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I have written a simple enumeration with some formulas inside. I don't understand why in some items the square parenthesis are bigger than in other items. How can I tell tex to make all of the same size? I prefer the smaller ones. I don't want to replace \left( \right) with \big( \big) or something similar, because I would like all parenthesis to behave the same way automatically.

Here the code

\begin{enumerate}
    \item se $x \in \left[\frac{1}{2},\frac{2}{3}\right]$, allora $F(x) \in A$;
    \item se $n \in \mathbb{Z}_{\geq 2}$ e $x \in \left]\frac{n}{n+1},\frac{n+1}{n+2}\right[$, allora $F(x) \in \left]\frac{1}{n+1},\frac{1}{n}\right[$;
    \item se $x \in \left[\frac{1}{3},\frac{1}{2}\right[$, allora $F(x) \in A$;
    \item se $n \in \mathbb{Z}_{\geq 3}$ e $x \in \left]\frac{1}{n+1},\frac{1}{n}\right[$, allora $F(x) \in \left]\frac{1}{n},\frac{1}{n-1}\right[$.
\end{enumerate}

And here is the output enter image description here

  • \left[...\right] adjusts automatically to the height of the contents, i.e. the fractions, but I wonder about the output of your code. Shouldn't the brackets around the fractions have the same size? – user31729 Mar 17 '16 at 17:34
  • you are right, I have uploaded the wrong screenshot, in the new one that I am going to upload you will se 1 the same as 3 but not like 2 and 4. I would like 1, 2, 3, 4 to be all equal. – Nisba Mar 17 '16 at 17:36
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The problem is using \left and \right, with the additional issue that + and - appear in the denominators, because they add depth and make TeX use the next size of delimiters.

For textual fractions, \big size is sufficient:

\documentclass[a4paper]{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[italian]{babel}

\usepackage{amsmath,amssymb}

\begin{document}

\begin{enumerate}
\item se $x \in \bigl[\frac{1}{2},\frac{2}{3}\bigr]$, allora $F(x) \in A$;

\item se $n \in \mathbb{Z}_{\geq 2}$ e $x \in \bigl]\frac{n}{n+1},\frac{n+1}{n+2}\bigr[$,
  allora $F(x) \in \bigl]\frac{1}{n+1},\frac{1}{n}\bigr[$;

\item se $x \in \bigl[\frac{1}{3},\frac{1}{2}\bigr[$, allora $F(x) \in A$;

\item se $n \in \mathbb{Z}_{\geq 3}$ e $x \in \bigl]\frac{1}{n+1},\frac{1}{n}\bigr[$,
  allora $F(x) \in \bigl]\frac{1}{n},\frac{1}{n-1}\bigr[$.
\end{enumerate}

\end{document}

enter image description here

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