# How to align two column equations inside an enumerate environment?

I want to align equations inside an enumerate environment or to get a similar result using another environment. Here's the actual code:

\documentclass{article}
\usepackage{amsthm}
\begin{document}
\begin{enumerate}
\item $\forall a,b \in K \qquad a + b \in K$
\item $\forall a,b \in K \qquad a \cdot b \in K$
\item $\forall a,b \in K \qquad a + b = b + a$
\item $\forall a,b,c \in K \qquad a + (b + c) = (a + b) + c$
\item $\exists n_{+} \in K \ \forall a \in K \qquad a + n_{+} = a$
\item $\forall a \in K \ \exists a_{+}' \in K \qquad a + a_{+}' = n_{+}$
\item $\forall a,b \in K \qquad a \cdot b = b \cdot a$
\item $\forall a,b,c \in K \qquad a \cdot (b \cdot c) = (a \cdot b) \cdot c$
\item $\forall a,b,c \in K \qquad a \cdot (b + c) = a \cdot b + a \cdot c$
\item $\exists n_{\cdot} \in K \ \forall a \in K\backslash\{n_{+}\} \qquad a \cdot n_{\cdot} = a$
\item $\forall a \in K\backslash\{n_{+}\} \ \exists a_{\cdot}' \in K \qquad a \cdot a_{\cdot}' = n_{\cdot}$
\end{enumerate}
\end{document}


I want the equations in the second column would be left aligned.

• Welcome to TSE. Please post a Minimal Working Example, instead of a code snippet. Sep 18, 2021 at 18:35
• Off-topic: You should use \setminus, not \backslash, in the final two rows.
– Mico
Sep 18, 2021 at 20:29

Here's one way to do it with a tabular which automatically numbers itself

\documentclass{article}
\usepackage{array}
\newcounter{rowcount}
\setcounter{rowcount}{0}
\begin{document}
\begin{tabular}{@{\stepcounter{rowcount}\therowcount. }ll}
$\forall a,b \in K$ & $a + b \in K$\\
$\forall a,b \in K$ & $a \cdot b \in K$\\
$\forall a,b \in K$ & $a + b = b + a$\\
$\forall a,b,c \in K$ & $a + (b + c) = (a + b) + c$\\
$\exists n_{+} \in K \ \forall a \in K$ & $a + n_{+} = a$\\
$\forall a \in K \ \exists a_{+}' \in K$ & $a + a_{+}' = n_{+}$\\
$\forall a,b \in K$ & $a \cdot b = b \cdot a$\\
$\forall a,b,c \in K$ & $a \cdot (b \cdot c) = (a \cdot b) \cdot c$\\
$\forall a,b,c \in K$ & $a \cdot (b + c) = a \cdot b + a \cdot c$\\
$\exists n_{\cdot} \in K \ \forall a \in K\backslash\{n_{+}\}$ & $a \cdot n_{\cdot} = a$\\
$\forall a \in K\backslash\{n_{+}\} \ \exists a_{\cdot}' \in K$ & $a \cdot a_{\cdot}' = n_{\cdot}$\\
\end{tabular}
\end{document}


I took the numbering system from this answer.

• +1 Surely. I have misunderstood....as usual :-( Sep 18, 2021 at 20:24

Does it have to be an enumerate environment? Might using an array environment be acceptable?

\documentclass{article}
\usepackage{array}
\newcounter{mycount}
\newcolumntype{Z}{>{\refstepcounter{mycount}\themycount.}r}
\newcolumntype{L}{>{\displaystyle}l}
\newcolumntype{R}{>{\displaystyle}r}

\begin{document}
$\renewcommand{\arraystretch}{1.5} \begin{array}{@{} Z R @{\qquad} L @{}} & \forall a,b \in K & a + b \in K \\ & \forall a,b \in K & a \cdot b \in K \\ & \forall a,b \in K & a + b = b + a \\ & \forall a,b,c \in K & a + (b + c) = (a + b) + c \\ & \exists n_{+} \in K \ \forall a \in K & a + n_{+} = a \\ & \forall a \in K \ \exists a_{+}' \in K & a + a_{+}' = n_{+} \\ & \forall a,b \in K & a \cdot b = b \cdot a \\ & \forall a,b,c \in K & a \cdot (b \cdot c) = (a \cdot b) \cdot c \\ & \forall a,b,c \in K & a \cdot (b + c) = a \cdot b + a \cdot c \\ & \exists n_{\cdot} \in K \ \forall a \in K\setminus\{n_{+}\} & a \cdot n_{\cdot} = a \\ & \forall a \in K\setminus\{n_{+}\} \ \exists a_{\cdot}' \in K & a \cdot a_{\cdot}' = n_{\cdot} \end{array}$
\end{document}


I often use multicol package in this simple case. The {2} corresponds to the alignment for two columns. I have used \usepackage{geometry} to get a correct spacing between two columns.

\documentclass{article}
\usepackage{amsthm}
\usepackage{multicol}
\usepackage{geometry}

\begin{document}
\begin{multicols}{2}
\begin{enumerate}
\item $\forall a,b \in K,\, a + b \in K$
\item $\forall a,b \in K \qquad a \cdot b \in K$
\item $\forall a,b \in K \qquad a + b = b + a$
\item $\forall a,b,c \in K \qquad a + (b + c) = (a + b) + c$
\item $\exists n_{+} \in K \ \forall a \in K \qquad a + n_{+} = a$
\item $\forall a \in K \ \exists a_{+}' \in K \qquad a + a_{+}' = n_{+}$
\item $\forall a,b \in K \qquad a \cdot b = b \cdot a$
\item $\forall a,b,c \in K \qquad a \cdot (b \cdot c) = (a \cdot b) \cdot c$
\item $\forall a,b,c \in K \qquad a \cdot (b + c) = a \cdot b + a \cdot c$
\item $\exists n_{\cdot} \in K \ \forall a \in K\backslash\{n_{+}\} \qquad a \cdot n_{\cdot} = a$
\item $\forall a \in K\backslash\{n_{+}\} \ \exists a_{\cdot}' \in K \qquad a \cdot a_{\cdot}' = n_{\cdot}$
\end{enumerate}
\end{multicols}
\end{document}


I have dealt with the issue using:

\documentclass{article}
\usepackage{tabto}
\begin{document}
\TabPositions{0.4\linewidth}
\begin{enumerate}
\item $\forall a,b \in K$ \tab $a + b \in K$
\item $\forall a,b \in K$ \tab $a \cdot b \in K$
\item $\forall a,b \in K$ \tab $a + b = b + a$
\item $\forall a,b,c \in K$ \tab $a + (b + c) = (a + b) + c$
\item $\exists n_{+} \in K, \forall a \in K$ \tab $a + n_{+} = a$
\item $\forall a \in K, \exists a_{+}' \in K$ \tab $a + a_{+}' = n_{+}$
\item $\forall a,b \in K$ \tab $a \cdot b = b \cdot a$
\item $\forall a,b,c \in K$ \tab $a \cdot (b \cdot c) = (a \cdot b) \cdot c$
\item $\forall a,b,c \in K$ \tab $a \cdot (b + c) = a \cdot b + a \cdot c$
\item $\exists n_{\cdot} \in K, \forall a \in K\backslash\{n_{+}\}$ \tab $a \cdot n_{\cdot} = a$
\item $\forall a \in K\backslash\{n_{+}\}, \exists a_{\cdot}' \in K$ \tab $a \cdot a_{\cdot}' = n_{\cdot}$
\end{enumerate}
\end{document}


However the solution provided by Willoughby looks better.