My OperaCantor's Theory Visualized (Updated Nov 2010)
Sunday, July 5, 2009 12:34:04 AM
Cantor's theory is one of the most misunderstood theories of all times. It's what happens when one does not understand what they are doing. By that, I mean that Cantor thought he was comparing real numbers and natural numbers when he was doing no such thing. Instead, he was comparing two different bases. Only problem is that he did not realize that enumerations also have a base. For this reason, the process of creating a diagonal is flawed.
One thing to remember when you create a diagonal is that both axis must be the same. If they are not, then you are skewing the results. Cantor completely disregards any notion or attempt to keep both axis the same. The x axis is base 2 (or another base). However, what base is the y axis? Most people don't even bother to think of this. In fact, they reject it because they do not want to think about it.
One thing to remember when you create a diagonal is that both axis must be the same. If they are not, then you are skewing the results. Cantor completely disregards any notion or attempt to keep both axis the same. The x axis is base 2 (or another base). However, what base is the y axis? Most people don't even bother to think of this. In fact, they reject it because they do not want to think about it.
What base does an enumeration have? I'm not talking about the representation of each item. I'm talking about the other axis, vertically. It is base 1. It's base 1 because we don't care what the row is. We only care that we can count them. In other words, we could replace each row with circles and we could still count them. As well, we will replace the binary representation with coins (Heads or Tails).
Note how the left side and the right side are exactly the same except that they use different representations. Hopefully, this is enough to show that we are indeed comparing bases. I can't stress enough the importance of this paragraph. BTW, I've just disproved Cantor's theory with the above image. However, more explanation may be in order.
Some people may still be confused about base 1. But it is used all the time. It simply means using ONE and only ONE symbol. In westerns, notches on a gun is a form of base 1. I'm sure you've all seen the following as well.
Base 1 is found everywhere.
Here is where Cantor went wrong. He tried comparing base 1 with base 2. In order to do this, he left base 2 horizontally. But he now needed a way to represent base 1 vertically. Well, that's not so hard to do. Using the first image as a reference, we no longer need to display circles horizontally. We can instead count them vertically. Wherever we stop, that is how many circles/rows we have at that point.
Here, we count vertically. Note the groupings. We can continue as far as we'd like. What I really want the reader to see is the last circle in any grouping. Note that the amount of circles upwards vertically or leftwards horizontally from that last circle are EXACTLY the same! This means we only need to keep ONE representation, the vertical one.
Since we don't need all the circles on the left, we remove them. Simply pick any row and count the circles all the way to the top.
Now Cantor wants to match the number of digits with the number of rows (or circles). In our example, we have 10 rows (or rather circles), so we also make 10 digits to have a 10 by 10 square.
What Cantor does at this point is remove the circles and then hopes no one sees the sleigh of hands. We don't need the circles anymore since we can use the rows themselves for counting.
(edit: For those that incorrectly believe this is a finite list, please consider that the above grid is only a portion of an infinite list of digits and rows.)
But now that we've established that the rows can be represented by circles or marks on a wall, I hope it's clear how base 1 is being compared to base 2. I also hope it's clear how ridiculous a notion it is to say that one base cannot be matched to another. For example, I can say that base 10 will always be able to produce more numbers than base 2 when using the same number of digits. But does that mean base 2 cannot be matched one to one with base 10? Of course not. And neither is it true of base 1 and base 2 in Cantor's theory. Base 1 is mapped one to one with the naturals and so is base 2 (and base 10 and every other base).
Cantor's theory is completely ridiculous and a waste of everyone's time. Programmers who deal with different bases (binary, octal, decimal and hexadecimal) should see right through Cantor's theory.
BTW, the scaling factor for Cantor's one to one mapping is nothing other than BN-N where B is the base used for the digits and N is the number of digits. Now that you have a one to one mapping, perhaps we can put Cantor's theory to rest once and for all. Even with all this said, I just know someone is dying to mention that it deals with real numbers.
UPDATE SEPT 2010
Want to post an update on my findings. I am correct that Cantor's diagonal argument is flawed. His first proof is also flawed. I can now better pinpoint the exact nature of said flaw. It is his use of incompatible generators. He never uses the entire list, or any entire list. He uses generators. Every infinite list has a way to generate more numbers as per Dedekind's theorem. The way in which you decide to generate new numbers is not related to the set, but rather to the generator you select. Cantor simply selected two different and incompatible generators.
He chose a bifurcation generator for the reals “between any two members there is another” and a sequential generator for the naturals. However, there is no need for different generators. We can use a bifurcation generator for both the reals and naturals.
Cantor was right on one thing. The use of a bifurcation generator and a sequential generator will not be one to one. But this gives us an opportunity to reduce the problem to one that uses two different and equivalent generators, but where it is known that the two sets DO map one to one. Such a reduction will prove that |N|=|R|. From there, the flaw in Cantor's proofs will become trivial.
Suppose we have a tree where we create a mapping that is not one to one. We create a 1 to n mapping. Each row will double the amount of nodes of the previous row. As such, since the first row is finite and doubling still results in a finite amount of nodes, all rows will have a finite amount of nodes. To state this more blatantly, there is no single row that has an infinite amount of nodes even if we have infinite rows. This is the same principle behind the natural numbers themselves. Even though we go to infinity, no single natural will use infinite digits. Here, we apply this principle to the rows. Even with infinite rows, we will have finite nodes on each one.
Here is a picture of such a tree. I only show the first few rows, but it continues on.
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Now we consider another such binary tree, but where each left child node is a 0 and each right child node is a 1.
The above tree can represent all real numbers via the infinite paths starting at the root node on downwards. BTW, note that each row represents a particular digit position. So we indeed have a 1 to n mapping here. Each row/digit maps to n paths/reals. This will become clearer later on. But this is in effect the flaw in Cantor's diagonal argument. It is not a one to one mapping as he had incorrectly assumed. He is using a SPECIFIC and ARBITRARY mapping. We continue in order to show that it is indeed arbitrary.
We now combine both trees together.
Now we see how it's possible for the naturals to map both to the rows and to all the nodes in the tree. The fact that both mappings are possible is fatal to Cantor's diagonal argument.
One mapping is that the naturals map one to one with the digits of the reals. The digits are represented by the rows. The rows are obviously countable since the progress sequentially. We can label them row 1, row 2, row 3, etc. very easily. But we can also map the naturals one to one with all the nodes of the tree. The first (and third) tree image shows this quite clearly. Since the naturals map to all the digits, then the naturals must also cross all the paths of every real. Therefore, |N| = |R|. Note that we did not map the naturals to the reals themselves, but rather their digits.
One objection I've seen to this is that it takes so many more sequential nodes (those mapped to the naturals) to reach any given digit down a path. To reach the third row, you'd be at the fourth sequential node. This is a difference of 1. But the further down you go, the larger this difference becomes until it is infinitely long. So the sequential nodes will never be able to reach those paths with infinitely many digits.
I'm paraphrasing as best I can. Some also use the notion of omega. I will call this and the above paragraph the omega theory.
This is a complete misunderstanding of infinity. There is no single row that has infinite digits. None! But the sequential nodes will indeed reach ALL the digits of ALL the reals. Infinity when using a sequential generator is not a speed contest. It's not who can reach infinity with the fewest amount of elements. The only requirement is that you reach it. Going down the rows is a sequential generator and so is going through all the nodes sequentially. Each new step is one more than the last. This is exactly like the Peano axioms and by definition, they will map one to one with each other.
To prove this, consider the situation where you would have only two nodes per row. It is no longer a tree, but that is irrelevant. The new structure is equivalent to mapping the odd numbers (first node of each row) to the naturals/rows. It also maps the even numbers (second node of each row) to the naturals/rows. However, we know that we can map all the nodes to all the rows one to one since all the nodes is simply the odd and even numbers together, aka the naturals. By definition the nodes must map one to one with the rows. But in its current arrangement within the structure, the mapping is 1 to 2. There are TWO nodes per row. That is a two to one ratio.
This means that any structure that has finite elements per row will indeed allow the mapping of sequential nodes one to one with the rows. The above tree qualifies.
The only way that a mapping cannot occur is if it can be proven that any given row has infinite nodes. If someone can prove this, they will also have proven that a given natural number has infinite digits. They are equivalent scenarios.
The explanation is quite simple. There is no last row. But there are infinitely many rows at the end of any infinite sequential list. So that is what is going on. No single row has infinitely many nodes. But infinitely many rows at the end of the tree has infinitely many nodes. The paradox disappears.
So we can safely conclude that |N| = |R|.
Where is the flaw in Cantor's diagonal?
It's the omega theory, albeit slightly covered up. To keep things simple, consider that each digit on the x-axis of Cantor's grid is equivalent to each row of our tree. And each row of Cantor's grid is equivalent to each path down the tree. In effect, Cantor is trying to flip vertically a single last row of the tree that has infinite nodes to be the rows in his grid. IOW, the nodes of the last row of the tree are the rows of Cantor's grid. But as I've just explained, there is no last row. There are infinitely many last rows. This is the omega flaw that also assumes that there is a single last row with infinitely many nodes.
The mapping in Cantor's grid is not one to one. Each digit maps to many rows exactly as each row/digit of the tree maps to many nodes/paths/reals.
The flaw is exactly as I've been saying all along. Cantor uses a very SPECIFIC and ARBITRARY mapping that is known to not be one to one.
As to Cantor's first proof, he simply uses two different generators as said before. The reals use a generator that produces new number between any two members. But the naturals use a sequential generator at the end. The location of the newly generated numbers is different for each generator and this is the contradiction shown in Cantor's first proof. But it says nothing about the sets themselves. If one uses the same generator, then both reals and naturals would be created the same way. The tree structure can be used to generate the naturals. The only difference is that the binary string not be converted to decimal. Instead, the frontmost digits up to the last 1 is what differentiates an element from all others. As such, the expansion of the list would clone each member and add a new digit which would alternate between 0 and 1. New numbers would appear between the existing ones. And they'd all be naturals. I can repeat this infinitely and I'd still only get finite digits for each member.
Some people incorrectly state that they would have infinitely many digits. Not so. This would be true if there was such a thing as one specific item at a single index of infinity. But infinity is not finite. It is simply not plausible to have a single point that is infinite. Again the omega theory rears its ugly head. The expansion continues forever.
I have another scenario, possibly a proof by contradiction about the omega theory. Suppose person A lays one tile at a time (one in front of the other). And immediately next to the tiles of person A, person B lays two tiles for each tile that person A lays down (again in a straight forward line). They both use identical tiles. If they both continue laying tiles infinitely, will there be any place where there is not a side by side pair of tiles?
The answer is simple. All tiles will be paired up. It is nothing more than an equivalent scenario to mapping odd numbers to the naturals. The speed at which you advance is not the issue. If both are sequential infinite lists, the elements will pair up one to one.
Now consider a slightly different scenario. Person A still lays tiles one at a time. But now Person B will double the amount of tiles laid every time. I ask the same question as before. Will there be any tiles not paired up?
Here is where the omega theory really falls apart. If you accept that Person B can only ever lay a finite amount of tiles at any given step, then all tiles will be paired up. OTOH, if you believe that Person B will eventually lay an infinite amount of tiles at a particular step, then this implies that at step X, you've reached infinity. That's a contradiction because the previous step, Person B laid a finite amount of tiles. Infinity/2 is not a finite number. And a finite number * 2 is not infinite.
Why is this important? Because it proves that all the nodes in the infinite binary tree can indeed map one to one with the rows of the tree. As such, it also proves that |N| = |R| and explains where the flaw is in Cantor's two proofs.
To those that are still unclear as to the flaw in Cantor's proofs, he doesn't use the list at all. He doesn't even look at the reals. He compares two different generators and proves that generated members appear in different places in the sets. But he does not prove that reals and naturals cannot use the same generator. Again, his proof is about generators, not sets. And certainly not about what generators can be used to produce what sets.
If you ask me for a list, you must use my generator. My generator is bound to the digits of the x-axis. Every digit you use generates more numbers in my list. Note the plural use of numbers. Hence the race and the flawed omega theory.
UPDATE NOV 2010
Here's a counterproof to Cantor's diagonal. I've always said that Cantor creates his own mapping. We'll see that right now.
We're supposed to be able to list all the reals, right? And Cantor is supposed to say that no matter what, he can create a new one in the list. Well, I've argued that he only uses a subset of any list you give him. Here's how.
Suppose you want to create the diagonal. It's fine and say that you're just going to take the x'th digit. But what does that really mean? I propose the creation of a virtual computer that can handle infinite length numbers. So now we can construct the new diagonal number based on arithmetic operations.
If we want the first digit of a real number, we can use AND and then XOR.
N1 = (X1 AND 0.1000...) XOR (0.1000...)
We can continue doing this for all digits.
N2 = (X2 AND 0.0100...) XOR (0.0100...)
N3 = (X3 AND 0.0010...) XOR (0.0010...)
N4 = (X4 AND 0.00010...) XOR (0.00010...)
At each step, we wouldn't save the N's. We would merge them together with an OR operation. N = N1 OR N2. Then N = N OR N2. N = N OR N3 and so on.
If you had infinite memory, you could certainly do everything in three steps. Do AND to extract a single digit from each real. Then XOR to flip each of those extracted bits. And finally OR to merge all those bits into the new real.
Here's the deal. Cantor already creates his own mapping. The numbers he uses for the AND and XOR operations are predefined. They are a subset of the rationals. No other numbers can be used. Hence, the mapping is already there. Any list we hand over will simply be ignored. Our list of X's which contain our reals don't even come into play. Note what happens if you use the same number for X's as the number it is being AND'd and XOR'd against (the 0.10, 0.010, 0.0010, etc.), call this B's. So X = B. The end result of the operations is 0, a number not part of the B's. And that's how Cantor creates his new number. Whatever is mapped to 0 will never be in the list. Heck, anything that isn't found in the list of B's will not be in the list either.
Sure, you can use different numbers for your list of reals, but all you're doing is swapping the entries in X from an earlier list when X = B. Nothing actually changes except for the representation. But no row is added or removed. You can only map to B's. There are no other entries. And we know that B is a subset of the rationals. So Cantor can only work with |B| rows. Not |N| and not |R|.
The digit selection is the flaw in his proof. It creates a unique and specific mapping as I've been saying all along.
QED.
edit: I thought I should add some additional explanation on possible arguments against this. Some might say that |B| = |N| because there are clearly |N| digits. So there has to be as many rows as there are digits. This is something else I've been saying all along where Cantor makes a mistake. Cantor use binary representation. So if you don't use binary, then certainly, you can create a one to one mapping. But if you DO use binary, then you're only mapping the numbers 1, 2, 4, 8, 16, 32, etc. All the other numbers from N are not mapped. So |N| > |B| in Cantor's grid. IOW, for equal amounts of digits in the binary representation, there are always more naturals than B's. This is critical and fatal to Cantor's diagonal argument.
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Unregistered user # Sunday, July 5, 2009 5:25:29 PM
Vorlath # Sunday, July 5, 2009 7:39:30 PM
It is an EXACT duplicate of Cantor's theory. EXACT! Look at the last image. It's identical.
He does no such thing. Go look again. If he did, then his theory would fall flat on its nose since it would be impossible to create new rows. We wouldn't even be here discussing this. It would have been laughed out of existence.
About being finite, I only duplicated what Cantor did. Cantor's grid is just as finite as mine. And Cantor's digits and rows are infinite just like mine are.
Why? As long as there is a square grid, that's good enough according to Cantor.
BTW, what you've said is MY argument and I agree with you completely. There SHOULD be 1024 rows. If there were, Cantor's theory would fall flat on its face.
This is handwaving without any proof. I've shown an equivalent situation to that of counting rows. So you cannot just wish it away. I've already shown it. You're too late to dismiss it. Doesn't matter if you find it troubling. Whatever else Cantor thinks he's doing, no one can say that he is not counting rows, especially when he's saying he can create a new one.
Re-asserting the original premise doesn't change the fact that I've shown an equivalent situation that compares bases. What Cantor thinks he's doing is completely irrelevant once I've shown a counter-example. That's the power of counter-examples. Once they're there, re-asserting the original premise is irrelevant.
Cantor does not use infinite lists. His grids are always finite. If you want to say his grids are infinite, the Cantor's theory is trivially wrong since infinities have different sizes and same sizes all at once (it's why ∞-∞ is indeterminate). Cantor would effectively be using a circular argument. I would LOVE that this is what Cantor's argument is, but it is not.
What Cantor THINKS he's doing is using finite grids within an infinite list. The infinite list of digits is the representation of reals and the finite grids are the finite numbers of the naturals. Right there, we're already into circular argument land. But it doesn't matter. I've already shown a counter-example. Not only that, but representations are arbitrary. Using an arbitrary representation to prove something is rather ridiculous in of itself.
Cantor falls for the same trap as the Halting Problem. The representation and the principal are two different beasts. What is true of the representation is not necessarily true of the principal. For example, PI is finite, but its base 2 representation is not. Now that I've shown a counter-example to PROVE that real numbers and their representations do not share the same properties, why should I believe any proof that depends on this false assumption? WHY? It just doesn't make sense to me.
Vorlath # Sunday, July 5, 2009 8:39:45 PM
Cantor is using representations to try and prove something about the elements. This is more commonly known as the fallacy of many questions where you assume a prior fact before proving it.
Is your brother still in the army?
A yes or no answer assumes you have a brother in both cases when you may not have a brother at all. With Cantor's theory, it is assumed that the representation has the same properties as the element when creating the grid. I've shown in the last comment how this is a false assumption. A number can be finite, yet have an infinite representation. This is trivially obvious and accepted by the mathematical community, but can be proven if need be.
I couldn't care less what he puts in the grid. If he doesn't prove that he's only using the properties of the elements, it's a fallacy of many questions. The fact that he's using digits of the representation to prove something about the actual elements shows that he's off his rocker and in over his head with fallacies of many questions. This also has the flavour of the fallacy of equivocation where he uses the number of digits and the number of rows to form a grid and yet both of these axis use a different base. Two different interpretations means you will get a false contradiction. The fallacy of equivocation is a well known fallacy.
Even if he were only using the properties common to both the representation and the element, I've shown how he ISN'T comparing reals to naturals at all, but is instead comparing base 1 to base 2. Cantor just SAYS he's comparing reals to naturals. Saying it doesn't make it so. That's the fallacy of begging the question.
If I tried, I could not make up a theory with more holes in it than Cantor's theory.
Cantor uses at least the following fallacies:
Fallacy of many questions.
Fallacy of begging the question.
Fallacy of equivocation.
There may be more.
Fallacy of many questions.
Assuming that proving something about digits of the representation will prove something about the principal.
Fallacy of begging the question.
Assuming that he's comparing naturals with reals.
Fallacy of equivocation.
Using different bases for the x and y axis while using the same number of elements in each axis.
Vorlath # Sunday, July 5, 2009 9:05:00 PM
Unregistered user # Sunday, July 5, 2009 11:07:49 PM
Vorlath # Sunday, July 5, 2009 11:57:13 PM
If he's using infinite grids, then he's using two different infinities to prove they are different. That's the fallacy of equivocation as well as the fallacy of begging the question and the fallacy of circular cause and consequence, etc...
Please understand that if he's truly using infinite grids, then his theory is trivially flawed. It's not even wrong. If you wish to go that way, then be my guest. But it's so trivial that trying to explain it would be somewhat absurd. I'll give it ONE go. If the digits are infinite, then he's assuming that
Infinite digits == log2(infinite digits)
Then he goes about showing how this is not true for finite cases (fallacy of composition), hoping for a contradiction. This is what's called the Texas sharpshooter fallacy. He sets up a bogus contradiction and then paints a bullseye around it. Only thing is that you cannot demonstrate properties of infinity with finite examples, so that makes it begging the question, circular reasoning and fallacy of equivocation all at once. Like I said, if Cantor is using infinities, then his argument enters complete absurdity every which way.
I did not bother much with the infinite grid case since Cantor does not use it and if he did, then it's a waste of everyone's time.
Vorlath # Monday, July 6, 2009 12:06:23 AM
Like this:
01.... 00.... 11.... 10.... 1100XX 1010XX .... .... .... ....The X's are your grid. Above it are the candidates you will chose to fill in the X's. On the left, are the associated row numbers for your candidates written in columns. Both rows and columns MUST use the same base though. In this case, we use base 2.
Pick any rows to fill in your grid and you will find exact matches in the left columns (you will need to rotate them to read them left to right). Likewise, any new item you create not found in the grid will be found in those columns.
You may choose any grid size that you please. As long as the bases are the same, then you will always get a one to one mapping. Only if you use different interpretations of the same data (ie. different bases) will you get a contradiction, but that would mean the use of the fallacy of equivocation.
Vorlath # Monday, July 6, 2009 1:06:30 AM
The above statement is only true for finite lists s1, s2, s3, etc.. AND only if you use a different base for the two axis. If one attempts to apply it to infinite lists, then this is using the fallacy of composition, fallacy of begging the question, fallacy of circular reasoning, fallacy of equivocation, etc..
Fallacy of composition
Just because each subset of an infinite list has certain properties, it does not mean that the infinite set has the same properties. Otherwise, they would be either all finite or all infinite.
Fallacy of begging the question
We are asked to believe that reals are compared to naturals just because we are told.
Fallacy of circular reasoning
Assume we have an infinite list (reals) and a finite one (grid), show that the infinite one is "larger" than the finite one for all specific cases.
Fallacy of equivocation
The diagonal matches rows with digits, but the digits are base 2 while the rows are base 1. Different interpretation of the same data (what each digit means) cannot be accepted in a proof.
Vorlath # Monday, July 6, 2009 3:27:06 AM
Cantor's theory is DEFINITELY suffering from the fallacy of Reductio ad absurdum. Just because you can create new rows does not imply that the natural numbers cannot be matched to it. It only says that you have more than n n-digit numbers for any base higher than 1. You can't draw a contradiction from that.
Vorlath # Monday, July 6, 2009 4:12:13 AM
Unregistered user # Monday, July 6, 2009 9:47:10 PM
Vorlath # Tuesday, July 7, 2009 12:21:26 AM
Um, he says it. At least Wikipedia does.
You can only compare finite lists, otherwise you're using the fallacy of composition, especially when trying to PROVE that infinities are different. In fact, it's easy to show that the above quote for wikipedia is only true for finite lists and only for different bases on each axis.
We are in agreement. But since it's not possible in the finite numbers, then Cantor's argument is flawed (fallacy of composition with a hint of circular reasoning mixed in).
Depends on interpretation. If you count them, then they are essentially base 1 since you don't care about what the symbol is. This is the step involved when FINDING each element of the diagonal. We match each digit to each row in order to find the diagonal. BTW, funny fact. There is no diagonal when using an infinite amount of digits.
And when you take the symbols into account, then it's base 2, so you have an intentional mismatch between axis. This leads to a Texas sharpshooter fallacy if you try to show that there are more than n n-digit numbers and use this fact for any sort of mismatch since Cantor intentionally put it in the premise to get his desired result.
If you're only counting (not caring what the symbols are), then there is no contradiction. Only when you take their base 2 properties into account does the Cantor's "contradiction" appear, and you must take them into account to produce a new number.
Assuming "inverse" means 1's to 0's and vice versa,
1) Sure.
2) The number is not infinite. It is finite (not sure if this is what you asked though). And you can find it in either the naturals or in the reals depending if you have the decimal point prefix with the usual qualifications that certain binary real representations are equal to each other.
3) Don't know that they're "wrong", whatever that would mean in this case, but they're definitely irrelevant.
You seem to be trying to show me how one fact follows to the next. While I admire your tenacity as well as your clear comments about the topic, the flaw lies in comparing base 1 to base 2. You're circling around it and I think you're VERY close to seeing what I'm talking about.
What most mathematicians tend to do is go back to the fact that the new number is not in the list. It doesn't matter. I don't have to use Cantor's mapping. He assumes that since he shows one example of how the mapping doesn't work, then there can be NO mapping that works. This is nothing more than the fallacy of proof by example.
Here is the example of "above the grid" again in visual form, but a much simpler variation.
Now, instead of using the representation in this grid for our naturals, we will instead use a ONE TO ONE mapping by using the row number. Please understand what we are doing here. We are mapping what Cantor used as a natural to a new list of naturals. Row numbers will be matched one to one to what Cantor uses for the naturals in the original grid, correct? We're in agreement on this? In fact, it has to be since Cantor uses this very argument when constructing the diagonal. We can switch between the two representations of naturals by simply switching grids and staying on the same row.
Cantor uses the representation in the first grid to form the new real (diagonal). But that's OK, because we can now use the same process in the second grid to create a new natural and keep the amount of reals and naturals the same.
Get it? With the ONE TO ONE mapping, no one can say I cannot do this since I'm just mapping one set of naturals to another set of naturals. And the reason there is no contradiction is because I'm now using base 2 instead of base 1 for listing the naturals. And to boot, with the ONE TO ONE mapping, it's EXACTLY the same as Cantor's diagonal argument. The new real is formed along with the new natural (and a new mapped row number).
And the really stupid thing is that what's good for Cantor is good for the rest of us. If he can create a new number by a certain process, then so can the rest of us, otherwise it's the fallacy of the double standard.
Cantor's mistake was assuming that we need to create our naturals from the list of digits/rows (COUNTING) when there are plenty of other representations that work just fine. That's why this is even more the Texas sharpshooter fallacy along with the fallacy of inconsistent comparison (it's not reals that there are more of, but rather that there are more base 2 numbers than base 1 numbers when using equal amount of digits).
Also, he's just using an example that SEEMS to show a contradiction. But you have to show that ALL representations display a mismatch. And they don't since I've shown a counter-example. There's only a mismatch when the bases are different. This isn't even the fallacy of hasty generalization or fallacy of composition since it's only ONE example (though the fallacy of composition applies to the construction of the diagonal). As a whole, Cantor's argument is the fallacy of proof by example.
BTW, what I've shown is a counter-example. That is fatal to Cantor's argument. No one can wish it away or dismiss it. Doesn't mean the opposite is true. It simply means that the "proof" is flawed for certain.
Unregistered user # Tuesday, July 7, 2009 1:01:15 AM
Vorlath # Tuesday, July 7, 2009 1:02:41 AM
Vorlath # Tuesday, July 7, 2009 1:22:10 AM
But this is Cantor's argument. If you don't agree with it, then we're done here.
But he only shows ONE mapping of base 1 vs base 2. Those aren't the only bases you can compare, you know!
I'm glad you said this because I'm going to hold you to it. You've just disproved Cantor's theory with this statement. It was the last piece I needed to convince you with my counter-example. Now, I don't have to bother with this part.
So have the reals.
BTW, how in the world can the naturals be exhausted before the reals if I'm using EXACTLY the same process to create both a new real and a new natural? It's impossible. Why is it that when Cantor creates a new real, it's ok. But when I go to create a new natural, I'm not allowed? BTW, I'm creating a new natural in the SECOND grid which is mapped ONE TO ONE with the first grid.
Just answer me why there is a double standard.
Vorlath # Tuesday, July 7, 2009 1:50:58 AM
But all his "proof" shows is that
infinity != log2(infinity)
which we both agree is incorrect.
1. infinity == log2(infinity)
2. Cantor's diagonal argument is true.
Pick one. They are mutually exclusive.
(And just in case, the infinity in #1 is the cardinality of the naturals on both sides of the equation).
BTW, if I can show that these two points are mutually exclusive, would you agree that Cantor's argument is flawed?
Vorlath # Tuesday, July 7, 2009 3:10:03 AM
Let's take the representation of reals as in Cantor's argument. He has an infinite amount of digits, right? But what cardinality is this list of digits? We can count them, first digit, second digit, etc. and they map one to one with the natural set. In fact, we cannot say the 2.6th digit, can we? So the amount of digits is the same as the cardinality of the naturals.
We agree so far?
The set of natural numbers has infinite elements. So how many base 2 digits would we need to represent them all? We're not talking about any given element, but rather the number of digits required to represent the entire set. We would likewise have an infinite amount of digits.
And since we've already agreed that log2(infinity) = infinity for countable infinities, then the fact that we need an infinite amount of digits to represent infinite countable elements should be trivial. (This applies to all bases 2 and above).
This means that we are using an equal amount of digits to represent both reals and naturals.
So if you use digits to represent reals, you may only represent as many as found in the set of naturals.
Note that I'm not saying the cardinality of reals is the same as that of the naturals. All I'm saying is that if you use digit representation, you can only list an equal amount to that of the set of naturals.
Since Cantor uses digits to represent his real numbers, only if infinity != log2(infinity) can Cantor's diagonal argument be true. Consider that the infinity on both sides of the equation is the cardinality of the naturals.
BTW, is it possible to represent PI with digits? Even if you had an infinite amount? Doesn't affect countability if it's not possible, but it'd still be cool to know in case it is possible. What's the currently accepted position about representing irrationals? I'm reading that irrationals can be represented by non terminating, non repeating decimals. If true, then the cardinality of reals and cardinals are the same since this would allow a one to one mapping.
Vorlath # Tuesday, July 7, 2009 3:54:42 AM
Unregistered user # Tuesday, July 7, 2009 6:01:48 PM
Vorlath # Tuesday, July 7, 2009 6:48:24 PM
Unfortunately, this is the fallacy of composition. He can only ever match a finite amount of digits with a finite amount of rows. Matching an infinite amount of digits with an infinite amount of rows has completely different properties.
Cantor's diagonal is an indeterminate form. For example, what is the ratio of digits vs. rows when forming the diagonal? For 10 digits and 10 rows, its 10/10 = 1. It has a ratio of 1:1. With 25 digits and 25 rows, it has a ratio of 25/25 = 1. So again, a ratio of 1:1.
But when you have an ∞ amount of digits and an ∞ amount of rows, it has ∞/∞ = indeterminate ratio.
The diagonal is an indeterminate form when you use an infinite grid. This is all explained in my counter-example with the two grids. This is quite fatal to Cantor's diagonal argument.
Cantor uses a FINITE grid. Otherwise, I want to know the ratio of digits vs. rows for his diagonal. Only a ratio of 1:1 (Cantor's OWN words of "ONE to ONE mapping") will yield a SINGLE diagonal the same length as the axis. This is how Cantor matches cardinality, is it not? Let's see it.
I'll just say it again. Cantor's diagonal has more holes than swiss cheese. It requires the use of a great deal of non sequitur's to make it work. The ONLY way Cantor's argument has ANY chance of working is if log2(∞)!=∞. This will let him create a grid of log2(∞) by log2(∞) with ∞-log2(∞) rows left over that can be created via the diagonal. But if log2(∞) is equal to ∞, then Cantor's argument falls apart because the ratio of the grid is indeterminate (and no new rows can be created).
Vorlath # Tuesday, July 7, 2009 9:06:00 PM
I'm going to take the diagonal top-right to bottom-left and flip the digits (1's to 0's and vice versa). Since the last digit will always be 1 after flipping, and since 1 can NEVER be in the set of even numbers, I can always create a new number that isn't in the set of even numbers no matter what. So the cardinality of naturals is larger than the cardinality of even numbers.
Stupid isn't it? Yet, it's equivalent to Cantor's theory. I simply used even set vs. naturals instead of Cantor's naturals vs. reals.
I know believers in Cantor don't like this, but the above 100% disproves Cantor's diagonal argument. Counter-examples are that powerful. Any attempt at reiterating the original argument is pointless because I'm doing the EXACT same thing as Cantor in the counter-example.
Heck, I could compare NATURALS to NATURALS and still find more rows not in the list. How's that happen?
Vorlath # Tuesday, July 7, 2009 9:25:58 PM
You missed the point. We know that logb(x) digits will create x numbers.
But if you have log(|N|) digits, then what you really have is |N| digits because |N| is infinity.
Since the reals and naturals both use an infinite amount of digits, specifically |N| digits, then only |N| numbers can be generated in both cases for representations of naturals and reals.
It doesn't mean that the cardinality of reals is the same as naturals UNLESS you can represent ALL reals using digits. That's why I asked if you can represent PI. If you can represent ALL irrationals using digits, then that covers ALL real numbers and this means that the cardinality of reals is the same as that of naturals.
Note that the representation of PI itself and the fact that it maps one to one with the naturals is completely irrelevant (unless you mean that the naturals can index into the digits of PI). What I care about is that |N| digits is enough to represent PI. If true for all irrationals, then
|N|==|R|
because all irrationals will already be included in the rows.
Vorlath # Wednesday, July 8, 2009 12:43:22 AM
There's no contradiction. He's using the same flaw as in the diagonal argument. Trying to create a unique item by matching two infinite sets in a finite way.
Vorlath # Wednesday, July 8, 2009 3:50:52 AM
The entire set needs an infinite amount of digits to represent all elements. If you only use a finite amount of digits, then I can simply create a new number by adding 1 to the largest possible number.
For the individual case, representations are completely arbitrary. I see no rule that restricts an element in the set of naturals from using an infinite representation. This has zero effect on the status of the element of remaining finite.
Anyways, if the popular belief is that naturals cannot have infinite digits, then the mathematical world will be in for one big shock one of these days.
Unregistered user # Wednesday, July 8, 2009 12:18:03 PM
Vorlath # Wednesday, July 8, 2009 6:49:30 PM
Fallacy of double standard. If I can't use it, then neither should Cantor. The concept of ratios is exactly what one to one relationship means, is it not? And it's quite a FINITE concept. But nowhere is it explained why a ratio of ∞:∞ should hold up in a finite sense.
I originally thought you were right, but you're not. The diagonal not being an even number is the whole point of the exercise in the same way Cantor's diagonal is not in the naturals. Cantor says he's using an infinite amount of rows created via an infinite amount of digits. This means only |N| rows will exist. That's the whole basis of Cantor's diagonal argument. Without this step, everything else falls apart. He believes he's mapping all naturals to SOME of the reals by placing a decimal point in front. In my example, I'm similarly mapping SOME of the naturals to the even numbers.
Since I'm listing the set of even naturals, then that's going to be the size of the set E, specifically |E|. By creating a new row, I'm showing that |N|>|E| according to Cantor.
Remember, the point in creating a "new real" is to show that the existing list has cardinality of |N| and that there must be a larger set since we create a "new real". When Cantor is creating a "new real", it is not found in the original list at all, neither as a real nor as a natural (as far as the argument goes). You said yourself he only attempts to list all the reals. And he uses the representation in the grid in two ways, as a natural and as a real. In my example, I'm pretty sure using the representation of even numbers as a subset of the naturals in an analogous way is trivial. Are the even numbers not found both in the even set and in the naturals set? Of course they are. The new number I'm creating CANNOT be in the set of evens (that's the point of the diagonal argument), but IS in the set of naturals.
Also, my second example still holds regardless. If you use the set of naturals, you can create a new natural in the same way Cantor creates a new real, thus "proving" that |N|>|N|. This is why I asked you about finite representations. It's the only counter to this I could find.
What I call the grid is the finite square used to build the diagonal. This square can never be infinite because then Cantor loses the one to one relationship he so desperately seeks.
By the same reasoning, he cannot create a "new real" if his square goes infinite. The only reason he can create a "new real" is because log(x)!=x. But when x=∞, it's IMPOSSIBLE to create a new real since every possible combination digit-wise has been included in the list.
Then there is the method used for creating the diagonal. That's a finite concept. It can't be done if the grid is infinite. In Cantor's diagonal argument, when he attempts this feat, he just does it. No explanation. No proof. Not even a reasoning as to how he's avoiding the fallacy of composition. Creating an infinite diagonal just because you can create a finite one is a non sequitur.
What's even funnier is that the finite version of Cantor's diagonal is quite bogus. He's not creating a new real. The disparity comes from using different bases.
About the finite digits issue, thanks for the answer. I was getting worried.
Vorlath # Thursday, July 9, 2009 3:17:04 PM
I just wanted to requote this because it's really a fantastic rebuttal to Cantor's diagonal argument. Note how you said I cannot add the new number to the list because it isn't even. I've explained how Cantor doesn't do this either, BUT he actually COULD do this, could he not? When he creates a new real, is it not also a new natural? Is the new number not composed with as many digits as the rows in the grid which can also be treated as naturals? So how can a new natural be created that's not already in the list of naturals? Ends up that my even number example is actually "better" than Cantor's.
And this is what Cantor doesn't realize. When he's creating a new real, he's also creating a new natural. And if he's creating a new natural, then |N|>|N|. This is a contradiction and Cantor's diagonal argument is flawed.
I don't see a way around this fatal flaw.
Unregistered user # Sunday, July 12, 2009 3:44:38 AM
Unregistered user # Sunday, July 12, 2009 3:49:32 AM
Vorlath # Sunday, July 12, 2009 4:00:36 PM
Cantor doesn't do this either. In fact, the diagonal ENSURES that he's not mapping every natural number because log(x)!=x.
(edit: Just have to say that your quote is a fallacy. If Cantor did try to match all members of the larger set, then he'd have to prove that he is using the BEST possible matching. This would lead to a circular argument of using an incomplete matching to show an incomplete matching.)
Cantor likewise lists members of the "smaller" set |N|.
And Cantor "proves" that reals have at least one more element (than N).
No, I've proven that |N| > |E| and it's not obvious. In fact, it's wrong. But if this is wrong, then so is |R| > |N| because I'm using the exact same technique as Cantor. You can't attack WHAT I'm using. You can only attack me if I'm using the technique differently.
Whut? |E|>=|N|??? Where'd that come from?
Ah, but this is not derived from the "proof", now is it. But I agree with you that |N|==|E|, hence a contradiction with "my proof", but also with Cantor's since they are identical. Once x is infinite, then you can only create |N| representations with |N| digits. Show one more and you've got a contradiction. This contradiction... this flaw... is the entire point of this exercise to show how Cantor's diagonal is flawed. Whatever applies to my example likewise applies to Cantor's diagonal argument. I'm not sure you quite realize that point. BTW, Cantor's argument is nothing more than the Grand Hotel paradox rephrased incorrectly. Cantor is trying to show this contradiction and arbitrarily say that reals have larger cardinality. It's bogus not just because of the paradox, but because he forgets that log(∞)==∞.
I'll just repeat. I WANT you to find a flaw in the "even number" example because I'm using proof by contradiction. If you do show a contradiction, then Cantor's technique is flawed since I'm using the exact same technique. Get it? That's what counter-examples do. Or the antagonist, as you call it. The ONLY way you can show that this counter-example is flawed is if I'm using a different technique than Cantor and I'm not.
Ummm.... NO!
Cantor is the one proposing the proof. At no point do I need to show a one to one mapping. In fact, I'm stating now for the record that one to one mapping is a flawed concept. It's a non sequitur. I've seen no proof that a one to one mapping means anything to infinite sets. I've shown how many one to one mappings are possible that leave out elements of one set and yet their cardinalities are the same. Finite one to one relationships are fine, but then applying that property to the entire set is a non sequitur. Plus, we know that a ∞:∞ relationship is indeterminate. Playing by Cantor's flawed rules is exactly how NOT to disprove Cantor's argument. You'll be stuck in the endless loop of circular argument.
On top of all that, I only need to show a contradiction in his "proof". That's all I need to do. And I've done so with the "even numbers" example (and you've shown this as well, but so far has failed to apply it to Cantor's diagonal argument in a fallacy of double standard).
I don't know where to start. It's all so very wrong. And if you were right, you would have disproved Cantor. I don't think you realize how my example is identical. Unfortunately, there's a flaw in your argument.
Let's just tackle the fallacy of composition. To represent ALL naturals, you need an infinite amount of digits. From these infinite digits, you can create all finite numbers. But you don't do that. You take the infinite property of the ENTIRE set of digits and apply it to the SINGLE diagonal. You can't do that. It's a non sequitur. All naturals are finite and were formed from those infinite digits. Regardless, what you're proposing is self-defeating. You're saying that |N| digits can form more than |N| representations. That's a contradiction and your counter-argument is flawed. IOW, you're using the exact same argument in Cantor's. Circular argument.
And then there's the fact that I've already asked you about any objection to naturals being able to use infinite representations. You had no problem with it. I said it was the only counter to my arguments. And now here you are presenting EXACTLY that bogus argument. Representations by themselves CAN NEVER have any implications on the finite (or infinite) status of a number. But like I said, that's for another topic.
Anyhow, you've reinforced my arguments (your arguments actually) against Cantor. Besides, all this is moot since log(x)==x when x=∞. Until someone shows me how this is not true, Cantor's diagonal argument is flawed.
Vorlath # Sunday, July 12, 2009 5:25:58 PM
Think of this. If |R|>|N|... if that is really true, then it means that a subset of R maps to N, right? So if I show that a subset of A maps to the entire set of B, does that show that |A|>|B|? No. I've shown this with even numbers which we know to have the same cardinality of natural numbers.
So if simply showing a mapping from one subset to the entire set of another is not enough, then what does it take? Showing that a particular mapping of a subset of reals to all naturals is possible and then creating a new real is rather mundane and devoid of anything interesting.
What does it take?
It means that we need to show how ALL possible one to one mappings of A to B will never include A in its entirety. ALL MAPPINGS! Or prove that one particular subset of A is the largest subset that will map to B.
So Cantor's diagonal argument is trivially flawed. He only shows one mapping (and a ridiculous one at that where he assumes log(x)!=x when x=∞).
The more I look at it, the more trivially flawed Cantor's diagonal argument is. (Oh, and my original objection, that log(x)==x when x=∞, is still valid. I just thought this new angle was worth mentioning as it might make the log(x) flaw irrelevant if this new angle is correct.)
(edit: This new angle would also counter Cantor's first uncountability proof.)
(edit2: Note that it's trivially obvious to generate both a new real and a new natural for any finite grid in Cantor's argument. And for the infinite case, you can likewise create a new natural through the use of Hilbert's Grand Hotel paradox. So Cantor's mapping is quite a weak one and nowhere near the best. He should at least start by getting a better mapping for the finite case by at least listing all combinations for all digits. Exhaust all finite combinations first.)
Vorlath # Sunday, July 12, 2009 8:28:35 PM
Been doing some more research on flawed proofs by contradiction.
Say I need to prove if x can be 4. Proof by contradiction says that we take x to be 4 and then see if a contradiction arises.
So I set x to be 4.
1. x = 5 (NOT a typo.)
Then I go about performing some actions.
2. x%2 = 5%2 = 1
3. But we know that four is even. CONTRADICITON! Therefore x cannot be four.
This is what Cantor does. He says he's ASSUMING the grid is infinite. But he then lists a grid that isn't. He's also assuming that one to one means something.
Also, his mapping between the reals and naturals can hardly be worse. In fact, only a subset of BOTH N and R are used. It is then trivial to create new numbers. And the infinite case is even more ridiculous since there is no longer any avenue to create a new number and the diagonal argument is void. Actually, there is ONE way to create a new number. By using Hilbert's Grand Hotel paradox, but you can do that for both reals and naturals.
And then you can simply ask someone to show a counter-example where 5 is even (or request a one to one mapping between naturals and reals).
Vorlath # Monday, July 13, 2009 5:38:30 AM
BTW Kyle, thanks for having this discussion. I understand the meaning of Hilbert's paradox much more clearly now. I now understand a new way of looking at flaws in proofs by contradiction when an assumption is stated, but given contradictory assumed properties. And I see what Cantor was after with his power sets (which unfortunately treats infinite sets as finite ones by simply not considering Hilbert's paradox when it comes to infinite sets). I find it funny that the person whose quote is most remembered about celebrating Cantor's proof is the person whose paradox will bring it down.
Unregistered user # Monday, July 13, 2009 8:53:15 AM
Vorlath # Monday, July 13, 2009 8:34:05 PM
Well, if I haven't proven that naturals are larger than even, then neither has Cantor proven that reals are larger than natural. Otherwise, you're using the fallacy of double standard.
The "larger" set is N in my case and the diagonal IS in N. What's the problem?
Also, I'm listing as many naturals as I can in my grid that maps to even numbers. Just like Cantor is listing as many reals as he can that maps to naturals (as the argument goes because he doesn't actually do this).
Everything is the same. Still want to use the fallacy of double standard? With all due respect, that seems to be all you've got against my disproof.
Sorry, but this is flat out wrong (or not the entire truth). Forget Cantor for a moment. Suppose you have a different proof about a different topic. Are you really suggesting that if the conclusion is correct, that the proof cannot be wrong?????? That makes no sense. There are plenty of proofs that end up being flawed even though the conclusion is correct. How do you disprove those proofs? By showing a logical fallacy. NOT by showing that the conclusion is flawed. That's impossible.
Back to Cantor. I do disagree with the conclusion (though there is one area unrelated to Cantor that I haven't discussed where I'm still analyzing), but showing a logical fallacy is still a proper method of demonstrating a flaw in a proof. So please stop this blatantly incorrect notion that I MUST show a one to one mapping, and that it MUST be between reals and naturals.
log(x)==x when x=∞.
The above line disproves Cantor's argument just fine, thank you. It's a contradiction to Cantor's diagonal and demonstrates the fallacy of composition.
However, I'm not one to disappoint. If you want a one to one mapping, I will give you one. But not on the reals vs. naturals (though I can supply that too), but rather naturals vs. evens. This shows the EXACT same argument, but with a conclusion opposite to Cantor's diagonal.
I've covered BOTH methods of countering a proof. Is this satisfactory?
This is trite. No one has ever been able to give a full expansion regardless of the set it's in. Fallacy of double standard.
You've proven that Hilbert's paradox is flawed? Seriously though, I'm not the one using it. It's Cantor. But he ends up using it wrong. Whether you move "all" elements up, or you flip "all" N digits, they're both operations on N elements in order to produce a "new" number. Not only is this the fallacy of double standard (using it for reals and not naturals), but it's also the fallacy of composition because Cantor's diagonal only works on finite grids.
Note too that the ONLY reason Cantor says it's not a natural is because there are |N| rows already listed. But if |R|==|N|, then his argument is flawed. The only reason he says it's a real is because it's unknown if |R|>|N|, so he just ASSUMES it in the conclusion. You're not supposed to assume anything in the conclusion, only in the premise. Cantor's argument is a circular argument.
Cantor is the one who's missing bn-n numbers. Not me. That's where his numbers for his diagonals are coming from. But use the same base and you lose the ability to create new numbers with finite grids. With infinite grids, log(x)==x when x=∞ and you again lose the ability to create a new number even when using different bases. This is in contradiction to Cantor's diagonal, so I MUST reject Cantor's argument. It's not a choice. I MUST reject it. So far, you've avoided responding to this.
Cantor only uses ONE and only ONE mapping, and quite a horrible one at that. I can create bn-n new naturals for each finite grid. Honestly, it doesn't get much worse than that. An intentional discrepancy of bn-n is TERRIBLE! I can create a better mapping blindfolded. At the very least, use the same amount of symbols per digit. Cantor doesn't even pretend to use the best mapping. He just blatantly uses the WORST mapping he could find. That you could suggest you can find a new number for all mappings is an insult. Use the same base and then show me. I dare you to find a new number when you use the same base for your mapping. Don't take my word for it. Try it for yourself.
Besides, creating a new number means nothing. Tons of subsets map one to one with infinite sets and still have the same cardinality. This is assuming Cantor COULD create a new real (which he does not, but even if he did, it's meaningless).
Vorlath # Monday, July 13, 2009 9:50:01 PM
Just had to requote this. I CAN in fact use Cantor's reasoning to any two sets. It's what he uses to demonstrate how one set has larger cardinality than the other. Yes, if I'm using the sets differently, then that would be a valid point. But I'm using them EXACTLY in the same manner. Cantor's argument should be able to tell me which set is larger for any two given sets, no? Otherwise, why is he allowed to use it on reals and naturals? Why would this not be the fallacy of double standard?
I think it would be interesting to see Cantor's diagonal argument applied to naturals and evens to see which one is larger, if any. Note that if they are equal, Cantor's argument MUST have a way to show how there are no new numbers.
Cantor's argument just keeps getting worse and worse. His argument has ZERO way of showing two sets being equal. No matter the sets you try, the outcome is always the same. This means that the sets don't even matter. And this is what I've been saying all along. It's the base difference. It always comes back to that. As long as one set is represented in a different base, you're going to get a discrepancy. Heck, did you know I can show that |N|>|R| with Cantor's argument?
Unregistered user # Monday, July 13, 2009 10:55:34 PM
Vorlath # Monday, July 13, 2009 11:12:16 PM
My diagonal only exists in the "larger" set of naturals and not the "smaller" one of evens exactly as you want. I don't understand your objection. Also, you shouldn't use the facts of which one is larger. We simply need to assume one is larger (or both equal). That's how proofs by contradiction work. You're putting the horse before the cart. (edit: Also, the evens aren't actually larger than naturals or vice versa. You know that right?)
(edit: Actually, the way you mention with E in the list would work just as good, but it's equivalent to comparing N with N. That's a good counter-example too.)
Saying that I'm limiting someone is funny. The list is composed of NATURALS, not even numbers. The naturals simply MAP to the evens as best I can in EXACTLY the same manner that Cantor tries as best he can to map the reals to natural numbers (as the argument goes since that's not what's going on). (edit: And I can't limit anything. They're both the same size.)
(edit2: Perhaps I should try and clarify my example. With Cantor, he simply lists |N| rows of real numbers. If you can create a new real, then obviously |R|>|N|, right? So what I want to do is list |E| rows of natural numbers and then create a new natural number. To ensure that only |E| rows are listed, I use natural numbers that map to E. If you don't like that last digit, you're free to start the diagonal on the second digit. The fact that we created a new number not found in the list of naturals [starting at position 2] means that |N|>|E| since there are already |E| rows listed).
You're getting close. Keep following your train of thought, but apply it to Cantor's argument and you'll see why it's flawed. You're SO SO VERY close. Remember, any objection you have to my disproof must also be applied to Cantor. So I WANT you to find a flaw in it. But so far, your arguments are what's been flawed. Find a REAL flaw in my argument and you find the EXACT flaw in Cantor. I've actually been DEFENDING Cantor. Once you find the true flaw, you will find a contradiction within Cantor's argument and you will have a proof by contradiction AGAINST Cantor's argument.
I'm not sure you realize yet that if you disagree with the even vs. natural example, you must also disagree with Cantor's diagonal argument because that is what I'm using.
C'mon... keep going. Your next objection should be that the representation (of naturals vs. reals) isn't what's being mapped. Am I right? So what is? When you state the objection, remember to apply it to Cantor's argument. There's only one more step after that.
NO NO NO... Don't change the topic. Answer the question I asked. C'mon. It's ok.
Apply Cantor's argument! Tell me the result.
And then note that m = 2n is a one to one relationship as requested.
R = {X >> (log2(|Z|)/2) | X ∈ Z}
Use the same algorithm. Just move the decimal point.
Vorlath # Monday, July 13, 2009 11:59:50 PM
With the infinite prime proof, if the new numbers are not prime, then no contradiction. So I want to know how it's possible for one to create a diagonal and NOT come up with a contradiction for ANY infinite set.
Vorlath # Tuesday, July 14, 2009 12:45:45 AM
I wanted to give you another one to one mapping.
That's three so far where two of them are between reals an naturals. Need more?
Vorlath # Tuesday, July 14, 2009 1:27:27 AM
This is incorrect. He ENSURES that such a mapping cannot be done by using different bases for each axis of the grid.
Unregistered user # Tuesday, July 14, 2009 10:09:39 AM
Vorlath # Tuesday, July 14, 2009 11:55:00 AM
But a one to one mapping is not the only way to refute a proof. And the diagonal is a flawed concept. So if I do show you a mapping, I bet you're going to show me this bogus diagonal, aren't you? So I have to show the logical fallacy.
(edit: All you're trying to do is keep me within the flawed rules created by Cantor. It would be RIDICULOUS for me to play by Cantor's flawed rules. What you suggest makes it IMPOSSIBLE to refute Cantor or ANY flawed proof by simply saying that if a proof is correct, then you MUST show a counter-example of what the proof proposes. That's circular logic of biblical proportions since the counter-example itself must play by flawed rules which makes the counter-example useless, and you've said as much by saying you would just take a flawed diagonal.
In French in the area where I'm from, we have an expression for what you're trying to do.
Loosely translated, it means "Let me go so I can hit you." That's what you're doing.
I'm going to repeat that you CANNOT show a new number. That process is FLAWED! I've proved it over and over again with log(x)==x when x=∞. So any argument that uses the "new" number angle is flawed and I automatically reject it. You can try to avoid it all you want, but Cantor's diagonal has been proven to be flawed. It only works with finite grids, and then that's only because of a base difference.
I repeat my challenge. USE THE SAME BASE AND SHOW ME A NEW NUMBER!)
These are naturals.
They're all there. Check again. And I didn't miss any.
It's a one to one mapping. Sorry for using s on the reals and r on the naturals. s is what wikipedia used.
R = {X >> log2(|N|) | X ∈ N}
Sure it is.
It's not supposed to converge. A number based on the entire set is not supposed to converge unless the range is bounded by finite numbers such as [0,1]. That's how infinity works with respect to REPRESENTATIONS, not elements (an element cannot converge). If I did define what say, 0.5 mapped to and showed you "ALL" digits, you could just double it and say that this new number is larger than (or equal to) |N| showing a contradiction. There's a reason it has infinite digits. It's so that no matter the size of |N| (if you add members, or partial finite representation), it will grow with it proportionally. It's still a finite number though by simply not being infinity.
Non sequitur. The representation has nothing to do with the finite status of an element. You've tried this before and I still haven't seen a proof of it. BTW, I've already analyzed this angle and it doesn't pan out.
(edit: If you have infinite naturals, you need infinite digits. Besides, I asked you about this before, and you agreed with this. And now you're repeating the opposite argument YET AGAIN! What gives? Where would you get such a crazy notion anyways? If you have finite digits, then you can list them ALL!!! You can write them out. Only with infinite digits can you have infinite elements. It's trivial.)
Unregistered user # Tuesday, July 14, 2009 2:35:58 PM
Vorlath # Tuesday, July 14, 2009 2:45:29 PM
Suppose a set has 2 items {0,1}. Now suppose you split it in half. The first element of the second subset is element 1, or n/2. With odd items, you can also consider (n-1)/2 for the middle element.
4 items, first element of second set is 2
6 items, first element of second set is 3
8 items, first element of second set is 4
3 items, middle element is 1
5 items, middle element is 2
7 items, middle element is 3
I can continue and the middle element is always a finite number in both case since its progression is mapped one to one with N and is always lower than |N| (definitely making it finite). Yet, it would have infinite digits since the progression never ends. IOW, the number of digits is based on |N|. The more digits you specify, the larger |N| becomes which is fine since |N| is infinite. And the more elements you add, the more digits will be needed which is also fine since adding digits to an infinite amount is also no problem.
Vorlath # Tuesday, July 14, 2009 2:50:25 PM
Forget it. You don't understand representations. The representation has nothing to do with an element being finite or not.
If this is your only argument (which is trivially false), then I've proven Cantor's argument to be flawed.
Vorlath # Tuesday, July 14, 2009 2:51:20 PM
I gave you TWO examples. It's your fault if you don't like it.
log(x)==x when x=∞.
This disproves Cantor's diagonal. Deal with it.
Unregistered user # Tuesday, July 14, 2009 7:22:28 PM
Vorlath # Tuesday, July 14, 2009 9:25:07 PM
I'm going to sum up the facts that are proven or trivially known to be true. If you wish to object to any of them, let's do it one at a time.
1. A representation, whether finite or infinite, has no bearing on the finite or infinite status of an element.
2. Cantor uses a finite grid.
3. Cantor's diagonal cannot be created with infinite grids.
4. Cantor can create a diagonal with finite grids ONLY because of a difference in bases.
5. Cantor *tries* to use a variation of Hilbert's Grand Hotel paradox to create his "new" real number.
6. A one to one mapping to disprove Cantor's diagonal argument is useless because you'll simply use circular argument and try to say you can create a diagonal (even though you can't).
7. A logical fallacy is fatal to a proof.
8. Cantor's diagonal suffers from the fallacy of composition when it uses a finite technique on a supposedly infinite grid.
9. To repeat, naturals CAN and DO have infinite digits. It's perfectly fine. There is nothing wrong with it. It doesn't break any rules or anything like that.
10. If the diagonal were to create a new real, then it would also create a new natural. Saying otherwise is the fallacy of double standard.
11. Cantor uses infinity as if it were finite. Not just in the grid or in the diagonal, but in the first uncountability argument and also in the power set argument.
12. I HAVE given you a one to one mapping between the naturals and reals. There is not a single number missing from either set. They are all there. No, I am not mistaken.
13. Cantor is trying to create more numbers than there are combinations allowed by the digits AND yet represents this new number using those very same digits. This alone defeats Cantor's argument.
14. Cantor's argument has no method of showing |R|==|N|. The only possible outcome, regardless of the sets is |X|>|N|. X can be anything. One can show |N|>|N|. The only way to counter this is with a priori knowledge thus invalidating the argument.
All of these are either trivially obvious or have been proven. Unfortunately, you have proven nothing. You have not shown any number to be missing. The requests for PI are trite (and I've given an answer multiple times). Your objections to the even example are confused and illogical.
I only have two requests.
1. USE THE SAME BASE!
2. Don't handwave this away.
Unregistered user # Wednesday, July 15, 2009 2:40:49 AM