# How to reduce space between $n\equiv 1\pmod{4}$

Judging from the italic font, the display is inside a theorem statement.

I would avoid the parentheses and the repetition of common parts:

```
\documentclass{article}
\usepackage{amsmath}
\newtheorem{theorem}{Theorem}
\DeclareMathOperator{\MSC}{MSC}
\begin{document}
\begin{theorem}
Blah, blah
\begin{equation*}
\MSC(C_n^2)=
\begin{cases}
1, & \text{if $n$ is odd and $d$ is even, i.e., $n\equiv 1$},\\
2, & \text{if $n$ is even and $d$ is even, i.e., $n\equiv 0$},\\
3, & \text{if $n$ is odd and $d$ is odd, i.e., $n\equiv 3$},\\
3, & \text{if $n$ is even and $d$ is odd, i.e., $n\equiv 2$},
\end{cases}
\end{equation*}
where congruences are modulo~$4$.
\end{theorem}
\end{document}
```

Some notes:

- “MSC” should be upright and treated as an operator;
- even in the context of italic font, math variables should be input in math mode;
`eqnarray`

is not to be used when`amsmath`

is available (and it should be loaded whenever a document has serious math in it;- use the proper environment, in this case
`equation*`

because you don't have an alignment.

Following Writing mod in congruence problems without leading space, use `\Mod`

as defined below:

```
\documentclass{article}
\usepackage{amsmath}
\newcommand{\Mod}[1]{\ (\mathrm{mod}\ #1)}
\begin{document}
\begin{align*}
MSC(C_n^2) &=
\begin{cases}
1, & \text{if $n$ is odd and $d$ is even ($n \equiv 1 \Mod{4}$)}, \\
2, & \text{if $n$ is even and $d$ is even ($n \equiv 0 \Mod{4}$)}, \\
3, & \text{if $n$ is odd and $d$ is odd ($n \equiv 3 \Mod{4}$)}, \\
3, & \text{if $n$ is even and $d$ is odd ($n \equiv 2 \Mod{4}$)}.
\end{cases}
\end{align*}
\end{document}
```

The LaTeX kernel borrows its definition of `\pmod`

```
\def\pmod#1{%
\allowbreak\mkern18mu({\operator@font mod}\,\,#1)}
```

(see `ltmath.dtx`

, code lines 39–40) from plain TeX, in which it reads

```
\def\pmod#1{\allowbreak\mkern18mu({\rm mod}\,\,#1)}
```

(see *The TeXbook*, p. 361); in both cases, the space that precedes the word “mod” is 1em wide. The examples of use given on p. 164 of *The TeXbook*, as well as the answer to Exercise 18.4, show that Knuth intended that that amount of space ought to be used also in in-line math formulas. Nonetheless, the `amsmath`

package modifies the above definition in such a way that a narrower space is employed when the formula is “not displayed”:

```
\newcommand{\pod}[1]{\allowbreak
\if@display\mkern18mu\else\mkern8mu\fi(#1)}
\renewcommand{\pmod}[1]{\pod{{\operator@font mod}\mkern6mu#1}}
```

(see also this answer to Writing mod in congruence problems without leading space, a question already linked from the accepted answer to this question). As you can see, the condition of “not being in display” is detected through the `\if@display`

switch.

Musing over this question, I began to wonder whether this could be a bug—or rather, a design flaw—in the package. Perhaps a better idea would be to have the amount of space change with the current math style, in a fashion similar to the way in which “big operator” symbols like `\sum`

or `\int`

change their size (and, to some extent, their shape) between display and non-display math. This behavior is easily achieved with the patch to the `\pod`

command (defined by the `amsmath`

package) that the following code shows:

```
% My standard header for TeX.SX answers:
\documentclass[a4paper]{article} % To avoid confusion, let us explicitly
% declare the paper format.
\usepackage[T1]{fontenc} % Not always necessary, but recommended.
% End of standard header. What follows pertains to the problem at hand.
% \usepackage{amsmath} % Required for what follows; but...
\usepackage{mathtools} % ... "mathtools" automatically loads "amsmath".
\makeatletter
\renewcommand*\pod[1]{%
\allowbreak
\mathchoice
{\mkern 18mu}%
{\mkern 8mu}%
{\mkern 8mu}%
{\mkern 8mu}%
(#1)%
}
\makeatother
\newtheorem{thm}{Theorem}
\begin{document}
In-line: \( 5\equiv 1 \pmod 4 \). In display:
\[ 5\equiv 1 \pmod 4 \]
Non-display inside display:
\[
f(x) =
\begin{cases}
4 & \mbox{if $x\equiv 0 \pmod 4$,} \\
x\bmod 4 & \mbox{otherwise.}
\end{cases}
\]
The same thing, but with \texttt{cases*} (requires \textsf{mathtools}):
\[
f(x) =
\begin{cases*}
4 & if $x\equiv 0 \pmod 4$, \\
x\bmod 4 & otherwise.
\end{cases*}
\]
\begin{thm}
All over again.
In-line: \( 5\equiv 1 \pmod 4 \). In display:
\[ 5\equiv 1 \pmod 4 \]
Non-display inside display:
\[
f(x) =
\begin{cases}
4 & \mbox{if $x\equiv 0 \pmod 4$,} \\
x\bmod 4 & \mbox{otherwise.}
\end{cases}
\]
With \texttt{cases*}:
\[
f(x) =
\begin{cases*}
4 & if $x\equiv 0 \pmod 4$, \\
x\bmod 4 & otherwise.
\end{cases*}
\]
\end{thm}
Nonetheless, inside an alignment, say
\begin{align*}
2 &\equiv 9 \pmod 7 \\
4 &\equiv 1 \pmod 3 \mbox{,}
\end{align*}
the output agrees with that of a displayed equation:
\[
5\equiv 1 \pmod 4 \mbox{.}
\]
\end{document}
```

This is the output the above code yields:

This modification could be an elegant and consistent way of achieving what the question asks for. If the `\mod`

command is used as well, it should be similarly modified too. Of course, one can’t introduce these changes in the code of `masmath`

package without disrupting all existing documents that use these features…

**Addition:**
Continuing to chew over this issue, I thought that it could even make sense to use the `\pmod`

command in super/subscripts (see example below); and in this case, it seems that the space before the left parenthesis should be made even thinner.

```
% My standard header for TeX.SX answers:
\documentclass[a4paper]{article} % To avoid confusion, let us explicitly
% declare the paper format.
\usepackage[T1]{fontenc} % Not always necessary, but recommended.
% End of standard header. What follows pertains to the problem at hand.
% \usepackage{amsmath} % Required for what follows; but...
\usepackage{mathtools} % ... "mathtools" automatically loads "amsmath".
\makeatletter
\renewcommand*\pod[1]{%
\allowbreak
\mathchoice
{\mkern 18mu}%
{\mkern 8mu}%
{\mkern 6mu}% "6mu" matches the space *after* the word "mod"
{\mkern 6mu}%
(#1)%
}
\makeatother
\newtheorem{thm}{Theorem}
\begin{document}
In-line: \( 5\equiv 1 \pmod{4} \). In display:
\[ 5\equiv 1 \pmod{4} \]
Non-display inside display:
\[
f(x) =
\begin{cases}
4 & \mbox{if $x\equiv 0 \pmod{4}$,} \\
x\bmod 4 & \mbox{otherwise.}
\end{cases}
\]
The same thing, but with \texttt{cases*} (requires \textsf{mathtools}):
\[
f(x) =
\begin{cases*}
4 & if $x\equiv 0 \pmod{4}$, \\
x\bmod 4 & otherwise.
\end{cases*}
\]
\begin{thm}
All over again.
In-line: \( 5\equiv 1 \pmod{4} \). In display:
\[ 5\equiv 1 \pmod{4} \]
Non-display inside display:
\[
f(x) =
\begin{cases}
4 & \mbox{if $x\equiv 0 \pmod{4}$,} \\
x\bmod 4 & \mbox{otherwise.}
\end{cases}
\]
With \texttt{cases*}:
\[
f(x) =
\begin{cases*}
4 & if $x\equiv 0 \pmod{4}$, \\
x\bmod 4 & otherwise.
\end{cases*}
\]
\end{thm}
Nonetheless, inside an alignment, say
\begin{align*}
2 &\equiv 9 \pmod{7} \\
4 &\equiv 1 \pmod{3} \mbox{,}
\end{align*}
the output agrees with that of a displayed equation:
\[
5\equiv 1 \pmod{4} \mbox{.}
\]
In super\slash subscripts:
\[
\int_{0}^{3\pi} \sin x\,dx
= -\cos x \biggr|_{0}^{3\pi}
= -\cos x \biggr|_{x\equiv 0 \pmod{2\pi}}^{x\equiv 3\pi \pmod{2\pi}}
= -\cos x \biggr|_{0}^{\pi}
= 1-(-1) = 2
\]
\end{document}
```

The output in this second case is