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By the Cantor slash argument, as explained for example here (at about 4:00), a new real number can always be generated out of any list of real number decimal expansions by taking the digits along the diagonal of the list and calling those digits a new number.

I am having a hard time understanding why Cantor's slash argument does not also apply to the rationals.

The rationals do not have infinite non-repeating decimal expansions, but they can be made arbitrarily large. Therefore we can always make list of rationals like the one you make for reals in the slash argument, with the one difference that the "..." at the end of each number is interpreted as "goes on to become arbitrarily large" instead of "goes on forever." On any such list, it will always be possible to generate a new number that is not on the list by taking the diagonals in the same way you do for reals. The rationals are thus unlistable or uncountable in the same way as the reals are. (Please point out the flaw in this reasoning.)

If I may also pose a very closely related follow-up:

If the Ford Circle algorithm is allowed to run infinitely, it will, in that limit, completely fill up every possible spot on the number line with rational numbers. Where on the number line, then, in that limit, is there any space for irrational and transcendental numbers? Or what am I missing here?

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    $\begingroup$ How do you ensure that the number you get produces a rational number. When you apply the diagonal argument (usual term, not "slash,") you'd have to ensure that the resulting number is still rational for it to work. We know the result is real, but ensuring it is rational is very very hard. $\endgroup$ – Thomas Andrews Oct 19 '15 at 1:42
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    $\begingroup$ On the followup (really an entirely different question): You can also "fill up" the real line with numbers of the form $a/2^n$ were $a,n$ are integers and $n\geq 0$, but those numbers aren't all rationals. So your intuition about what it means to be able to "fill up" the real line is just wrong. That happens with infinity a lot in the early going, because we think intuitions we understand from the finite world apply. $\endgroup$ – Thomas Andrews Oct 19 '15 at 1:45
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Actually, to expound on T. Andrews' comment, by applying the Cantor diagonal procedure to a complete listing of rational numbers (which exists, by Cantor's "anti-diagonal" or "short diagonal" enumeration of $N$x$N$) you will obtain a non-rational number. Guaranteed.

The original listing of rationals was complete (by assumption), and the construction builds a decimal expansion different from all others in the list. Thus the number obtained is irrational.

Your error is that you start with a very small (negligible, cardinal wise) subset of the reals, and your procedure takes you out of this subset into the much larger (humongous really) set of irrational reals. This happens no matter what the exact starting point and the details of your procedure (what enumeration of rationals is used, the digit modification procedure).

edit add-on In contradistinction the result of Cantor's diagonal procedure was in the same set/class, namely the non-terminating decimals, as the original objects of the procedure. end add-on


Actually I bet one can obtain any one real number from doing the Cantor diagonal procedure on an adequate listing of rationals.

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  • $\begingroup$ As to Ford Circles, there are 2<sup>n</sup> new circle-line tangent points in the n-th step? Such union is still very much countable. It is a beautiful construction, but still from the set theory POV the generator of a countable set of circles (and tangent points). $\endgroup$ – Dacian Bonta Nov 9 '15 at 21:44
  • $\begingroup$ Also, unless I mis-read the comment, but the numbers of form a/(2^n) where a, n are integers are rational (ratio of two integers). As such these numbers are very much countable as an (infinite) subset of the rationals. $\endgroup$ – Dacian Bonta Nov 10 '15 at 1:48

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