Trapping a Transcendental (take 2 - sorry)

penzba.co.uk

17 points·RiderOfGiraffes··3 comments

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

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?

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.

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