0

I have this table on Word and I would like to convert it to a latex table. The only thing is I don't know how to create rows that have 2 lines.

If someone could just give me the code to make this table a latex table, I'd really appreciate it!

enter image description here

\begin{center}
\begin{tabular}{ |c|c|c| } 
 \hline
 $(x)_2 = n$; $n\%5$  & $(x0)_2 = 2n$; $2n\%5$ & $(x1)_2 = 2n + 1$; $(2n+1)\%5$ \\ 
 \hline
 0 & 0 & 1 \\ 
 1 & 2 & 3 \\
 2 & 4 & 0 \\
 3 & 1 & 2 \\
 4 & 3 & 4 \\
 \hline
\end{tabular}
\end{center}
3
  • 2
    If you really want to duplicate that layout, just leave off the \hline between rows. But I'd recommend getting rid of most of the rules entirely. See texdoc.net/pkg/booktabs Commented Apr 27, 2016 at 1:54
  • OK, so please give us the code you've got for the table and somebody will be happy to adapt it to show you how to use multirow or p{} or similar. But @PaulGessler 's suggestion is better (either of them) in this case.
    – cfr
    Commented Apr 27, 2016 at 1:54
  • @cfr Alrighty. I've added the code
    – 14wml
    Commented Apr 27, 2016 at 2:06

2 Answers 2

1

The easiest option is to split the heading into two rows with no line between them:

enter image description here

\documentclass{article}

\begin{document}
\begin{center}
\begin{tabular}{ |c|c|c| } 
 \hline
 $(x)_2 = n$ & $(x0)_2 = 2n$ & $(x1)_2 = 2n + 1$ \\ 
$n\%5$  & $2n\%5$ & $(2n+1)\%5$ \\ 
\hline
 0 & 0 & 1 \\ 
 1 & 2 & 3 \\
 2 & 4 & 0 \\
 3 & 1 & 2 \\
 4 & 3 & 4 \\
 \hline
\end{tabular}
\end{center}
\end{document}
2

Many are likely to suggest that you should follow the advice from the booktabs package and dispense with the vertical rules and most of the horizontal rules, which aren´t doing much more than distracting the eyes in so simple a table. Note also how the spacing between rows tends to be better by default as well (especially around the \hlines).

\documentclass{article}
\usepackage{booktabs}
\begin{document}
\begin{center}
\begin{tabular}{ ccc } 
 \toprule
 $(x)_2 = n$ & $(x0)_2 = 2n$ & $(x1)_2 = 2n + 1$ \\ 
$n\%5$  & $2n\%5$ & $(2n+1)\%5$ \\ 
\midrule
 0 & 0 & 1 \\ 
 1 & 2 & 3 \\
 2 & 4 & 0 \\
 3 & 1 & 2 \\
 4 & 3 & 4 \\
 \bottomrule
\end{tabular}
\end{center}
\end{document}

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .