Alan Quatermain

Alan Quatermain

The Tumblog of one Jim Dovey, iOS Software Chief Architect at Kobo in Toronto, Ontario.
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Edit: Per the addendum to the previous article, I’ve updated the result here to use abs(y) instead of just (y) So, after trying to type math properly using HTML for the last post, I really got to missing LaTeX. So I did the same thing in LaTeX too—and this is the source for it all: \documentclass{amsart} \usepackage{amssymb} \begin{document} Pythagoras' theorem: $x = \sqrt{y^{2} + z^{2}}$ So where $y = z$ we have: \begin {eqnarray*} x = \sqrt{y^{2} + y^{2}} \\ \therefore x = \sqrt{2y^{2}} \\ \therefore x = \sqrt{2} \sqrt{y^{2}} \\ \therefore x = \sqrt{2} abs(y) \end{eqnarray*} \end{document} Lovely, isn’t it?

Edit: Per the addendum to the previous article, I’ve updated the result here to use abs(y) instead of just (y)

So, after trying to type math properly using HTML for the last post, I really got to missing LaTeX. So I did the same thing in LaTeX too—and this is the source for it all:

\documentclass{amsart}

\usepackage{amssymb}

\begin{document}

Pythagoras' theorem: $x = \sqrt{y^{2} + z^{2}}$

So where $y = z$ we have:
\begin {eqnarray*}
	x = \sqrt{y^{2} + y^{2}} \\
	\therefore x = \sqrt{2y^{2}} \\
	\therefore x = \sqrt{2} \sqrt{y^{2}} \\
	\therefore x = \sqrt{2} abs(y)
\end{eqnarray*}

\end{document}

Lovely, isn’t it?