1

I wanted to draw the working column for this method of finding square roots: http://youtu.be/UnshaiwljGY

Here is an image example of the process:

image

But I am unable to draw it because I am new to LaTeX tabulars, arrays and stuff.

Also, in the classic style of it, in the 2nd step, instead of underlining the number we have to guess is put in a circle. How do I circle a number?

Thank you in advance for your help.


Edit:

Here is another image:

another one

The columns can be achived through latex tables

Perhaps TikZ has to be used for other stylizations.

8
  • Welcome to TeX.SX! Can you provide a direct link to the question on the other site and explain what you mean by "compatible" Nov 13, 2014 at 14:07
  • There are known limitations on the latex that can be used on Math.SE and I assure you, a link to that question won't help you because all it asks is "Find the square root of 1000" which I want to demonstrate elaborately.
    – Nick
    Nov 13, 2014 at 14:10
  • MathJax is a limited TeX emulator. So if someone is going to provide a solution it is most often with additional packages and that will render your intention useless. There are not many macros available in MathJax so usage of graphical components is (practically ) impossible. You can fake it via tabulars but I don't think they are also implemented anyways.
    – percusse
    Nov 13, 2014 at 14:53
  • @percusse: Exactly, I'm asking how to fake it on there but if that's too difficult then fine. I'll be happy with code that I can compile on TexMaker. I'll just add it as an image file.
    – Nick
    Nov 13, 2014 at 15:18
  • To whomever it may concern, I have removed my requirement for the code to be compliant with MathJax. I now only require a solution that can be compiled in TexMaker
    – Nick
    Nov 13, 2014 at 17:00

1 Answer 1

1

It is fairly simple with TikZ. It will not work in MathJax though.

\documentclass[tikz,margin=10]{standalone}
\usetikzlibrary{tikzmark}
\usetikzlibrary{shapes.geometric}
\begin{document}
\sffamily
\begin{tikzpicture}[remember picture]
  \foreach \i in {0,-1,-2,-3,-3.5} \draw (0,\i) -- (-1.5,\i);
  \draw (-1.5,0) -- (-1.5,-3.5);
  \path (0,-.5) node[above left] 
    {10\subnode{4-a}{4}\subnode{9-a}{9}\subnode{7-a}{7}\subnode{6-a}{6}} 
    node[below left] {9\hphantom{0000}}
    (0,-1.5) node[above left] {1\subnode{4-b}{4}\subnode{9-b}{9}\hphantom{00}}
    node[below left] {124\hphantom{00}}
    (0,-2.5) node[above left] {25\subnode{7-b}{7}\subnode{6-b}{6}}
    node[below left] {2576}
    (0,-3.5) node[above left] {0}
    (-1.5,-.5) node[above left] {3}
    (-1.5,-1.5) node[above left] {62}
    (-1.5,-2.5) node[above left] {64{\color{red}4}}
    (0,.1) node[above left,xshift={-width("000")}] (x) {32\color{red}4};
  \node[draw=green!50!black,ellipse,inner sep=.2em] (y) at (x) {\phantom{324}};
  \draw[green!50!black,-stealth] 
    (y) -- ++ (1.5,0) node[right,draw] {Square root};
  \foreach \i in {4,9,7,6} \draw[red,-stealth] (\i-a) -- (\i-b);
\end{tikzpicture}
\end{document}

enter image description here

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .