0

The align environment is not advancing the equation counter. Here's the code that produced the output below:

\documentclass{report}

\usepackage{mathtools,amsthm}



\begin{document}
 \numberwithin{section}{chapter}
    \numberwithin{equation}{section}
    \let\realequation\equation
    \def\equation{\setcounter{equation}{\arabic{subsection}}%
       \refstepcounter{subsection}%
       \realequation}

    \theoremstyle{definition}
    \newtheorem{thm}[subsection]{Theorem}
    \newtheorem{lemma}[subsection]{Lemma}
    \newtheorem{corol}[subsection]{Corollary}
    \newtheorem{defn}[subsection]{Definition} % definition numbers are dependent on theorem numbers
    \newtheorem{defns}[subsection]{Definitions} % definition numbers are dependent on theorem numbers
    \newtheorem{prop}[subsection]{Proposition} % proposition numbers are dependent on theorem numbers
    \newtheorem{exmp}[subsection]{Example} % same for example numbers
    \newtheorem*{remark}{Remark}
    \newtheorem{enum_remark}[subsection]{Remark}
    \newtheorem{exercises}[subsection]{Exercises}
    \newtheorem*{opex}{Opening Exercises}

        \begin{align} \label{nchoosed}
            \begin{split}
            \prescript{}{n}C_d = \binom{n}{d} 
                &=  \frac{n \cdot (n-1) \cdot (n-2) \cdots (n - d + 1)}{d!} \\
                &= \frac{n!}{d! (n-d)!}
            \end{split}
        \end{align}

some text
    \begin{equation}
        a = b
    \end{equation}
\end{document}

enter image description here

The document is set up so that a single counter runs through each chapter and gets advanced by subsections, theorems, equations, etc.

  • the numbering is determined by code you are not showing so it is impossible to guess what is wrong. Please edit your example so that it is a complete small document that shows the problem. – David Carlisle Jun 13 at 15:11
  • @DavidCarlisle thanks I've updated the code – jtbrasel Jun 13 at 15:40
  • well it wasn't runnable as posted but I guessed some needed packages, it doesn't make the posted image but i suppose now shows the issue – David Carlisle Jun 13 at 15:45
  • align advances the equation counter but you have redefined equation to ignore that and use the subsection counter, so naturally they both use 1 – David Carlisle Jun 13 at 15:48
  • ahhh, ok, so I need to redefine align so that it also uses the subsection counter? – jtbrasel Jun 13 at 15:50
2

enter image description here

align advances the equation counter but you have redefined equation to ignore that and use the subsection counter, so naturally they both use 1

This makes subsection and equation counter the same, so if any construct increases one, the value of the other will have changed.

\documentclass{report}

\usepackage{mathtools,amsthm}

 \numberwithin{section}{chapter}
 \numberwithin{equation}{section}

% make equation counter an alias for subsection
\makeatletter
\let\c@equation\c@subsection
\makeatother



    \theoremstyle{definition}
    \newtheorem{thm}[subsection]{Theorem}
    \newtheorem{lemma}[subsection]{Lemma}
    \newtheorem{corol}[subsection]{Corollary}
    \newtheorem{defn}[subsection]{Definition} % definition numbers are dependent on theorem numbers
    \newtheorem{defns}[subsection]{Definitions} % definition numbers are dependent on theorem numbers
    \newtheorem{prop}[subsection]{Proposition} % proposition numbers are dependent on theorem numbers
    \newtheorem{exmp}[subsection]{Example} % same for example numbers
    \newtheorem*{remark}{Remark}
    \newtheorem{enum_remark}[subsection]{Remark}
    \newtheorem{exercises}[subsection]{Exercises}
    \newtheorem*{opex}{Opening Exercises}

  \begin{document}

first align
      \begin{align} \label{nchoosed}
            \begin{split}
            \prescript{}{n}C_d = \binom{n}{d} 
                &=  \frac{n \cdot (n-1) \cdot (n-2) \cdots (n - d + 1)}{d!} \\
                &= \frac{n!}{d! (n-d)!}
            \end{split}
        \end{align}

some text
    \begin{equation}
        a = b
    \end{equation}

second align
      \begin{align} \label{nchoosedz}
            \begin{split}
            \prescript{}{n}C_d = \binom{n}{d} 
                &=  \frac{n \cdot (n-1) \cdot (n-2) \cdots (n - d + 1)}{d!} \\
                &= \frac{n!}{d! (n-d)!}
            \end{split}
        \end{align}

some text
    \begin{equation}
        a = b
    \end{equation}
\end{document}
  • Thanks so much! that's the fix. – jtbrasel Jun 13 at 16:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.