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Sorry but this is probably a naive question.

Why can't you generate real numbers by a*10^b, the same way as rational numbers by a/b? a and b could be integers so that you would start counting real numbers like:

 a\b   0     1    -1     2    -2
 0     0     0     0     0     0
 1     1    10   0.1   100  0.01
-1    -1   -10  -0.1  -100 -0.01
 2     2    20   0.2   200  0.02
-2    -2   -20  -0.2  -200 -0.02

That would just take all of the integers and also apply a decimal point anywhere on those integers, thus making the real numbers no? Which ones would be missing?

Plus I don't understand the diagonal argument because the real number set is infinite, so surely the diagonal would just go on forever so you can never check them all since there will be more and more, never ending.

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    $\begingroup$ All irrational numbers will be missing. $\endgroup$
    – njguliyev
    Commented Oct 16, 2013 at 15:59
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    $\begingroup$ Make an estimate on which position is $\pi$; or even $1/3$. $\endgroup$
    – Carlos Eugenio Thompson Pinzón
    Commented Oct 16, 2013 at 15:59
  • $\begingroup$ I thought about the case of 1/3 for example. I guess the reason this method does not work is because to get 1/3 you would need "a" to be an integer of an infinite amount of 3s. In which case you would need to move the 3s over by an infinite amount, thus you can't do it? $\endgroup$
    – Matt Mitchell
    Commented Oct 16, 2013 at 16:00
  • $\begingroup$ This way you will define a subset of rational numbers, i.e. any of your numbers can be written as $\frac{a}{10^{(-b)}}$ (or plainly as $a*10^b$ if $b \geq 0$). $\endgroup$
    – dtldarek
    Commented Oct 16, 2013 at 16:01
  • $\begingroup$ How would this generate numbers like $\sqrt2$ or $e$ or many other numbers that are in the Reals because it is complete yet may not be Rationals? $\endgroup$
    – JB King
    Commented Oct 16, 2013 at 16:04

3 Answers 3

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The numbers that would be missing from this scheme would include every single irrational number, and many rationals, too. Specifically, this scheme includes only the rationals with terminating decimal representations.

The upshot of the diagonal argument has nothing to do with checking. What we do is take a countable list of real numbers, and then construct a real number that is not on that list (it will be real by completeness of the reals), by making sure that it fails to match every number on the list in (at least) one decimal place. This argument works for any countable list of real numbers, meaning that no countable list of real numbers will include every real number, meaning that the set of all reals is uncountable.

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  • $\begingroup$ I see what you mean about it being the case for any countable list. $\endgroup$
    – Matt Mitchell
    Commented Oct 16, 2013 at 16:09
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    $\begingroup$ I'm glad you made the point that diagonalization works for any countable list of reals. I'd add that the new number you are constructing has to be real by the completeness of the reals, and this is why you can't do diagonalization on the rationals; i.e. the constructed number is only guaranteed to be real, not rational. Making these two points explicit I think makes the proof a bit clearer to beginners. $\endgroup$
    – user452
    Commented Oct 16, 2013 at 16:36
  • $\begingroup$ @trb456: That is an excellent point. I will add that to my answer. $\endgroup$
    – Cameron Buie
    Commented Oct 16, 2013 at 16:47
  • $\begingroup$ I don't follow why it would necessarily omit irrational and repeating numbers. The reason being, as a and b increase without bound, irrational numbers get approximated more and more closely. In the limit of a and b going to infinity, it seems like it would reach them. And until a and b reach infinity, we haven't counted all the integers. $\endgroup$
    – Michael
    Commented Feb 4, 2014 at 20:35
  • $\begingroup$ With regards to non-terminating rationals, it occurs to me that, for instance, 1/3 in base 3 is 0.1 which terminates. $\endgroup$
    – Michael
    Commented Feb 4, 2014 at 20:39
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Because $\pi$ (or $1/9$) doesn't have an index in this counting scheme. If this were a way to count the reals, each real would have an index.

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  • $\begingroup$ I think the idea I had was that π with the decimal point removed would just be an integer and you could place the decimal point with b. But I realised that b would have to be infinite to place a decimal point on an infinitely long integer this way. $\endgroup$
    – Matt Mitchell
    Commented Oct 16, 2013 at 16:11
  • $\begingroup$ @MattMitchell FWIW (probably nothing), it's only infinite in one direction... $\endgroup$
    – Michael
    Commented Feb 4, 2014 at 20:40
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You would be missing all of the irrational numbers and also all non-terminating rational numbers.

For your question on Cantors argument. It does not matter that we can't possibly check everything, we just know that the $n^{th}$ digit of our new sequence is different from the $n^{th}$ element of the $n^{th}$ sequence. Hence, they cannot be equal.

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