Why Cantor's Diagonalization Proof is Flawed.

[Divine Insight]

Here is a Youtube video, under 10 minutes, of Cantor's diagonalization argument.

Ok, I've seen this proof countless times.

And like I say it's logically flawed because it requires the a completed list of numerals must be square, which they can' t be.

~~~~

First off you need to understand the numerals are NOT numbers. They are symbols that represent numbers. Numbers are actually ideas of quantity that represent how many individual things are in a collection.

So we aren't working with numbers here at all. We are working with numeral representations of numbers.

So look at the properties of our numeral representations of number:

Well, to begin with we have the numeral system based on ten.

This includes the ten digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.

How many different numbers can we list using a column that is a single digit wide?

Well, we can only list ten different numbers.

0
1
2
3
4
5
6
7
8
9

Notice that this is a completed list of all possible numbers. Notice also that this list is not square. This list is extremely rectangular. It is far taller than it is wide.

Let, apply Cantor's diagonal method to our complete list of numbers that are represented by only one numeral wide.

Let cross off the first number on our list which is zero and replace it with any arbitrary number from 1-9 (i.e. any number that is not zero)


[strike]0[/strike]
1
2
3
4
5
6
7
8
9

Ok we struck out zero and we'll arbitrary choose the numeral 7 to replace it.

Was the numeral 7 already on our previous list? Sure it was. We weren't able to get to it using a diagonal line because the list is far taller than it is wide.

Now you might say, "But who cares? We're going to take this out to infinity!"

But that doesn't help at all.

Why not?

Well what happens when we make the next step? We need to make the list 2 digits wide now.

What happens?

Here is a 2-digit list of all possible numbers represented by 2 numerals.

00
01
02
03
04
05
06
07
08
09
11
12
13
14
15
.
.
.

95
96
97
98
99

What happened? Well, our completed list of possible numerals that is two digits wide has incrusted in vertical height exponentially. This list is now 100 rows tall and only 2 column wide.

Now let's cross off the first two digits of our list and replace them with arbitrary numerals.

[strike]0[/strike]0
0[strike]1[/strike]

Ok, for the first zero being stuck off the list, I'll chose to arbitrarily replace that with a 5. For the second digit being struck off the list I'll replace that arbitrarily with a 7.

My new number is 57.

Is 57 already on my completed list? Yes. It's just further down the list where I couldn't possibly reach it by drawing a diagonal line.

Now you might say, "But who cares? We're going to take this out to infinity!"

But duh? We can already see that in a finite situation we are far behind where we need to be, and with every digit we cross off we get exponentially further behind the list.

Taking this process out to infinity would be a total disaster.

You could never claim to have "completed" this process because you can't move down the list fast enough using a diagonal line that crosses off each digit diagonally.

The very nature of our system of numerical representation forbids this. You can't complete this process in a finite situation, and it gets exponentially worse with every digit you add to the width, then you could never claim to have completed this process by claiming to have taken it out to infinity.

"Completed Lists" of numerical representations of numbers are NOT SQUARE.

Yet Cantor claims to be creating a "Completed List" here. It's a bogus proof that fails. Cantor didn't stop to realize that our numerical representations of number do not loan themselves to nice neat square competed lists. And that was the flaw in his logic.

By the way you can't even do this using binary representations of numbers.

In Binary Representation

A completed list of binary numbers 2 digits wide:

00
01
10
11

It's not square. It's twice as tall as it is wide.

Add another digit it gets worse:

000
001
010
011
100
101
110
111

There is no way that a completed list of numbers can be represented numerically in square lists.

Yet Cantor's diagonal argument demands that the list must be square. And he demands that he has created a COMPLETED list.

That's impossible.

Cantor's denationalization proof is bogus.

It should be removed from all math text books and tossed out as being totally logically flawed.

It's a false proof.

Cantor was totally ignorant of how numerical representations of numbers work. He cannot assume that a completed numerical list can be square. Yet his diagonalization proof totally depends on this to be the case.

Otherwise, how can he claim to have a completed list?

If he's standing there holding a SQUARE list of numerals how can he claim that he has a completed list?

Yet at what point does his list ever deviate from being square?

It never deviates from being square. It can't because he's using a diagonal line to create it. That forces his list to always be square.

Georg Cantor was an idiot.

He didn't even understand how numerical representations of numbers work.

His so-called "proof" doesn't prove anything. It's totally bogus.

He can't claim to have a "completed list" by the way he is generating his list. Claiming to take this out to infinity doesn't help. With every new digit he creates he falls exponentially behind where he would need to be to create a "Completed List".

Yet that's what he claims to have: A Completed List.

It's a bogus proof, and I'm shocked that no mathematicians have yet recognize this extremely obvious truth.

They keep publishing this proof and teaching it like as is it has merit when in fact it's totally bogus.

[Divine Insight]

You can't stick these numbers just anywhere, or you will lose your one to one correspondence. These numbers will be on the list, but they will be an infinite distance down the list.

You'll need to tell that to Georg Cantor. The real number decimal expansions he uses in his so-called "proof" are totally random without any order at all.

Therefore, according to you, he cannot be claiming to be making any sort of one-to-one correspondence anyway.

[olavisjo]

.

Therefore, according to you, he cannot be claiming to be making any sort of one-to-one correspondence anyway.

That is a good point, the thing that he set out to disprove (there is no possible one to one correspondence between the natural and real numbers) is the very thing that he ignored in his proof.
His proof relies on our ability to manipulate numbers at the infinith place, and when you do that you must be very careful or you will create contradictions and paradoxes, as he so unfortunately did.
We have to also remember that Cantor was not computer literate, if he had been he would have easily seen the folly of his theorem.

The only questions I now have is, why did it take us a hundred years to figure this out and where do we go with this 'new' knowledge?

[Divine Insight]

The only questions I now have is, why did it take us a hundred years to figure this out and where do we go with this 'new' knowledge?

Unfortunately the mathematical community hasn't figured it out yet.

Where we can go with this new knowledge is truly fascinating. But we can't even begin that journey until the mathematical community wakes up to this mistake and backtracks far enough to repair it. The longer they go before they repair it the further they are going to need to backtrack when they finally do realize the folly.

It's not just Cantor's proof. It's the whole treatment of irrational relationship like as if they can be treated as cardinal quantities. So they've got their work cut out for them. It could actually prove to be a very exciting time in mathematics when they finally realize that they had made a very wrong turn.

At first they will be devastated, but they will soon realize that making the right turn is actually far more interesting and productive.

[olavisjo]

.

Unfortunately the mathematical community hasn't figured it out yet.

I asked a mathematician to look at our proof and he said that I don't understand Cantor's diagonal proof.


1) i_1,2 i_1,3 i_1,4 i_1,5 ... i_1,n

[row]1 )[col] i[sub]1,1[/sub] [col] i[sub]1,2[/sub] [col] i[sub]1,3[/sub] [col] i[sub]1,4[/sub] [col] i[sub]1,5[/sub] [col] ... [col] i[sub]1,n[/sub]

[Friedrich]

I have to say that you are very patient. Very few can go this many rounds with me without pulling out their hair.

Edgar Nackenson said “but you will never have a mapping from the naturals to the reals."

We don't need to find a mapping from the naturals to the reals. In my first post I said "Cantor could have used the Schröder–Bernstein theorem. Which says, for two sets to be equipotent, you need only show that there exists injective functions f : A → B and g : B → A  then |A| = |B|." An injection from the naturals to the irrationals is trivial, so we need only find an injection from the irrationals to the naturals and we are done. Do you agree with this?

Cantor mapped his argument with an arbitrary function. There was no reason why any natural was mapped to any real. This is a problem and a straw man doomed to failure. Let me demonstrate.

If we map the irrationals to the naturals with an arbitrary function, we get something like this.

0.31415927... -> 1
0.32415927... -> 2
0.31515927... -> 3
0.51435927... -> 4
0.51415927... -> 5
0.61415127... -> 6
0.61415928... -> 7
etc...

We have mapped an irrational number to every natural number, and we still have irrational numbers that are not on the list but we have nothing to map those irrationals to. We have too many irrationals, so no injection.

To reduce the quantity of irrationals on our list we could remove all the irrationals that begin 0.5 but we get the same result.

0.31415927... -> 1
0.32415927... -> 2
0.31515927... -> 3
-strike_0.51435927... -> 4_strike-
-strike_0.51415927... -> 5_strike-
0.61415127... -> 4
0.61415928... -> 5
0.62415928... -> 6
etc...

We have mapped an irrational number to every natural number, and we still have irrational numbers that are not on the list (the diagonal numbers and the numbers that begin with 0.5) and we still have nothing to map those irrationals to. We have too many irrationals.

We will try the same thing with our mirror function. So our mapping would look like this.

0.31415927... -> 3
0.32415927... -> 23
0.31515927... -> 13
0.51435927... -> 5
0.51415927... -> 15
0.61415127... -> 6
0.61415928... -> 16
etc...

I will repeat something in a previous post, because I don't think you understood me. Cantor's assumption that all the reals were on his list is only useful if all the naturals are on the list, then finding a real not on the list produces a contradiction. But if all the naturals are not on the list then there is no useful contradiction, since we also assumed that all the naturals were on the list. Think about mapping the real numbers to the power set of the naturals, and concluding that the reals are larger because you can find a real number not in the list. Finding a number not on the list is useful only if the other list is complete. Which is not the case with the natural numbers, when mapped with our mirror function.

Now we will remove all the irrationals that begin with 0.5, like this.

0.31415927... -> 3
0.32415927... -> 23
0.31515927... -> 13
-strike_0.51435927... -> 5_strike-
-strike_0.51415927... -> 15_strike-
0.61415127... -> 6
0.61415928... -> 16
etc...

Now our mapping will not include any natural numbers with a 5 in the units position. This is important because it tells us that if something is missing in our list of irrational numbers then there will be something missing in our list of natural numbers as well. This would not be true with Cantor's arbitrary function, but it is true if the naturals are dependent on the list of irrationals, as is the case with our mirror function.

Even if we diagonalize our list like this.

0.31415927... -> 3
0.32415927... -> 23
0.31715927... -> 13
0.61415127... -> 6
0.61515928... -> 16
0.61315928... -> 316
0.613259287... -> 2316
etc...

We get d = 0.3271287... change the digits d' = 0.7737773...

As you can see 7 is not on our list, so we can map d' -> 7. If 7 had been on the list we could map d' -> 77 or d' -> 377 or d' -> 7377 etc. If we can't find a natural number to map to our diagonal, then d' would be a number already in the list of irrational numbers, so it would already be mapped.

Your objection was that as soon as you say we have a mapping, we can find an irrational that is not on the list. And that is true, but we can also at that exact same time find a natural not on the list to pair with it.

I hope that you can see, that we can find a partner to every irrational number on the list as well as all the irrational numbers that are not on the list. Therefore, Cantor's argument is moot.

[Friedrich]

They keep publishing this proof and teaching it like as is it has merit when in fact it's totally bogus.

It is not totally bogus, it just depends on the axioms that you use. The problem is that there is no infinity in the real world, so you have a counterfactual like this, if false (infinity exists) then true (anything). In other words falsity implies anything.

At the risk of bumping a very old thread, I will show you a very simple construction to form a one to one correspondence between the natural numbers and the real numbers. It is the best anti Cantor argument that I have seen. If anyone has seen a better one, I would love to see it. And that goes either way, pro or con.

First we list rows 1 and 2,

row 1 -> 0.0
row 2 -> 0.1 this is all binary numbers between 0 and 1 with one digit past the point.

Then we form rows 1 to 4

row 3 -> 0.1 copy row 2 to make row 3
row 1 -> 0.0
row 2 -> 0.1
row 4 -> 0.0 copy row 1 to make row 4

Then append a 0 to the top half of the list and a 1 to the bottom two rows.

row 3 -> 0.10
row 1 -> 0.00
row 2 -> 0.11
row 4 -> 0.01 this is all binary numbers between 0 and 1 with two digits past the point.

Now we form rows 1 to 8

row 7 -> 0.11 rows 7 and 5 come from rows 2 and 4
row 5 -> 0.01
row 3 -> 0.10
row 1 -> 0.00
row 2 -> 0.11
row 4 -> 0.01
row 6 -> 0.10 rows 6 and 8 come from rows 3 and 1
row 8 -> 0.00

Then we append 0s to the top 4 rows and 1s to the bottom 4 rows.

row 7 -> 0.110
row 5 -> 0.010
row 3 -> 0.100
row 1 -> 0.000
row 2 -> 0.111
row 4 -> 0.011
row 6 -> 0.101
row 8 -> 0.001 this is all binary numbers between 0 and 1 with three digits past the point.

Next we form rows 1 to 16

row 15 -> 0.111 rows 15 to 9 come from rows 2 to 8
row 13 -> 0.011
row 11 -> 0.101
row 9 -> 0.001
row 7 -> 0.110
row 5 -> 0.010
row 3 -> 0.100
row 1 -> 0.000
row 2 -> 0.111
row 4 -> 0.011
row 6 -> 0.101
row 8 -> 0.001
row 10 -> 0.110 rows from 10 to 16 come from rows 7 to 1
row 12 -> 0.010
row 14 -> 0.100
row 16 -> 0.000

Then we append 0s to the top 8 rows and 1s to the bottom 8 rows.

row 15 -> 0.1110
row 13 -> 0.0110
row 11 -> 0.1010
row 9 -> 0.0010
row 7 -> 0.1100
row 5 -> 0.0100
row 3 -> 0.1000
row 1 -> 0.0000
row 2 -> 0.1111
row 4 -> 0.0111
row 6 -> 0.1011
row 8 -> 0.0011
row 10 -> 0.1101
row 12 -> 0.0101
row 14 -> 0.1001
row 16 -> 0.0001 this is all binary numbers between 0 and 1 with four digits past the point.

As you can see we now have a list of all possible numbers made of 4 binary digits to the right of our binary point. And if we repeat this procedure an infinite number of times, we can see that row two will have an infinite number of 1s. And all the rows below it will have an infinite number of significant digits. This will include all the irrational numbers and the rational numbers that repeat. The top of the list will have all the rational numbers that terminate with one or more repeating 0s.

This is a list of all the real numbers. It is a two ended list for construction purposes, but can easily made one ended.

We can try to diagonalize our list.

If we take the first digit in row 1 and the second digit in row 2, we get 0.01 we can flip the digits and get 0.10 and we know this number can't be in rows 1 or 2, but the first 2 digits in row 3 is 0.10

If we diagonalize 3 rows, we get 0.101 which is the first 3 digits of row 6.
If we diagonalize 4 rows, we get 0.1010 which is the first 4 digits of row 11.
If we diagonalize 5 rows, we get 0.10101 which is the first 5 digits of row 22.
If we diagonalize 6 rows, we get 0.101010 which is the first 6 digits of row 27.

Or generally if our row is odd then our special number will be in row 2^((n+4)/2)-5 if even then row 2^((n+5)/2)-10

Since there is no row infinity we can only diagonalize rows 1 to n, where n is some natural number, there will always exist some row where we can find the anti diagonal number. And this is true for the reason that Divine Insight gave regarding the list not being "square".

Like I said there exists no infinity, except in our imaginations, so we can imagine it to be whatever we want it to be.

[micatala]

.

The above is flawed in that it includes the incorrect assumption that all numbers are listable.

All numbers are listable.

Here is a list of all real numbers R; where 0.0 < R <= 1.0
And they are in a one-to-one correspondence with the natural numbers.

1 - 0.1
2 - 0.2
3 - 0.3
...
9 - 0.9
10 - 0.01
11 - 0.11
12 - 0.21
...
99 - 0.99
100 - 0.001
101 - 0.101
102 - 0.201
...


Similarly, the entire set of real numbers can be placed in a one-to-one correspondence with the natural numbers.

http://debatingchristianity.com/forum/v ... 630#607630

Your list has no irrational numbers in it. Every single one is a finite decimal number and therefore rational. It is true the numbers on the list have more and more digits as you go, but they are all rational. Thus, the square root of 2 divided by 10, for example, is not on the list.

[micatala]

[Replying to post 73 by Friedrich]

This argument suffers from essentially the same problem as olavisjo's. All the numbers in your list of binary numbers have finitely many digits and so are rational. The list does not contain any irrational numbers.

[micatala]

Here is a Youtube video, under 10 minutes, of Cantor's diagonalization argument.

Ok, I've seen this proof countless times.

And like I say it's logically flawed because it requires the a completed list of numerals must be square, which they can' t be.

[Friedrich]

[Replying to post 75 by micatala]

Look at row number 2 and tell me how many 1s it has. If it is finite you should be able to give me an exact number.

Every time we repeat the procedure we will get one more 1 in row 2. If we repeat an infinite number of times then we will have an infinite number of 1s. And every row has the same number so they will all have an infinite number of 0s and or 1s.

Consider this construct

row 1 -> 0.1
then the next iteration

row 1 -> 0.11
row 2 -> 0.1

then the next iteration

row 1 -> 0.111
row 2 -> 0.11
row 3 -> 0.1
then the next iteration

row 1 -> 0.1111
row 2 -> 0.111
row 3 -> 0.11
row 4 -> 0.1

And repeat an infinite number of times. Every row will end up with an infinite number of 1s, even though the first row seems to have more than all the others.

In the same way the construct in my first post has an infinite number of digits in each row.