
Douglas asks,
You have a numberline, and on the numberline you have plotted two distinct irrational numbers, which forms a line segment. You bisect that segment to form two more, bisect each of those, ad infinitum.
Question: Will every irrational number between the two endpoints be plotted eventually?
Answer: No. Here's why...
Let S be the set of numbers that are plotted in this manner. I will demonstrate that every element of S can be reached by a finite number of binary choices. The first two elements of S are the endpoints of the first segment. The next element of S is the midpoint of the first two points. From there, each Every element of S can be reached by a path of choices between the "left" and "right" segments.
To make this clear, I will identify the elements of S as a_{bitstring}, where the bitstring gives the sequence of choices needed to reach this element by successive bisecting of segments. I'll let the first two elements of S be a_{10} and a_{11}. For convenience, I'll say that trailing zeros in the bitstring don't matter. That is, a_{101}=a_{1010}=a_{10100}, etc.
With this in mind, the third element of S is
a_{101} = (a_{10}+a_{11})/2
Then, the next two elements of S are
a_{1001} = (a_{100}+a_{101})/2
a_{1011} = (a_{101}+a_{110})/2And then,
a_{10001} = (a_{1000}+a_{1001})/2
a_{10011} = (a_{1001}+a_{1010})/2
a_{10101} = (a_{1010}+a_{1011})/2
a_{10111} = (a_{1011}+a_{1100})/2and so on.
We will continue defining all the elements of S as a_{bitstring} in this way, so that every element of S has such a representation. And since each of these representations contains a counting number in binary format, it is clear that the number of elements of S is countable  that is, it can be put in onetoone correspondence with the counting numbers.
However, the number of irrational numbers in any closed interval with distinct endpoints is uncountable, which can be shown using the Cantor diagonal argument. Therefore, there are lots and lots of irrational numbers that are never reached by any finite number of successive bisections.
Thanks to William Cushing for help correcting errors that appeared on this page.
Definition of Interval
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