2 Questions about Cantor's Diagonal Argument

[Mates]

This is why I ignored this. My understanding is that the rule applies to the construction of the enumeration S1, S2, S3, .... Not to the creation of the "new number" S from what is an arbitrary enumeration. You need two rules for that. In this example they are:

• If Sn(n) is not 3, then S(n)=3.
• If Sn(n)=3, then S(n)=????? . You don't specify.

Is Sn(n) the nth number listed, and S(n) the number that cannot be listed?

If the above is correct, then I want Sn(n) to not equal 3, and S(n) to always equal 3 (with both n's being equal).

The problem is that you don't get for control the Sn. Some digits that you apply a diagonalizing rule to will be 3s - for example, if there is an Sn that already is 1/3.

But this still doesn't make sense. So maybe of you explain what "the rule is not 3" means? What does it apply to?

I want to control what S(n) equals. I want it to be 0.33333 ....

 

[Mates]

The problem is that I have no clue what your argument is. It seems to wander all over the place.

In my mind it is very clear, so I must be conveying it poorly.

To the extent that your argument is that by some more specific rule, you can create 1/3 through diaganolization of the list of all rational numbers, it is necessarily true that diagonalization creates a real number that is not in the given list and that 1/3 is in the given list so there is no specific rule that can create 1/3.

Then why don't you have the same issue with Cantor's argument? Assume Cantor's argument generates the number x. Why wouldn't you say that x should have been in the list all along since it was a given that the list were a list of all the reals. It seems like my argument is not doing anything different in this particular issue that you have with my argument.

If your argument is that you can create some unknown rational number by diagonalizing the list of all rational numbers, that argument fails for the exact same reason. The only numbers you can construct by diagonalizing the rational numbers are irrational numbers. That is what distinguishes the implication of applying diaganolization to the rational and to the real numbers. Diaganolization creates a real number that is not in the list. That is not a problem if the list is of rational numbers. Nor is it a problem if the list contains only some real numbers but not all.

You are saying that it can't generate a rational number, and I know it can't, but I still don't see a reason why it can't be done.

It creates a huge problem for any claim that the list contains all real numbers.

Isn't that what Cantor's argument does? It starts off assuming that n can list all of the reals? And then of course it fails.

 

[Dr.Peterson]

@Mates:
I don't like to jump into a discussion as long as this, since I can't read everything and be sure exactly what I am answering. But since you seem to be at an impasse, I'd like to see if I can help. In order to do that, I'd ask you to state, concisely, what your (clearly false) proof of the uncountability of the rational numbers would be, so we can talk directly about it and point out what is wrong. (In particular, there are many variations of Cantor's proof, so we need to see exactly how you want to express yours.)

In addition, I am reminded of the following post on my site, where we look at old questions and answers from Ask Dr. Math:

Frequently Questioned Answers: Uncountable Infinities​

The last section is titled "Doesn’t the same proof apply to rationals?", and should be at least partly relevant to your question. This is preceded by a couple different presentations of Cantor's proof.

 

[pka]

@Mates, I too find this to a most confusing & frustrating thread.
I spent most of this afternoon looking for a reference that I know that I have and I found it.
It is Foundations for Advanced Mathematics by Carol Avelsgaard. Prof. Avelsgaard is one of clearest writers of text material I have ever encountered. I found the text available for @&20 U.S. But if you happen to have use of a good mathematics library, were I you that is where I would look first. In Carol's text it is to be found in chapter four.
On the other hand, for a deep look into this question use Set Theory by Felix Hausdorff.

 

[Steven G]

@Mates, I too find this to a most confusing & frustrating thread.
I spent most of this afternoon looking for a reference that I know that I have and I found it.
It is Foundations for Advanced Mathematics by Carol Avelsgaard. Prof. Avelsgaard is one of clearest writers of text material I have ever encountered. I found the text available for @&20 U.S. But if you happen to have use of a good mathematics library, were I you that is where I would look first. In Carol's text it is to be found in chapter four.
On the other hand, for a deep look into this question use Set Theory by Felix Hausdorff.

Prof. Avelsgaard books are clear than I.N. Herstein? I need to read some of her books. Thanks for the info.

 

[Mates]

@Mates:
I don't like to jump into a discussion as long as this, since I can't read everything and be sure exactly what I am answering. But since you seem to be at an impasse, I'd like to see if I can help. In order to do that, I'd ask you to state, concisely, what your (clearly false) proof of the uncountability of the rational numbers would be, so we can talk directly about it and point out what is wrong. (In particular, there are many variations of Cantor's proof, so we need to see exactly how you want to express yours.)

In addition, I am reminded of the following post on my site, where we look at old questions and answers from Ask Dr. Math:

Frequently Questioned Answers: Uncountable Infinities​

The last section is titled "Doesn’t the same proof apply to rationals?", and should be at least partly relevant to your question. This is preceded by a couple different presentations of Cantor's proof.

Thanks, I see the responses to my exact question. It seems to say that we cannot get a rational because we can't prove that the number that is unlistable is rational or not.

But then I have to ask, why can't we just have rational numbers be in the argument? And wouldn't it even be more parallel to Cantor's argument to use the naturals to only list rationals?

I would think that should be something that has been done many times before, to "double check" if the set of naturals really are the same size as the rationals.

 

[Dr.Peterson]

Thanks, I see the responses to my exact question. It seems to say that we cannot get a rational because we can't prove that the number that is unlistable is rational or not.

But then I have to ask, why can't we just have rational numbers be in the argument? And wouldn't it even be more parallel to Cantor's argument to use the naturals to only list rationals?

I would think that should be something that has been done many times before, to "double check" if the set of naturals really are the same size as the rationals.

Please try writing out the proof you imagine, as I asked. Often you don't see what's wrong with your thinking until you write it out and see it all in front of you.

To put it another way, whenever a student asks, "why can't you do this?", I tell them "Try it and see!" Sometimes you find that you actually can; other times you see for yourself why not, and you don't have to ask the question again.

 

[Mates]

Please try writing out the proof you imagine, as I asked. Often you don't see what's wrong with your thinking until you write it out and see it all in front of you.

To put it another way, whenever a student asks, "why can't you do this?", I tell them "Try it and see!" Sometimes you find that you actually can; other times you see for yourself why not, and you don't have to ask the question again.

Thank you, that is great advice.

In this diagram, I would only allow each row to be rational if that is allowed . The argument assumes that all rational numbers (between 0 to 1) can be listed using the set of natural numbers. Looking at it, I can imagine each rational number generating some rational number that would not be on the list, just like the original argument does with the reals.

What would be wrong with this?

 

[Dr.Peterson]

In this diagram, I would only allow each row to be rational if that is allowed . The argument assumes that all rational numbers (between 0 to 1) can be listed using the set of natural numbers. Looking at it, I can imagine each rational number generating some rational number that would not be on the list, just like the original argument does with the reals.

What would be wrong with this?

View attachment 35300

Imagining is not enough. How do you plan to prove that the number you get is rational? Or, how do you plan to select digits that will force it to be rational?

I'd say, to the contrary, that what others have said proves that it must be irrational.

 

[JeffJo]

Is Sn(n) the nth number listed, and S(n) the number that cannot be listed?

Sn is the nth decimal representation in your list.

Sn(n) is the nth character of Sn.

If the above is correct, then I want Sn(n) to not equal 3, and S(n) to always equal 3 (with both n's being equal).

You can't. From Cantor (he used different symbols):

• "If E1, E2, …, Ev, … is any simply infinite series of elements...

From Wikipedia:

• "If s1, s2, ... , sn, ... is any enumeration of elements ...

The point is that the conclusion applies to any listing of (in your case) rational numbers that exists. You can't control it. And even if you could choose the order, your rule can't allow both 2/15=0.133333... and 7/30=0.23333... in such a list.

I want to control what S(n) equals. I want it to be 0.33333 ....

You can't.

 

[NotJeffM]

There is a PROOF that all the rationals are listable. There is no proof that all the reals are listable. That is the first crucial difference.

What is common to any list of decimal expansions of any type of real number is that diagonalization of the list creates a REAL number that is not in the list. There is no guarantee that the number created will be rational.

So if we diagonalize the list of all rational numbers, we get a real number that is not listed. You say you imagine that it might be rational. PROVE THAT. (You can‘t of course because the list contains all the rational numbers. But the number you created is only guaranteed to be real. And there are real numbers that are not rational, namely the irrationals. No contradiction entailed.)

Unlike the rational numbers, we have no assurance that a list of all reals in (0, 1) is feasible in principle. But if such a list is feasible, diagonalize it. You get a real number that is in the same interval but not in the list, which contradicts the supposition that all reals in that interval are in the list.. So, our supposition was wrong.

If you ignore the necessary details, for example the difference between unique representations of numbers and numbers themselves, the difference you have been asking about comes down to this obvious point. It is possible to have a real number that is not a rational number, but it is impossible to have a real number that is not a real number. That is the second crucial difference.

 

[pka]

After fifty replies does this thread remined anyone else of the saying:
"It is useless to beat a dead horse" ?

 

[NotJeffM]

After fifty replies does this thread remined anyone else of the saying:
"It is useless to beat a dead horse" ?

I do

 

[lookagain]

After fifty replies does this thread remined anyone else of the saying:
"It is useless to beat a dead horse" ?

Of course. And, I was responsible for asking that the other thread by this same OP get closed because it got repetitive and not progressing. Super
moderator stapel closed that other thread.

 

[Mates]

Imagining is not enough. How do you plan to prove that the number you get is rational? Or, how do you plan to select digits that will force it to be rational?

Ok, I think the concept is settling in now. Thanks for your help (and not trying to make me look like an idiot. I appreciate that.)