizes]{book}
\usepackage[margin=10pt,paperwidth=496pt, paperheight=588pt]{geometry}
\usepackage[fleqn]{amsmath}
\usepackage{multicol,array,collcell,stix}
\usepackage[shortlabels]{enumitem}
\usepackage[dvipsnames]{xcolor}
\everymath{\displaystyle}
\usepackage{setspace, etoolbox}
%\raggedbottom
\begin{document}

\section*{Converting repeating decimals to fractions}
\vspace{-1\baselineskip}

\begin{enumerate}[\color{MidnightBlue} 1.]\setstretch{3}%
\item Express $0.\dot{3}$ as a fraction. \\
Let $x=0.\dot{3}$ \\
then $10x=3.\dot{3}$ \\
$10x-x=3.\dot{3}-0.\dot{3}$ \\
$9x=3$ \\
$9x=\frac{3}{9}$
\item Express $0.\dot{1}\dot{2}$ as a fraction. \\
Let $x=0.\dot{1}\dot{2}$ \\
then $100x=12.\dot{1}\dot{2}$ \\
$100x-10x=12.\dot{1}\dot{2}-0.\dot{1}\dot{2}$ \\
$90x=12$ \\
$x=\frac{12}{90}$ \\
$x=\frac{2}{15}$
\end{enumerate}

\end{document}


• thanks bernard, I will read through this and try to understand! – user1448093 May 27 '14 at 17:20