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I am new to LaTeX and I am trying to get those three parts of the Number horizontally aligned but it seems I can't use {align} around {enumerate}. I tried to play around with {tabulator} a bit but it doesn't help me with aligning the top number with the Overbraces.

Does someone have a smart idea how to solve this the best way?

Below a screenshot and my code belonging to it.enter image description here

\begin{enumerate}
\item  \begin{tabular}{llcc}
         $\overbrace{0}^\text{Vorzeichen}$
         &$\overbrace{1000 0000}^\text{Exponent (8Bit)}$ 
         &$\overbrace{1000 0000}^\text{Mantisse}$
         &$\longrightarrow + 2^1*(2^0 + 2^{-1}) = 3$
       \end{tabular}
\item  \begin{tabular}{rrrcccc} 
         & 1 &&& 10000001 && 00000000 
       \end{tabular} 
       $\longrightarrow + 2^2*(2^0) = 4$
\item  \begin{tabular}{rrrcccc} 
        & 0 &&& 01111110 && 10000000 
       \end{tabular} 
       $\longrightarrow + 2^{-1}*(2^0 + 2^{-1}) = 0,75$
\item  \begin{tabular}{rrrcccc} 
         & 0 &&& 10000010 && 01000100 
       \end{tabular} 
       $\longrightarrow + 2^3*(2^{-2} + 2^{-6}) = 10,125$
\end{enumerate}
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    – Werner
    Dec 18, 2014 at 3:22

1 Answer 1

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Here it would be best to define an environment that handles some of the alignment setup. Additionally one should consider using a fixed-width column if you want entries to line up across list elements:

enter image description here

\documentclass{article}
\usepackage{amsmath,array}
\newcolumntype{C}[1]{>{\centering\arraybackslash$}p{#1}<{$}}
\newenvironment{numbers}
  {$\begin{array}[b]{C{4em}C{6em}C{5em}>{\longrightarrow{}}l}}
  {\end{array}$}
\begin{document}

\begin{enumerate}
  \item  \begin{tabular}{llcc}
           $\overbrace{0}^\text{Vorzeichen}$
           &$\overbrace{1000 0000}^\text{Exponent (8Bit)}$ 
           &$\overbrace{1000 0000}^\text{Mantisse}$
           &$\longrightarrow + 2^1*(2^0 + 2^{-1}) = 3$
         \end{tabular}
  \item  \begin{tabular}{rrrcccc} 
           & 1 &&& 10000001 && 00000000 
         \end{tabular} 
         $\longrightarrow + 2^2*(2^0) = 4$
  \item  \begin{tabular}{rrrcccc} 
          & 0 &&& 01111110 && 10000000 
         \end{tabular} 
         $\longrightarrow + 2^{-1}*(2^0 + 2^{-1}) = 0,75$
  \item  \begin{tabular}{rrrcccc} 
           & 0 &&& 10000010 && 01000100 
         \end{tabular} 
         $\longrightarrow + 2^3*(2^{-2} + 2^{-6}) = 10,125$
\end{enumerate}

\begin{enumerate}
  \item  \begin{numbers}
           \overbrace{0}^\text{Vorzeichen\vphantom{()}} &
           \overbrace{1000 0000}^\text{Exponent (8Bit)} &
           \overbrace{1000 0000}^\text{Mantisse\vphantom{()}} &
           + 2^1 \times (2^0 + 2^{-1}) = 3
         \end{numbers}
  \item  \begin{numbers}
           1 & 1000 0001 & 0000 0000 &
           + 2^2 \times (2^0) = 4
         \end{numbers}
  \item  \begin{numbers}
           0 & 0111 1110 & 1000 0000 &
           + 2^{-1} \times (2^0 + 2^{-1}) = 0.75
         \end{numbers}
  \item  \begin{numbers}
           0 & 1000 0010 & 0100 0100 &
           + 2^3 \times (2^{-2} + 2^{-6}) = 10.125
         \end{numbers}
\end{enumerate}

\end{document}

Here is a highlight of the modification I suggest:

  • Use array to define a new C-style paragraph column, which centres its contents;

  • A new environment numbers is defined that sets up the 4-column array;

  • \vphantoms ensure that the vertical alignment across the headers for each \overbrace line up; and

  • \times looks better than * in math mode.

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  • thank you so much, it's the perfect solution :) just out of interest: in [b]{C{4em}C{6em}C{5em} what does the [b] and 4em,6em,5em stand for? I assume C stands for "Center".
    – Cold_Class
    Nov 11, 2014 at 17:20
  • @Cold_Class: I've created the C-column. It's the same as a p-column (or paragraph column) of fixed width with \centering added to every cell in that column (that's what the >{\centering\arraybackslash} notation means. [b] makes the array align at the bottom of the last row (usually it is vertically centred; you can also use [t] for top-alignment). The fixed-width columns use widths 4em (roughly the width of 4 m's), 6em (2 m's wider than 4em) and 5em.
    – Werner
    Nov 11, 2014 at 18:27

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