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