5
\begin{equation}\label{eq:OrdenaAlpha}
    \begin{array}{lrl}
        \alpha^{0} = \alpha & = & (+6 \quad -1 \quad +4 \quad -5 \quad \underline{-2} \quad \underline{-3}) \\
        \alpha^{1} = \alpha^{0} \circ \tau(5,6,7) & = & (+6 \quad -1 \quad  \underline{+4 \quad -5 \quad -3 \quad -2}) \\
        \alpha^{2} = \alpha^{1} \circ \rho(3,6) & = & (\underline{+6 \quad -1} \quad \underline{+2} \quad +3 \quad +5 \quad -4) \\
        \alpha^{3} = \alpha^{2} \circ \tau(1,3,4) & = & (+2 \quad \underline{+6} \quad \underline{-1} \quad +3 \quad +5 \quad -4) \\
        \alpha^{4} = \alpha^{3} \circ \tau(2,3,4) & = & (+2 \quad -1 \quad +6 \quad \underline{+3} \quad \underline{+5} \quad -4) \\
        \alpha^{5} = \alpha^{4} \circ \tau(4,5,6) & = & (+2 \quad -1 \quad +6 \quad +5 \quad +3 \quad -4)
    \end{array}
\end{equation}

enter image description here

Would like to use /underline and maintain alignment.

  • 2
    Please don't change the question to invalidate existing answers. – egreg Sep 26 at 19:56
  • Only use the additional brace group {} when you want to make the + or - a binary operator. BTW, are those signs supposed to be binary operators representing plus or minus operations, or do they indicate a negative or positive number? Want to make sure you are getting the correct spacing? For the cases where you don't have a sign, but want to align with a sign above, you need to use \underscore{\hphantom{-}...}. – Peter Grill Sep 26 at 19:57
6

Use \undeline{{} ...} when you want the binary spacing. Otherwise, the + (similarly a -) in \undeline{+5}, etc is treated as a unary operator.

enter image description here

Code:

\documentclass{article}

\begin{document}
\begin{equation}\label{eq:OrdenaAlpha}
    \begin{array}{lrl}
        \alpha^{0} = \alpha & = (+6 \quad -1 \quad +4 \quad -5 \quad \underline{{}-2} \quad \underline{{}-3}) \\
        \alpha^{1} = \alpha^{0} \circ \tau(5,6,7) & = (+6 \quad -1 \quad  \underline{{}+4 \quad -5 \quad -3 \quad -2}) \\
        \alpha^{2} = \alpha^{1} \circ \rho(3,6)   & = (\underline{+6 \quad -1} \quad \underline{{}+2} \quad +3 \quad +5 \quad -4) \\
        \alpha^{3} = \alpha^{2} \circ \tau(1,3,4) & = (+2 \quad \underline{{}+6} \quad \underline{{}-1} \quad +3 \quad +5 \quad -4) \\
        \alpha^{4} = \alpha^{3} \circ \tau(2,3,4) & = (+2 \quad -1 \quad +6 \quad \underline{{}+3} \quad \underline{{}+5} \quad -4) \\
        \alpha^{5} = \alpha^{4} \circ \tau(4,5,6) & = (+2 \quad -1 \quad +6 \quad +5 \quad +3 \quad -4)
    \end{array}
\end{equation}
\end{document}
3

I suggest a more complicated input, but with better output:

\documentclass{article}
\usepackage{amsmath,array,booktabs}

\begin{document}

\begin{equation}\label{eq:OrdenaAlpha}
\setlength{\arraycolsep}{0pt}
\setlength{\aboverulesep}{-3pt}
\renewcommand{\arraystretch}{1.5}
\newcolumntype{f}{>{\kern0pt\quad\kern0pt}c}
\begin{array}{
  l                % the powers
   >{{}}r<{{}}     % the equals sign
  r<{\,}           % the parenthesis
  *{5}{rf}         % the first five values
  r                % the last value
  >{\,}l           % the parenthesis
}
\alpha^{0} = \alpha & = &
  (& +6 && -1 && +4 && -5 && -2 && -3 &) \\
\cmidrule{12-12}\cmidrule{14-14}
\alpha^{1} = \alpha^{0} \circ \tau(5,6,7) & = & 
  (& +6 && -1 &&  +4 && -5 && -3 && -2 &) \\
\cmidrule{8-14}
\alpha^{2} = \alpha^{1} \circ \rho(3,6) & = &
  (& +6 && -1 && +2 && +3 && +5 && -4 &) \\
\cmidrule{4-6}\cmidrule{8-8}
\alpha^{3} = \alpha^{2} \circ \tau(1,3,4) & = &
  (& +2 && +6 && -1 && +3 && +5 && -4 &) \\
\cmidrule{6-6}\cmidrule{8-8}
\alpha^{4} = \alpha^{3} \circ \tau(2,3,4) & = &
  (& +2 && -1 && +6 && +3 && +5 && -4 &) \\
\cmidrule{10-10}\cmidrule{12-12}
\alpha^{5} = \alpha^{4} \circ \tau(4,5,6) & = &
  (& +2 && -1 && +6 && +5 && +3 && -4 &)
\end{array}
\end{equation}

\end{document}

tabs

The intercolumn space is set to zero, but between any two of the values there is a phantom column 1em wide; this allows for getting the precise length of the underlines. By setting \aboverulespace to a negative value, we get it nearer the numbers. A thin space is added after ( and before ) to avoid conflicts.

The values are in columns 4, 6, 8, 10, 12 and 14.

For unsigned values, here's a possible variation using center alignment.

\documentclass{article}
\usepackage{amsmath,array,booktabs}

\begin{document}

\begin{equation}
\setlength{\arraycolsep}{0pt}
\setlength{\aboverulesep}{-3pt}
\renewcommand{\arraystretch}{1.5}
\newcolumntype{f}{>{\kern0pt\quad\kern0pt}c}
\begin{array}{
  l                % the powers
   >{{}}r<{{}}     % the equals sign
  r<{\,}           % the parenthesis
  *{5}{cf}         % the first five values
  c                % the last value
  >{\,}l           % the parenthesis
}
\alpha^{0} = \alpha & = &
  (& 2 && 4 && 6 && 1 && 5 && 3 &) \\
\cmidrule{4-4}
\alpha^{1} = \alpha^{0} \circ \rho(1,1) & = & 
  (& -2 && 4 && 6 && 1 && 5 && 3 &) \\
\cmidrule{12-14}
\alpha^{2} = \alpha^{1} \circ \rho(5,5) & = &
  (& -2 && 4 && 6 && 1 && -5 && 3 &) \\
\cmidrule{14-14}
\alpha^{3} = \alpha^{2} \circ \rho(6,6) & = &
  (& -2 && 4 && 6 && 1 && -5 && -3 &) \\
\cmidrule{10-12}
\alpha^{4} = \alpha^{3} \circ \rho(4,5) & = &
  (& -2 && 4 && 6 && 5 && -1 && -3 &) \\
\cmidrule{4-4}\cmidrule{6-6}\cmidrule{8-8}
\alpha^{5} = \alpha^{4} \circ \rho(2,4) & = &
  (& -2 && -5 && -6 && -4 && -1 && -3 &) \\
\cmidrule{4-8}
\alpha^{6} = \alpha^{5} \circ \rho(1,3) & = &
  (& 6 && 5 && 2 && -4 && -1 && -3 &) \\
\cmidrule{10-14}
\alpha^{7} = \alpha^{6} \circ \rho(4,6) & = &
  (& 6 && 5 && 2 && 3 && 1 && 4 &)
\end{array}
\end{equation}

\end{document}

enter image description here

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