It seems to me that the Digit-Matrix (the list of decimal expansions) in Cantor's Diagonal Argument is required to have at least as many columns (decimal places) as rows (listed real numbers), for the argument to work, since the generated diagonal number needs to pass through all the rows - thereby allowing it to differ from each listed number. With respect to the diagonal argument the Digit-Matrix then would ideally have a square shape.

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But then again, how can it have a square shape?

With every additional column (decimal place) you necessarily multiply the number of rows (listed real numbers) by 10 (or b in a b-based number system), just like our decimal system allows for 10^availableDigits natural numbers (including zero), i.e. 10 numbers with one digit, 100 numbers with 2 digits and 1000 numbers with 3 digits.

Now this should result in a vertically stretched, rectangular-shaped Digit-Matrix measuring an area of numberOfDecimalPlaces * 10^numberOfDecimalPlaces, where the first factor would be the number of columns and the latter the number of rows.

Having a DigitMatrix with so many more rows (listed real numbers) than columns (decimal places) would mean that the diagonal argument wouldn’t work, because the generated diagonal number would then only differ from the numbers in the square-area it crosses but not necessarily from all the numbers in the much longer part of the list below it.

I understand that Cantors Argument is about real numbers, which have infinitely many digits. Therefore one might argue that infinity is not a number and that infinite many decimal places might just be as infinite as a list with 10^infinity many entries.

Does this mean, that the Digit-Matrix in Cantor's Diagonal Argument can be considered an infinite square, even though the number of rows (length) clearly exceeds the number of columns (width) when regarding any finite number of decimal places of that diagonal number?

If not: How can the diagonal argument be applicable to a Digit-Matrix that is longer than wide, due to more real numbers than decimal places?

@Greg Martin:

To 1: As stated in my question: „I understand that Cantor’s Argument is about real numbers, which have infinitely many digits.“ I only mention those finite row counts to point out a general property of positional number systems: n-Digits allow for base^n unique numbers. This is what I have on my mind when I look at the first n columns of Cantor’s DigitMatrix: the decimal system with n decimal places.

And yes, this means indeed, that I am ignoring the infinite decimal expansions of the reals - but only for the moment. Only to emphasise that any sufficiently completed list is much longer than only those first n numbers that differ from the generated n-digit diagonal number, in fact so long that it actually includes the diagonal number somewhere deeper down, which means that Cantor’s argument does not work for any finite number n of decimal places.

But this is where I acknowledge that „…Cantors Argument is about real numbers, which have infinitely many digits.“

To 2: I understand that the DigitMatrix in Cantor’s argument is a static mathematical object. But anyway, here you are answering my question. You are answering it with „yes“, by calling both sides of the DigitMatrix infinite and the resulting shape a „square“ which works fine with the diagonal argument. I have expected this answer but I still have my difficulties to understand how that inequality: n digits != 10^n rows ,which we observe for any finite number of columns, can turn into an equality like: infinite many digits = infinite many rows.

To 3: I am not proposing a specific way of ordering the real numbers. I understand that the order might be, as you describe it: „someone’s unexpected attempt to list all these real numbers“ but while I don’t care about the order I still expect the list to be as complete as possible. In case you are referring to the picture: the digits are named by indices i.e. a23 (3rd digit of second number) but they are still totally variable in their values.

  • 2
    $\begingroup$ There seems to be a pretty serious misunderstanding of the argument, here. The idea of Cantor's argument is that we can show that any function from $\mathbb{N}$ to $\mathbb{R}$ cannot be surjective. Given a function $f : \mathbb{N} \to (0,1)$, we construct a number $x$ which is not in the image of $f$. To do this, we take the $n$-th decimal digit of $x$ to be distinct from the $n$-th decimal digit of $f(n)$. As $x$ is different from $f(n)$ for every possible $n$, it cannot be in the image of $f$. I don't need a "digi-matrix" to build the argument. $\endgroup$
    – Xander Henderson
    Jan 28, 2022 at 18:12

1 Answer 1


There are a few misunderstandings here:

  1. First, from your description of how row are added, it seems that you are thinking of real numbers as having only finitely many (nonzero) decimal places. (At least that's the only way I can make sense out of the proposed counts of how many new rows are added.) That's not the case: real numbers can and do have infinite decimal expansions.
  2. Second, each row of the table is infinite, and there are infinitely many rows. That's as "square" as one can hope for in this context, and it doesn't depend on what "speed" we add rows or columns. The list is a static mathematical object, not a running process.
  3. Most importantly: you have proposed one specific way of ordering these real numbers into a list. But that's not how Cantor's proof works. We are given an arbitrary list of real numbers (perhaps someone's unexpected attempt to list all these real numbers) and the diagonal argument shows that it must omit some real number.
  • $\begingroup$ I have added some comments to your answer directly below my question. $\endgroup$ Jan 28, 2022 at 23:19

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