When Cantor published his seminal diagonal argument, he used binary code and did not distinguish the real numbers 0.1000… and 0.0111… Later, Koenig found a way to save the diagonal argument in binary digits (a reference would be appreciated). Meanwhile, the most popular form is to use decimal digits.

My main question: Who was the first person to mention that in the diagonal argument digits should not be replaced by 9's because of 0.1999⋯ = 0.2000… ?


2 Answers 2


The first mention of the "nine-problem" was by Felix Klein, Vorträge über ausgewählte Fragen der Elementargeometrie, Teubner 1895, p. 42. http://quod.lib.umich.edu/cgi/t/text/text-idx?c=umhistmath;idno=ACV2370.0001.001

  • $\begingroup$ I was unable to find the relevant section on the page you cite; Can you reproduce it? $\endgroup$
    – Danu
    Dec 3, 2015 at 10:08
  • 2
    $\begingroup$ Klein writes on p. 42: 1 = 0.99...9.... Um der dadurch bewirkten Unbestimmtheit zu entgehen, setzen wir ein für alle Mal fest, dass wir unendliche Neuner-Reihen vermeiden wollen. $\endgroup$
    – Franz Kurz
    Dec 3, 2015 at 15:16

When Cantor published his seminal diagonal argument, he used binary code and did not distinguish the real numbers 0.1000… and 0.0111…

This is not relevant for the proof.

See :


For detailed discussion of Cantor's proofs:

1873 proof: Dauben page 50 and Ferreiros page 179.

1877 proof: Ferreiros page 190-2. See page 190, for the presumibely source of the misconception.

1891 proof (the diagonalization one): Dauben page 165 and Ferreiros page 287.

For an early textbook version of the diagonal proof, see:

"we now define a number [...] $b=0.b_1b_2b_3 \ldots$, where $b_r$ is never the same as $a_{rr}$."

Regarding the purported "mistake" in the diagonal proof, see e.g. Robert Gray Georg Cantor and Transcendental Numbers, The American Mathematical Monthly, vol. 101, 1994, page 823.

The "general" theorem consider an enumeration of an infinite sequnce of infinite strings of symbols, with only two symbols : $m$ or $w$.

The proof does not rely on any property of real numbers (or their representation).

It is easily "applied" to the real numbers through their binary representation.

In the case of number with a "dual" representation, both representation must be included into the enumeration.

We can consider a simplified case :

$$a_1=0.000 \ldots$$

$$a_2=0.011 \ldots$$

$$a_3=0.100 \ldots$$


and apply Young's procedure; we will get :

$$b=0.101 \ldots$$

because: $b_1 \ne a_{11}=0$, $b_2 \ne a_{22}=1$, $b_3 \ne a_{33}=0$.

The presence of $a_2$ and $a_3$, that are two "dual" representation of the same number, does not invalidate the diagonalization.

With this simple example we see that also if two members of the list, say $a_n$ and $a_m$, are two distinct representations of the same number we have that :

$b \ne a_n$, because $b_n \ne a_{nn}$

and :

$b \ne a_m$, because $b_m \ne a_{mm}$

and this "works" irrespectively of the fact that $a_n=a_m$ or not.

  • $\begingroup$ You are wrong. See Cantor's original paper [G. Cantor: "Über eine elementare Frage der Mannigfaltigkeitslehre", Jahresbericht der DMV I (1890-91) 75-78]. He used binary digits m and w and did not bother whether mwww... or wmmm... are identical. Further I have reason to believe that Young is not the first source of the nine-problem. $\endgroup$
    – user3292
    Nov 28, 2015 at 8:27
  • $\begingroup$ @Mauro Allegranza: Gray is completely wrong. Cantor's diagonal number differs from all $n$-digit numbers for every $n$ but does never complete a digit sequence of a transcendental number. Proof: For every digit $d_n$ we can prove that it is not completing an infinite sequence. $\endgroup$
    – Franz Kurz
    Jun 10, 2017 at 21:14

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