5 ms·

This a very terse proof. It is stated "The same method can also be used to prove that e cannot be a root of a second degree equation with rational coeﬃcients",

This a very terse proof. It is stated "The same method can also be used to prove that e cannot be a root of a second degree equation with rational coeﬃcients", but if I recall correctly, the generalization for the irrationality of $e^2$ isn't entirely straightforward, mainly due to the presence of powers of 2 in the terms of the expansion.

conistonwater 10y ago
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.
- siegelzero 10y ago
  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 $e^{-1}$.
  - pmalynin 10y ago
    It can be shown by clever application of Taylor Expansion that e^2 is irrational.
arglebarnacle 10y ago
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?
- siegelzero 10y ago
  It's simply a habit from LaTeX :-)
- ajross 10y ago
  Math people can read TeX without compilation.
- stablemap 10y ago
  I don't have any recommendations, but they do exist: http://mathoverflow.net/questions/22141/how-do-i-see-latex-math-on-any-web-page-and-in-email http://mathoverflow.net/questions/22141/how-do-i-see-latex-m...
- simplicio 10y ago
  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.
  - nikdaheratik 10y ago
    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.
- mixedmath 10y ago
  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 http://davidlowryduda.com/static/chatjax.html
stablemap 10y ago
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.

conistonwater 10y ago

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.

siegelzero 10y ago
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 $e^{-1}$.
- pmalynin 10y ago
  It can be shown by clever application of Taylor Expansion that e^2 is irrational.

arglebarnacle 10y ago

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?

siegelzero 10y ago
It's simply a habit from LaTeX :-)
ajross 10y ago
Math people can read TeX without compilation.
stablemap 10y ago
I don't have any recommendations, but they do exist: http://mathoverflow.net/questions/22141/how-do-i-see-latex-math-on-any-web-page-and-in-email http://mathoverflow.net/questions/22141/how-do-i-see-latex-m...
simplicio 10y ago
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.
- nikdaheratik 10y ago
  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.
mixedmath 10y ago
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 http://davidlowryduda.com/static/chatjax.html

stablemap 10y ago

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.