This is perhaps a 2-parter in that 1) I'd like to have certain equations numbered using the theorem counter (and its associated "style") rather than that built into \begin{equation}. Per MWE, equation (1.1) would then be displayed as 1.3 (w/o parenthesis and bold). And 2) is it possible to display the theorem numbering "split" or centered about a series of equations, much like \split does with equation numbering? I've hacked the \newtheoremstyle to provide the appropriate numbering (as displayed as 1.2 in MWE), but this numbering is outside of the aligned equations. I'm not sure what/where I need to make adjustments. Any ideas would be greatly appreciated.


    \usepackage{amsmath, mathtools,amssymb,amscd,amsthm,amstext}

    % HACKING \adjustwidth so that it has equal vertical whitespace above/below


    % Indentation

    % \swapnumbers puts number ahead of heading - as in 3.4 Definition, rather than Definition 3.4

    % Redefine theorem style
    \newtheoremstyle{mytheoremstyle} % name
        {\topsep}                    % Space above
        {\topsep}                    % Space below
        {\itshape}                   % Body font
        {5mm}                 % Indent amount
        {\bfseries}                  % Theorem head font
        {.}                          % Punctuation after theorem head
        {0.5em}                      % Space after theorem head
        {}                           % Theorem head spec (can be left empty, meaning ‘normal’)

    % Creates a theorem-numbered entry
       { }


    The next result is also an almost immediate consequence of the preceding theorem.

    % Corollary 2.5
    The additive inverse of an element $a$ of a ring $R$, whose existence is
    asserted by Property \textup{\textbf{P}}$_4$, in unique.

    \noindent\textsc{proof}. To prove this statement, suppose that $a+x= 0$ and that
    $a+y=0$. Then $a=x=a+y$, and one of the cancellation laws of addition shows
    at once that $x=y$.

    \noindent Here's some statements:

          &(\textup{i})   &      -(-a) &= a,       \\
          &(\textup{ii})  &     -(a+b) &= -a-b,    \\
          &(\textup{iii}) &     -(a-b) &= -a+b,    \\
          &(\textup{iv})  &   (a-b)-c  &= a-(b+c). \\

    Let us prove the second of these ...
       a = -(-a)
  • Do you really want to use the theorem counter, or use the look and feel of the theorem counter for the equation number?
    – Bernard
    Feb 5 '16 at 20:27
  • I am trying to replicate the style found in an algebra book I'm working in, which assigns theorem-like numbering to significant non-theorem elements. Since these elements would necessarily increment the counter, subsequent theorems/corollaries/lemma/etc. would be appropriately numbered. So I'm thinking its more than look & feel.
    – Will Tech
    Feb 6 '16 at 0:47
  • Then you can simply use the \newtagform and \usetagform commands defined by mathtools (see § 3.2 of the documentation).
    – Bernard
    Feb 6 '16 at 0:53
  • Those two commands plus redefining theorem using \numberwithin{equation}{chapter} followed by \newtheorem{theorem}[equation]{Theorem} did the trick. Thanks, @Bernard.
    – Will Tech
    Feb 6 '16 at 6:25
  • @bernard -- answer, please. (this is a nice question.) Feb 6 '16 at 15:01

Sorry for the rather non-minimal MWE but I wanted to show a somewhat comprehensive example. My desire was to have theorems, corollaries, definitions, and specific equations work off of the same counter throughout all the sections of one chapter, and then reset in the next. I've attempted to self-document the example to show the significant parts of the code, as I'm probably the last one to explain the details of Latex. Both of my aforementioned questions in the original post are addressed here. Thanks to those whose comments made this work as I had envisioned.

  \documentclass[12pt, leqno]{book}
  \usepackage{changepage} % Use the \adjustwidth environment

  %% Set equation numbering to chapter, resetting
  %% on next chapter

  % Define theorem style - indentation + italics
  \newtheorem{theorem}[equation]{Theorem}      %% Theorems and Corollaries are 
  \newtheorem{corollary}[equation]{Corollary}  %% now tied to equation numbering

  % Define definition style - indentation only
   \newtheorem{definition}[equation]{Definition}  %% Defn tied to equation numbering

  %% Define tagform to remove brackets/braces around equation tag

  %% Apply tagform to document

  %% Inserts spaces, as in "\blank{3cm}"

  \section{Basic Concepts}

  Start with an equation:
  \begin{equation} %% Equation 1.1
     1 = 1 + 0
  We next prove the following theorem.

  \begin{adjustwidth}{.6cm}{.6cm} %% Theorem 1.2
     \textsc{(Cancellation Laws of Addition)}. If $a$, $b$, and $c$ are elements of a ring $R$, 
             the following are true:
     \item If $a+c = b+c$, then $a=b$,
     \item If $c+a=c+b$, then $a=b$.
  \noindent \textsc{proof}. We proceed to prove the first statement of this theorem. Let us therefore assume that 

  \usetagform{default} %% Reset equation numbering to "default" behavior
     a + c = b + c. \tag{1} %% Tag should be generated autom. but for now hard-coded
  By \textbf{P}$_4$, there exists an element $t$ of $R$ such that 
     c + t = 0. \tag{2}
  Now if follows from Equation (1) that ...
  \noindent This leads to the following corollary.

  \usetagform{nobrackets} %% Future equations now will follow "modified" behavior

  \begin{adjustwidth}{.6cm}{.6cm} %% Corollary 1.3
     Given ring $R$, then for all $a,b,c \in R$, then we have the following:
        &\text{(i)}   & \blank{1cm} -(-a)    &= a, \blank{6cm}  \\
        &\text{(ii)}  & \blank{1cm} -(a+b)   &= -a-b,           \\ 
        &\text{(iii)} & \blank{1cm} -(a-b)   &= -a+b,           \\ 
        &\text{(iv)}  & \blank{1cm} (a-b)-c  &= a-(b+c).        \\

  \begin{adjustwidth}{.6cm}{.6cm} % Definition 1.5
     Let $a$ be an element of a ring $R$ with unity $e$. If there exists an element
     $s$ of $R$ such that 
        as = sa = e,
     then $s$ is called the \textit{multiplicative inverse} of $a$.

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.