Adjusting the height of math floor symbol

Question

Recently I started to play with LaTeX again just to train for when I have to use it again. I like to write mathematical formulas and I am trying to nail down floor and ceil symbols with fractions. However the results I get are kind of weird, in my opinion. Example code below:

    $\left\lfloor\dfrac{2}{3}\right\rfloor$
    $\left\lfloor\dfrac{a}{n}\right\rfloor$ 
    $\left\lfloor\dfrac{A}{n}\right\rfloor$
    $\left\lfloor\dfrac{a}{N}\right\rfloor$

This code generates this output (compiling with PDFLaTeX):

The height of the floor symbol is inconsistent, it is smaller when the fraction contains a lowercase letter in the numerator and larger when the fraction contains numbers or uppercase letters in the numerator. Why is that the case? How can I produce floor symbols that are always the larger size shown in the picture?

Answer 1

TeX uses a discrete set of five bracket sizes (not counting the massive brackets around matrices, etc.). Using \left and \right invokes an algorithm that calculates which to use, based on what appears inside. Sometimes this algorithm results in a choice that is not visually appealing.

If you want to create a sequence of objects contained in brackets of the same size, it's simplest to choose the size yourself, using \bigl, \Bigl, \biggl or \Biggl along with the associated commands ending in r. Another option is to influence \left and \right by inserting vertical rules (struts). Note that the optional argument to \rule is used to move it up and down, whereas the mandatory arguments are the width followed by the height. You can change the width to 0pt once you are happy with the result.

\documentclass{article}
\usepackage{mathtools}
\begin{document}
$\biggl\lfloor\dfrac{2}{3}\biggr\rfloor$
$\biggl\lfloor\dfrac{a}{n}\biggr\rfloor$
$\biggl\lfloor\dfrac{A}{n}\biggr\rfloor$
$\biggl\lfloor\dfrac{a}{N}\biggr\rfloor$

\bigskip

$\left\lfloor\rule[-8pt]{1pt}{20pt}\dfrac{2}{3}\right\rfloor$
$\left\lfloor\rule[-8pt]{1pt}{20pt}\dfrac{a}{n}\right\rfloor$
$\left\lfloor\rule[-8pt]{1pt}{20pt}\dfrac{A}{n}\right\rfloor$
$\left\lfloor\rule[-8pt]{1pt}{20pt}\dfrac{a}{N}\right\rfloor$
\end{document}

Answer 2

You can use \overwithdelims TeX primitive:

\let\ds=\displaystyle
$\ds {2\overwithdelims\lfloor\rfloor 3}$
$\ds {a\overwithdelims\lfloor\rfloor n}$
$\ds {A\overwithdelims\lfloor\rfloor n}$
$\ds {a\overwithdelims\lfloor\rfloor N}$

Answer 3

You can use \DeclarePairedDelimiter from the mathtools package to create your own \floor command that lets you freely choose between inferred and manually-specified delimiter sizes:

\usepackage{mathtools}

\DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor}

Use \floor as \floor*{...} for interfered delimiter sizes or as \floor[\SIZE]{...} for \SIZE-sized delimiters.

\documentclass{article}
\usepackage{enumitem}  
\usepackage{mathtools}

\DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor}

\begin{document}
\begin{description}[labelwidth=\widthof{\bfseries Inferred:}]
    \item[Inferred:]
    $\floor*{\dfrac{2}{3}}$
    $\floor*{\dfrac{a}{n}}$ 
    $\floor*{\dfrac{A}{n}}$
    $\floor*{\dfrac{a}{N}}$

    \item [\texttt{\textbackslash{}big}:]
    $\floor[\big]{\dfrac{2}{3}}$
    $\floor[\big]{\dfrac{a}{n}}$ 
    $\floor[\big]{\dfrac{A}{n}}$
    $\floor[\big]{\dfrac{a}{N}}$
    
    \item [\texttt{\textbackslash{}Big}:]
    $\floor[\Big]{\dfrac{2}{3}}$
    $\floor[\Big]{\dfrac{a}{n}}$ 
    $\floor[\Big]{\dfrac{A}{n}}$
    $\floor[\Big]{\dfrac{a}{N}}$
    
    \item [\texttt{\textbackslash{}bigg}:]
    $\floor[\bigg]{\dfrac{2}{3}}$
    $\floor[\bigg]{\dfrac{a}{n}}$ 
    $\floor[\bigg]{\dfrac{A}{n}}$
    $\floor[\bigg]{\dfrac{a}{N}}$
    
    \item [\texttt{\textbackslash{}Bigg}:]
    $\floor[\Bigg]{\dfrac{2}{3}}$
    $\floor[\Bigg]{\dfrac{a}{n}}$ 
    $\floor[\Bigg]{\dfrac{A}{n}}$
    $\floor[\Bigg]{\dfrac{a}{N}}$
\end{description}
\end{document}