I have been writing

$a \equiv r (\mod n)$

but this puts a space between the ( and mod. For example, r ( mod n ) instead of r (mod n). So, how do you write r (\mod n) without that space? What would be the correct way to write $a \equiv r (\mod n)$?


The spacing issue is an obvious indication that it's not meant to be used that way. However, you could wing your own, depending on the sophistication you're after:

enter image description here


\newcommand{\Mod}[1]{\ (\mathrm{mod}\ #1)}


  a &\equiv r \mod n \\
  a &\equiv r \pmod{n} \\
  a &\equiv r \Mod{n}

  • This is great! I like how \Mod{n} looks. The only problem I'm encountering with \Mod{n} is that the "mod" becomes italicized in theorem/lemma/example/proof/etc. environments. – Mathemanic Aug 31 '14 at 19:12
  • 1
    @Mathemanic: You can try using \textup (or perhaps \mbox) instead of \text. – Werner Aug 31 '14 at 19:38
  • 4
    This should be \mathrm{mod}. – egreg Mar 9 '17 at 13:23
  • @egreg why is it \mathrm over \operatorname? – It'sNotALie. Nov 3 at 22:59
  • @It'sNotALie. Because this is not an operator. – egreg Nov 3 at 23:27

I think $a \equiv r \;(\bmod\; n)$ does what you want (as mentioned by jfbu in the comment).


Both the latex kernel and amsmath provide the command \pmod. In the latex kernel \pmod provides a fixed amount of space before (mod ...). The amsmath version is slightly more sophisticated, using a smaller amount of space when not in a display. So one way is to define a new command that temporarily switches off the display mode:



Sample output





Text (no change): \( a\tpmod{b} \) vs.\ \( c\pmod{d} \).

  a\tpmod{b} \\


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