Off topic, but can you tell me whether you're using a browser extension to render $e^2$ (and if so which one!), or if you're simply using the $ character as a delimiter for humans who read and write a lot of LaTeX code?

Math people can read TeX without compilation.

It's simply a habit from LaTeX :-)

I always found it weird that despite being developed at CERN and most widely used in the early years by universities and research labs, HTML never developed a decent standard for transmitting/rendering math equations (I know that there's a math tag now, but it doesn't seem to get much use). Especially given that Latex would seem to provide a pretty widely known de-facto standard for the syntax.

1) HTML is derived directly from SGML which was used to handle similar markup issues in printed texts. One of the reasons why HTML was successful was because there were already similar tools in place from SGML that could be quickly applied to handle processing and correctness. Adding math would require building a number of additional tools.

2) TeX by itself doesn't resolve very well into the DOM tree which is what parsers use to organize the markup data internally. You need a markup language that is similar to HTML if you want to make it easy for browsers, parsers, and libraries to adopt the new standard.

This may be a little ridiculous, but I have a bookmarklet that I click on sometimes to render most LaTeX in the current window.

I use this because I'm a mathematician and I chat with some other mathematicians through Slack frequently, and this doesn't have good LaTeX support. But by clicking this, it loads MathJax from the MathJax CDN to render. We also use this on the chatrooms associated to Math.StackExchange and MathOverflow, which is why it's called ChatJax.

If this is something that interests you, you can find it on this semi-hidden page on my website: http://davidlowryduda.com/static/chatjax.html

I don't have any recommendations, but they do exist:

http://mathoverflow.net/questions/22141/how-do-i-see-latex-m...

As pointed out in the annotation, isn't the case $e^2$ just exactly the same proof but for the special case $c=0$? And in any case, if $e^2$ were rational, then $e$ could only be rational or a quadratic irrationality.

Yeah, this proof seems to work. I was thinking of the case where you attack $e^2$ directly, instead of going after both $e$ and $e^{-1}$.

It can be shown by clever application of Taylor Expansion that e^2 is irrational.

Maybe it's better to write his equation as ae² - ce + b = 0. The proof would say that e isn't degree ≤ 2 over the rationals.