2

I do not like the spacing printed by sqrt. In the following code, which not takes care of empty optional argument for the moment, I break the good spacing before the sqrt symbol.

\documentclass[12pt,a4paper]{article}

\let\stdsqrt\sqrt
\renewcommand\sqrt[2][]{\stdsqrt[#1\,\,]{#2\,}}


\begin{document}

Bla, bla $\sqrt{2} = \stdsqrt{2}$ et $\sqrt[2]{3} = \stdsqrt[2]{3}$ .

Bla, bla $\stdsqrt{2} = \sqrt{2}$ et $\stdsqrt[2]{3} = \sqrt[2]{3}$ .

\end{document}
  • Might be an idea to include an image as well – daleif Sep 26 '19 at 18:25
5

You're better using \leftroot:

\documentclass[12pt,a4paper]{article}
\usepackage{amsmath}

\begin{document}

Bla, bla $\sqrt{2} = \sqrt{2}$ et $\sqrt[\leftroot{1}3]{3} = \sqrt[3]{3}$.

Bla, bla $\sqrt{2} = \sqrt{2}$ et $\sqrt[\leftroot{3}3]{3} = \sqrt[3]{3}$.

Bla, bla $\sqrt{2} = \sqrt{2}$ et $\sqrt[3\,\,]{3} = \sqrt[3]{3}$.

\end{document}

enter image description here

I've never seen the index just above the left margin of the radical.

  • Thanks for pointing me to \leftroot such as to not have no more problem with \sqrt left unsolved. – projetmbc Sep 26 '19 at 20:43
0

Indeed the problem comes from the missing treatment of the first empty argument. The following code seems to do the job.

\documentclass[12pt,a4paper]{article}

\let\stdsqrt\sqrt
\renewcommand\sqrt[2][]{\if\relax\detokenize{#1}\relax\stdsqrt{#2\,}\else\stdsqrt[\!\!#1\,\,]{#2\,}\fi}


\begin{document}

Bla, bla $\sqrt{2} = \stdsqrt{2}$ et $\sqrt[2]{3} = \stdsqrt[2]{3}$ .

Bla, bla $\stdsqrt{2} = \sqrt{2}$ et $\stdsqrt[2]{3} = \sqrt[2]{3}$ .

\end{document}

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.