0

Why doesn't this code compile?

\usepackage{tikz}
\begin{document}

We will let $P_2\textit{n}+1 = v_0v_1v_2…v_2\textit{n}$ be a path of 
length $2\textit{n}+1$. We will plant an end vertex of a path $P_a$ 
of length \textit{a} to $v_4\textit{i}-3$ and an end vertex of a path 
$P_\textit{a}+2$ of length $\textit{a}+2$ to $v_4\textit{i}-1$ for 
$\textit{i}$ = $1,2,…,\textit{n}$. 

\end{document}
  • 1
    You are missing a \documentclass{...} command – Andrew Nov 1 '16 at 3:27
  • You don't need TikZ, though. – cfr Nov 1 '16 at 3:29
  • 1
    Why do you use \textit in math mode? Variables are set in italics by default in math – siracusa Nov 1 '16 at 4:39
  • I don't get any compilation errors if I use the article document class. (Of course, one shouldn't be using \textit for math-mode material, but that doesn't generate a compilation error.) Which document class do you use? – Mico Nov 1 '16 at 6:13
3
\documentclass{article}

\begin{document}

We will let $P_{2n+1} = v_0v_1v_2 \dots v_{2n}$ be a path of length
$2n+1$. We will plant an end vertex of a path $P_a$ of length $a$ to
$v_{4i-3}$ and an end vertex of a path $P_{a+2}$ of length $a+2$ to
$v_{4i-1}$ for $i = 1, 2, \dots, n$.

\end{document}

enter image description here

  • 1
    +1. I believe you identified the OP's issue, which was not a compilation error per se but, rather, a failure to use grouping to typeset the subscript terms correctly. (Separately, you also -- correctly! -- removed the \textit wrappers.) It would be useful if you wrote up in words what exactly you've done. – Mico Nov 1 '16 at 6:19
1

This one works fine:

\documentclass{article}

    \begin{document}

    We will let $P_2 n+1 = v_0v_1v_2…v_2 n$ be a path of 
    length $2n+1$. We will plant an end vertex of a path $P_a$ 
    of length $a$ to $v_4i-3$ and an end vertex of a path 
    $P_a+2$ of length $a+2$ to $v_4i-1$ for 
    $i$ = $1,2,\ldots,n$. 

    \end{document}

By placing $..$ within the math equations will be italic mode. So you do not need \textit{}

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.