50

The way I currently do it is $x \not | y$, which looks awful. There's got to be something better available.

Charles Stewart
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Qiaochu Yuan
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5 Answers5

63

An alternative to \nmid is to use the \centernot command from the centernot package. The resulting \centernot\mid symbol aligns perfectly with \mid and has a more pronounced slash than \nmid:

alt text

(On the right, the image shows how the commands behave in sub/superscript.)

lockstep
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Philippe Goutet
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60

$x\nmid y$ saves the day.

Harald Hanche-Olsen
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29

Another good looking (best to me) and easy option is to use the command \notdivides from the mathabx package. The code

\documentclass{article}
\usepackage{mathabx}
\begin{document}
\[ \prod_{a \notdivides b}^{a \notdivides b} a \notdivides b \]
\end{document}

creates the output

a \notdivides b

The negating line is longer than \nmid's but shorter than \centernot's.

Rayllum
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sebastian
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6

$x \bmod y \ne 0$ ;)

Or, as is more commonly written in mathematics: $x \not\equiv 0 \pmod y$ or $x \ne 0 \pmod y$.

Emil Jeřábek
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everybodyelse
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    That's a programmer's answer. Mathematicians would probably write it $x \neq 0 (\mathrm{mod} y)$ – Phil Miller Oct 25 '10 at 02:53
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    We'd both be wrong in an algebraist's eyes because a congruence is technically in order: $x \ncong 0 \left( \mathrm{mod} y\right)$. Thanks for pointing that out. – everybodyelse Oct 25 '10 at 04:34
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    The TeXnically correct way to do either of those is $x \ncong 0 \pmod{y} :) – Ryan Reich Nov 21 '10 at 08:59
  • To get the spacing right, the mod binary operator needs to be written as \bmod, not \mod (the latter is part of the congruence notation a\equiv b\mod c, which is more commonly seen with parentheses). Also, modular congruences are written with \equiv, not \cong (the latter is completely inappropriate in this context; it denotes the geometrical notion of congruence). – Emil Jeřábek May 04 '21 at 11:11
2

One possibility to assert, in symbols, that "a divides b" would be to use the MnSymbol package and then use $a \divides b$ (or $a \ndivides b$ for doesn't divide).

As I am typing a good deal of ring theory, I'm using those all the time.

rbrito
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