

Trapping a Transcendental (take 2 - sorry) - RiderOfGiraffes
http://www.penzba.co.uk/Writings/TrappingATranscendental.html?HN

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RiderOfGiraffes
Continuing the series started in response to this:
<http://news.ycombinator.com/item?id=1070604>

This was my first response: <http://news.ycombinator.com/item?id=1071734>

And this newest item lays the groundwork for talking about the questions
raised here: <http://news.ycombinator.com/item?id=1071748>

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BearOfNH
I don't get why the interval is divided into three pieces. Wouldn't the same
reasoning work if the interval were divided into just two pieces and you
always select the sub-interval not containing the Nth algebraic number?

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RiderOfGiraffes
The algebraic might be the mid-point, and hence, in some sense, in both. Note
that I have included both endpoints of each division to avoid the problems
that the infinite intersection of half-open intervals can create.

In fact I've recently thought that one could divide it into 5. Then you always
discard the outer divisions, and choose one of the inner divisions that
doesn't contain the algebraic you're avoiding. In this way the leftmost
endpoints of the chosen intervals form a strictly increasing sequence that's
bounded above, and the rightmost endpoints of the chosen intervals form a
strictly decreasing bounded sequence. That helps you to see that you really
want these things to have limits.

Does that help? If you ask more questions I can improve the article.

Thanks.

