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I have read other answers about this same subject, however, no answers seem to formally prove, why the Cantor Diagonal Argument applied on a list of all the rational does not create another rational as Cantor Diagonal Argument applied on the Reals create another real.

Thank you

Michael Hardy
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Beaba
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    The premise is that the argument produces something different from every element of the list that is fed into the argument. If the list is all rational numbers, it will necessarily produce something that is not a rational number, because at each step of the chain it differentiates the sequence produced in at least one way from each element on the list. People do formalize the argument in different ways but that is the underlying premise. The decimal expansions of rationals are also periodic, and the Cantor argument does not seem to produce that, but that's another issue. – leslie townes Jan 31 '21 at 18:34
  • See this answer of mine, starting at the third paragraph ("That said, ..."). – Noah Schweber Jan 31 '21 at 19:45

3 Answers3

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Because if you have a sequence $(d_n)_{n\in\Bbb N}$ of digits, there is some real number whose decimal expansion is $0.d_1d_1d_3\ldots$ So, if you apply the Cantor diagonal argument to the numbers of $[0,1]$, you will get a number from $[0,1]$. But if you apply it only to the rationals form $[0,1]$, it may well happen that the number that you created is not rational, since most sequences $(d_n)_{n\in\Bbb N}$ of digits lead you to a number $0.d_1d_1d_3\ldots$ which is irrational.

José Carlos Santos
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    In fact, if your sequence contains all the rationals (after all, $|\mathbb N|=|\mathbb Q|$), then the number produced by Cantor's diagonal argument is necessarily not rational (!). – Derive Foiler Jan 31 '21 at 18:43
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If you mean the version of the diagonal argument that deals with decimal digits, or with binary digits (etc.), then the answer is that we know how to create an enumeration of the rationals, i.e. a sequence in which for every rational there is some finite number of steps after which we will reach that rational in the list, and the diagonal argument shows that the real number that emerges differs from every number in the list.

Michael Hardy
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Cantor's diagonal argument on a given countable list of reals does produce a new real (which might be rational) that is not on that list.

The point of Cantor's diagonal argument, when used to prove that $\mathbb{R}$ is uncountable, is to choose the input list to be all the rationals. Then, since we know Cantor produces a new real that is not on that input list, that real can't be rational.

Patrick Stevens
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