Blank space with \left[ and \right]

Question

I have the following code:

\documentclass{article}
\usepackage{amsmath}
\begin{document}
    $c_{1}=\left[\dfrac{m\left(\dfrac{N}{q}-1\right)}{N-1}\right]$
\end{document}

which gives:

Why do I have this blank space which is colored by red? Is there any way to remove it?

Because the \left...\right mechanism demands vertical symmetry about the math axis — Steven B. Segletes, Oct 25 '17 at 15:54
See https://tex.stackexchange.com/questions/59747/proper-display-of-fractions — Steven B. Segletes, Oct 25 '17 at 15:57
note the gap is larger than it would be as you are (for some reason?) forcing display fractions in inline math. (\dfrac rather than the usual \frac) — David Carlisle, Oct 25 '17 at 19:40

score 8 · Answer 1 · answered Oct 25 '17 at 16:43

Like Mico I would try to avoid that the upper part is so much larger.

\documentclass{article}
\usepackage{amsmath}



\begin{document}
    $c_{1}=\left[\dfrac{m\left(\frac{N}{q}-1\right)}{N-1}\right]$\quad     
    $c_{1}=\left[\dfrac{m\left(N/q-1\right)}{N-1}\right]$
\end{document}

score 8 · Answer 2 · answered Oct 25 '17 at 17:33

My two cents:

\documentclass{article}
\usepackage{amsmath,delarray}

\begin{document}

\section{Ugly}
\[
c_{1}=\left[\dfrac{m\left(\dfrac{N}{q}-1\right)}{N-1}\right]
\]

\section{Bad}
\[
c_{1}=
\sbox0{$\displaystyle\frac{m\biggl(\dfrac{N}{q}-1\biggr)}{N-1}$}
\raisebox{\dimexpr\depth-\dp0}{%
  $
  \begin{array}\lbrack{@{}c@{}}\rbrack
  \displaystyle\frac{m\biggl(\dfrac{N}{q}-1\biggr)}{N-1}
  \end{array}
  $%
}
\]

\section{Good}
\[
c_{1}=
\left[\frac{1}{N-1}\cdot m\biggl(\frac{N}{q}-1\biggr)\right]
\]

\end{document}

Steven B. Segletes · Accepted Answer · 2017-10-25T16:13:52.350

6

Just \fixit[]{}{}{}! The syntax is

\fixit[<mathstyle>]{<left-delim>}{<content>}{<right-delim>}

The MWE:

\documentclass{article}
\usepackage{amsmath}
\newcommand\fixit[4][\displaystyle]{
  \setbox0=\hbox{$#1#3$}
  \setbox2=\hbox{$\vcenter{\copy0}$}
  \raisebox{\dimexpr\ht0-\ht2}{$#1\left#2\copy2\right#4$}
}
\begin{document}
    $c_{1}=\fixit{[}{\dfrac{m\left(\dfrac{N}{q}-1\right)}{N-1}}{]}
    \ne\fixit{[}{\dfrac{N-1}{m\left(\dfrac{N}{q}-1\right)}}{]}$

    $c_{1}=\fixit[\textstyle]{[}{\frac{m\left(\frac{N}{q}-1\right)}{N-1}}{]}
    \ne\fixit[\textstyle]{[}{\frac{N-1}{m\left(\frac{N}{q}-1\right)}}{]}$
\end{document}

edited Oct 25 '17 at 16:13

answered Oct 25 '17 at 16:08

Steven B. Segletes

237,551

2

This looks very odd, as if the math is drunk ;-) – Ulrike Fischer Oct 25 '17 at 16:21
@UlrikeFischer The alternative is like Sandy's solution, where the vinculum is misaligned. One must appreciate that such an application is the exception and not the norm, and so it would be, I think, unlikely to see a result in the actual form I present. – Steven B. Segletes Oct 25 '17 at 16:24
@UlrikeFischer On further consideration, lock me up! I'm guilty of drinking and deriving! – Steven B. Segletes Oct 25 '17 at 16:26

Sandy G · Answer 4 · 2017-10-25T16:27:02.937

5

You can use bmatrix and consider the expression to be an entry in a 1x1 matrix.

\documentclass{article}
\usepackage{amsmath}
\begin{document}
$c_{1}=\begin{bmatrix}\dfrac{m\left(\dfrac{N}{q}-1\right)}{N-1}\end{bmatrix}$
\end{document}

This will also vertically center the = sign (which may or may not be desirable).

edited Oct 25 '17 at 16:27

answered Oct 25 '17 at 16:08

Sandy G

42,558

Mico · Answer 5 · 2017-10-25T16:17:52.340

(Edited the answer to incorporate the information, provided by the OP, that the square brackets are supposed to denote the "integer part" of the argument.)

I truly think your best option is to rewrite the term so as to avoid having to display a three-level fraction. In the following screenshot, the first term is from your code, the second is from @jfbu's answer, and the third implements the suggested rewrite, where I'm using a macro called \floor for the sake of variety of appearances.

\documentclass{article}
\usepackage{mathtools,mleftright}
\DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor}
\begin{document}
\[
c_{1}
=\left[\frac{m\left(\dfrac{N}{q}-1\right)}{N-1}\right]\\
=\raisebox{\dimexpr.25\height+1pt}{$\displaystyle\left[\raisebox{-.25\height}{$\dfrac{m\left(\dfrac{N}{q}-1\right)}{N-1}$}\right]$}\\
=\floor*{\frac{m}{N-1}\mleft(\frac{N}{q}-1\mright)}
\]
\end{document}

This is not mathematical correct because [ ] is used for the integral part. — Vassilis Chasiotis, Oct 25 '17 at 16:10
@VassilisChasiotis - Oh, I didn't realize what the [ and ] square brackets were for. I will edit my answer accordingly. — Mico, Oct 25 '17 at 16:13

score 2 · Answer 6 · 2017-10-25T16:07:48.147

2

Quick hack

\documentclass{article}
\usepackage{amsmath}
\begin{document}
    $c_{1}=\raisebox{\dimexpr.25\height+1pt}{$\displaystyle\left[\raisebox{-.25\height}{$\dfrac{m\left(\dfrac{N}{q}-1\right)}{N-1}$}\right]$}$
\end{document}

The cause of your troubles is that TeX positions delimiters symmetrically with respect to math axis. I did not make the utmost effort here to get the fraction rule exactly at math axis after my surgery.

More detailed and complete answers at Proper display of fractions

edited Oct 25 '17 at 16:07

answered Oct 25 '17 at 15:57

actually this looks quite similar to https://tex.stackexchange.com/a/59750/4686 pointed out by @StevenSegletes. And https://tex.stackexchange.com/a/251919/4686 possibly better yet. – Oct 25 '17 at 16:04

Blank space with \left[ and \right]

6 Answers6

Linked