7

In handwritten mathematics it is common for the sake of clarity to repeat the last mathematical operator in those line breaks that occur in the middle of a formula. I mean something like:

We compute 1 + 2 + 3 + 4 + 5 +
+ 6 + 7 + 8 to know the answer.

In LaTeX I always do this repetition in displayed environments such as align or multline by manually adding the symbol after the \\ but this is much more difficult in inline formulas because the breaking point may change during different compilations. In this case I only do it occasionally for the multiplication \cdot. Anyway, this makes my mathematical style not coherent, so I would like to know if there is any package to implement this option.

If you think that this practice is bad I would like to read the reasons.

8

I have always considered this repetition a peculiarity of Russian typesetting, see e.g. http://www.tug.org/TUGboat/Articles/tb17-4/tb53grin.pdf. The article http://www.tug.org/TUGboat/Articles/tb16-4/tb49olga.pdf claims existence of a package you are looking for, but it's nowhere to be found.

Anyway, one Denis Ryabov took this into his hands and translated Grinchuk's ideas from the articles above into code: http://d-ryabov.livejournal.com/723.html. Copy the text from the page, paste into a file named rumathbr.sty, include it by \usepackage{rumathbr} and you are ready to go.

  • 3
    Thank you. I have found a newer version of this package in joomup.com/rumathbr.sty and I have uncommented a line to make the \cdot a breakable symbol. In the comment he says that he wants to implement it so that the \cdot becomes a \times in the case of a line break, but I do not follow this practice and I always use a repeated \cdot. – smp Jan 6 '11 at 20:41
  • By the way, I am Catalan and I did not know the Russian origin of this tradition. I know many people in here that use it in handwritten mathematics, but I cannot say that they form a vast majority. – smp Jan 6 '11 at 20:50
  • Current version at github.com/dryabov/rmathbr – AVB Jan 22 '17 at 20:36
2

A lightweight solution from Cherepanov's russadd.sty:

\def\hm#1{#1\nobreak\discretionary{}{\hbox{\m@th$#1$}}{}}
% usage: $a\hm+b\hm=c$
  • (Of course, one has to add \hms by hand...) – Grigory M May 23 '11 at 19:01

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