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How to prove that the new number produced by the Cantor's diagonalization process applied to $\Bbb Q$ is not a rational number ?

Suppose, someone claims that there is a flaw in the Cantor's diagonalization process by applying it to the set of rational numbers. I want to prove that the claim is false by showing that the new number produced by this process is not rational. How to prove this ? Hope I have made my question clear.

Marc van Leeuwen
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user103623
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    It can be. It depends on which numbers you put into your list of real numbers to begin with. – Samuel Nov 04 '13 at 13:33
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    @Samuel: not real, rational. – user103623 Nov 04 '13 at 13:56
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    The rationals are countable since we can construct an explicit bijection with the positive integers. This proof doesn't depend on any diagonal argument. Now if you list all rationals and apply the diagonal argument, the resulting number cannot be on the list, so it cannot be rational. – Ayman Hourieh Nov 04 '13 at 14:30
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    If the other person wants to show that the proof is flawed by applying it to a list of all rationals, this other person would have to show that the produced number is rational. – Carsten S Nov 04 '13 at 14:58
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    @AymanHourieh The OP knows for a fact that Cantor's argument doesn't flaw in rational number. The reason, as OP claimed, is because the resulted number is not rational. But why is the result? How can it be derived. You mentioned and I quote "...the resulting number cannot be on the list..." is instead an argument supporting "Cantor's diagonal proof is flawed", which is opposite of what the OP is doing--Cantor's diagonal proof isn't flawed on rational number. Please correct me if I am wrong. – Andes Lam Jan 17 '21 at 07:33
  • CarstenS's explanation completely convinced me, and I believe it is the key to the problem. – if_ok_button May 15 '23 at 00:06

2 Answers2

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This has very little to do with rational numbers themselves. You apply the diagonal argument to construct a number that is not on the list. Now the only reason that you must have produced an irrational number is that all the rational numbers are on the list, so it cannot be any of them. There is nothing intrinsic the construction that favours irrational numbers: if just one rational number were missing from the list, it might be that you have constructed just that number.

Added. Actually, now that I think about that, it is not true: in order for given some rational number to be the result of the diagonal procedure, there are a great many (other) rational numbers that must be absent from the list. Assuming decimal notation and for simplicity a rational number without terminating decimal representation, there are for every digit position $9$ rational numbers that differ from it only in that digit position. Of those, only one can be on the list, and only if it occurs at the exactly right position. So an infinite number of those "neigbours" need to be absent from the list. And even if one assumes binary notation (with only one neighbour per digit position), there must be one neighbour absent for every non-neighbour that is on the list (because the position of the former is taken), still making for a requirement of infinitely many rational numbers from the list. All this is related to the fact that modifying a rational number in a single position will not make it irrational. Curious!

Marc van Leeuwen
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  • A possibly interesting concrete example of this is to explicitly enumerate all the terminating rational numbers and then apply the diagonalization argument, which evidently generates a nonterminating number, probably something of the form $\frac19 + \frac{m}{10^{n}}$. – MJD Nov 04 '13 at 15:00
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Suppose the constructed number $a$ belongs to the set of rational numbers $Q$. In that case, $a$ will necessarily start repeating from the $s$th digit, with a repeating sequence denoted as $U$. The premise of the diagonal argument is that we can always find a digit $b$ in the $x$th element of any given list of $Q$, which is different from the $x$th digit of that element $q$, and use it to construct $a$.

However, when there exists a repeating sequence $U$, we need to ensure that $b$ follows the pattern of $U$ after the $s$th digit. This constraint hinders the operation of finding a different digit and makes it impossible to continue at some point after the $s$th digit.

Harish Chandra Rajpoot
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if_ok_button
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