How to typeset the house of an algebraic number?

Question

I would like to reproduce the following text

but I couldn't find the house symbol neither through Detexify nor by looking through the Comprehensive LaTeX Symbol List. I also tried combining vertical bars and overline like

\left\lvert\overline{ \alpha }\rvert\right
\overline{\left\lvert \alpha \rvert\right}

to less than desirable results.

Answer 1

There are some issues to consider:

Line width

• The standard line width in TeX is 0.4pt, it is hardcoded in TeX as 26214sp.

• \overline uses a different line width. It is taken from font parameter 8 of math font family 3 in TeX.

  The standard math fonts (Computer Modern) are using 0.39998pt in all font sizes. But it is scalable in other fonts, e.g. package mathptmx with 10pt as class option:

  • 0.3999pt in \textfont (\textstyle/\displaystyle)
  • 0.29591pt in \scriptfont (\scriptstyle)
  • 0.23993pt in \scriptscriptfont (\scriptscriptstyle)

\overline

• TeX constructs a \overline{...} the following way with θ as line width (pseudo code):

  \vbox{
    \kern θ
    \hrule height θ
    \kern 3*θ
    \hbox{$...$}
  }
  

  Especially there is white space above the rule. This is useful, if the overlined symbol is put in the denominator of a fraction, for instance.

• TeX uses cramped math style for the math expression of \overline. This means, the superscripts are moved down a little, for a better visual appearance.

Thus \overline looks like a good choice for the "house" symbols. However, the vertical lines should use the same line width as used by \overline.

Spacings

• TeX inserts 3 * θ (θ = line width of \overline) between the formula and the line in \overline. This cannot be easily changed without loosing cramped style. (AFAIK only LuaTeX can set cramped style explicitly.)

• Extra space is added above the line in \overline, see above.

• Usually symbols have side bearings, space at the left and right side of the symbol. The exact amount is usually not available in TeX. The example uses 2 * θ as horizontal space between the vertical lines and the formula. Also a horizontal space 1.5 * θ is put outside the vertical lines. Some kind of side bearings of the new symbol to avoid touching of housed math expressions.

• Since the OP wants longer lines (see comment), the example adds 2 * θ as additional depth. (3 * θ seems too large, because there is not a underline that needs some distance from the formula.)

The full example also takes into account the different math styles, thus macro \house respects the current math sizes.

\documentclass{article}
\usepackage{mathptmx}

\makeatletter
\newcommand*{\house}[1]{%
  \mathord{%
    \mathpalette\@house{#1}%
  }%
}
\newcommand*{\@house}[2]{%
  % #1: math style
  % #2: expression that gets the "house"
  % get the line width of `\overline' in the current math font size
  \dimen@=\fontdimen8 %
      \ifx#1\scriptscriptstyle\scriptscriptfont
      \else\ifx#1\scriptstyle\scriptfont
      \else\textfont\fi\fi
      3 %
  \sbox0{%
    $#1%
      \vrule width\dimen@\relax
      \overline{%
        \kern2\dimen@
        \begingroup % to keep changes of \dimen@ in #2 local
          #2%
        \endgroup
        \kern2\dimen@
      }%
      \vrule width\dimen@\relax
      \mathsurround=1.5\dimen@ % outside margin
    $%
  }%
  % TeX adds an empty space above `\overline', it needs to be
  % removed to get the correct height for the `\vrule's
  \ht0=\dimexpr\ht0-\dimen@\relax
  \dp0=\dimexpr\dp0+2\dimen@\relax
  \vbox{%
    \kern\dimen@ % reinsert previously removed space
    \copy0 %
  }%
}

\begin{document}
\[
  \textstyle % same size as \displaystyle
  \house{\alpha + \beta}
  \stackrel{\mbox{\tiny T}}{=}
  \house{\alpha} + \house{\beta}^n
\]
\[
  \scriptstyle
  \house{\alpha + \beta}
  \stackrel{\mbox{\tiny S}}{=}
  \house{\alpha} + \house{\beta}^n
\]
\[
  \scriptscriptstyle
  \house{\alpha + \beta}
  \stackrel{\mbox{\tiny SS}}{=}
  \house{\alpha} + \house{\beta}^n
\]
\[
  \def\x{\frac{\house{\alpha + \beta}}{\house{\alpha} + \house{\beta}}}
  \x, \frac{\x}{\x} 
\]
\[
  \underbrace{\overline{\alpha^2}, \house{\alpha^2}}_{\mbox{\tiny cramped}} \ne
  \overline{\mbox{$\alpha^2$}}, \house{\mbox{$\alpha^2$}}
\]
\end{document}

Answer 2

Here is a plain TeX solution. The downside is, that \everymath is executed inside the "house". If you want every house to have the same height and depth, add a strut or a phantom inside it.

\nopagenumbers% for cropping
\def\house#1{{%
    \setbox0=\hbox{$#1$}
    \vrule height \dimexpr\ht0+1.4pt width .4pt depth \dp0\relax
    \vrule height \dimexpr\ht0+1.4pt width \dimexpr\wd0+2pt depth \dimexpr-\ht0-1pt\relax
    \llap{$#1$\kern1pt}
    \vrule height \dimexpr\ht0+1.4pt width .4pt depth \dp0\relax
}}


$\house{\alpha + \beta} \leq \house{\alpha} + \house{\beta}$
\medskip
$\house{\strut\alpha + \beta} \leq \house{\strut\alpha} + \house{\strut\beta}$
\bye

Bonus question: Is there a plain TeX dimen, which holds the default linewidth of .4pt? I only know about \p@=1pt and \z@=0pt.

Answer 3

With the horizontal bar slightly protruding:

\documentclass[a4paper,12pt]{article}
\usepackage{array}

\newcommand\house[1]{%
  \begingroup\setlength\arraycolsep{0pt}
  \begin{array}[t]{@{\mkern1mu}c@{}|c|@{}c@{\mkern1mu}}
  \firsthline
  &\;#1\;{}&
  \end{array}
  \endgroup
}


\begin{document}

$\house{\alpha + \beta} \leq \house{\alpha} + \house{\beta}$

\end{document}

The protrusion can be controlled by increasing the amount given with \mkern; this is with \mkern2mu

As barbara beeton notes in a comment, the bar shouldn't protrude; for this just remove the \mkern instruction:

\newcommand\house[1]{%
  \begingroup\setlength\arraycolsep{0pt}
  \begin{array}[t]{@{}c@{}|c|@{}c@{}}
  \firsthline
  &\;#1\;{}&
  \end{array}
  \endgroup
}

Answer 4

An implementation with tikz, shamefully adapted from this answer

\documentclass{standalone}
\usepackage{tikz}

\usetikzlibrary{calc}
\newcommand{\house}[1]{%
  \tikz[baseline]{\node[anchor=base,inner sep=0.3ex](mynode){\ensuremath{#1}};
  \draw(mynode.south west)--(mynode.north west)--(mynode.north east)--(mynode.south east);
  \path[use as bounding box]($(mynode.south west)+(-0.3ex,-0.3ex)$)rectangle($(mynode.north east)+(0.3ex,0.3ex)$);}
}

\begin{document}
  $\house{\alpha+\beta}\leq\house{\alpha}+\house{\beta}$
\end{document}

---

Answer 5

Never seen it either. However, here is a way of doing things: put everything into an array, with hhline to have a clean connection of vertical and horizontal lines. I define a normal and a bold version I don'tknow how to declare it, so as to have only one command for both versions):

\documentclass[a4paper,12pt]{article}

\usepackage{mathtools}
\usepackage{hhline, bm}
 \newcommand\house[1]{\mkern2mu\begin{array}{|@{\:}c@{\:}|}\hhline{|-|}#1\end{array}\mkern1mu}
 \newcommand\boldhouse[1]{\setlength{\arrayrulewidth}{0.6pt}\mathversion{bold}%
 \mkern2mu\begin{array}{|@{\:}c@{\: }|}\hhline{|-|}#1\end{array}\mkern1mu}%

\begin{document}

 $ \house{α + β} \leq \house{α} + \house{β}$\\

 $\mathversion{bold} \boldhouse{α + β} \leq \boldhouse{α} + \boldhouse{β} $

\end{document} 

The result, in my opinion, is better than the OP's image (same height for all symbols):

There should remain to define a bold version.

Answer 6

\newcommand{\house}[1]{
  \setbox0=\hbox{$#1$}
  \rule{.4pt}{\dimexpr\ht0 + 1.6pt}
  \overline{\mkern+1mu
    #1
  \mkern+1mu}
  \rule{.4pt}{\dimexpr\ht0 + 1.6pt}
}