Re: Para-consistent logic
Posted: Mon May 12, 2014 2:04 pm
Mike, that's true, but I think everyone has been using the proof by contradiction route in this thread.
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I defined the number L as "the largest number", and if you find a number that verify the propriety "is the largest number", then this number is, by definition, LPippen wrote:You assumed a "largest number" and you could indeed prove that this very "largest number" is not the largest, because of the successor of it. So your "largest number" was not the largest number right from the beginning, as the proof shows. Therefore we are now open to look for another assumed largest number. What about "largest number's successor"? Obviously the "largest number's successor" wasn't covered by "largest number", so we could now think of this number to be the new "largest number" and here we go again with the proof and it goes on and on....
The discussion here has relatively little to do with "number theory."SmoothOper wrote:The generative approach to number theory is all well and good, but people have to understand its limitations, mainly that there is only one object and the further removed you get in notation from that one object the more difficult it becomes to discourse, quickly degenerating to let's see who can make the biggest number... Which is disallowed under the axiom of determination.
Another approach would be to state properties of the sets as a whole.
The generative approach is all well and good, but it is disallowed under the axiom of degeneration.RBerenguel wrote:The discussion here has relatively little to do with "number theory."SmoothOper wrote:The generative approach to number theory is all well and good, but people have to understand its limitations, mainly that there is only one object and the further removed you get in notation from that one object the more difficult it becomes to discourse, quickly degenerating to let's see who can make the biggest number... Which is disallowed under the axiom of determination.
Another approach would be to state properties of the sets as a whole.
We do accept this axiom, don't we?Bill Spight wrote:The generative approach is all well and good, but it is disallowed under the axiom of degeneration.RBerenguel wrote:The discussion here has relatively little to do with "number theory."SmoothOper wrote:The generative approach to number theory is all well and good, but people have to understand its limitations, mainly that there is only one object and the further removed you get in notation from that one object the more difficult it becomes to discourse, quickly degenerating to let's see who can make the biggest number... Which is disallowed under the axiom of determination.
Another approach would be to state properties of the sets as a whole.
Let's break down assuming a largest number. We assume that a largest natural number exists. If it does exist, there is only one of them, and we can give it a name, so that we can designate it. We do not have to actually find the largest number. We can talk about it, anyway. Just like they talk about the UNSUB on "Criminal Minds" without knowing who it is.Pippen wrote:You assumed a "largest number" and you could indeed prove that this very "largest number" is not the largest, because of the successor of it. So your "largest number" was not the largest number right from the beginning, as the proof shows.
No, because we never looked for one in the first place. It is not like we found one and then said, this is not it. We simply talked about one as though it existed. Once we show that one does not exist, what is there to look for?Therefore we are now open to look for another assumed largest number.
I have seen some old problems that do not use variables. They are often not easy to understand. Even the easy ones are like, "Heap, its third, . . ."How does a variable work?
Suppose that something is true for each member in a set. We can then use "x" to denote some member of the set (without necessarily knowing which one) and prove something else about it, using only what is true for each member of the set. Then we know that what we have proved about it is true for each member of the set. It is not that "x" covers all members of the set, it is that we restricted our proof to what is true about each member of the set.How can an "x" stand for one object out of a set, but at the same time cover all object of a set, so that if you prove something for x it is proven for all objects of the set?
Well, it was the only book in the library about go that wasn't obviously trivial.RBerenguel wrote:
Good for Berlekamp and Wolfe that they got a sale for the book. There's a lot of knowledge and content between the "yeahs." And I can recommend many books where you may think you understood something and the rest is "yeah yeah," want a list? They obviously will make no sense for you.
How about that?Bill Spight wrote:I know I kind of went around the bush there, but I wanted to be clear that we did not identify the largest natural number, we simply talked about it as though it existed.
No , I proved that L does not exist. Assuming that it does exist leads to a contradiction. I proved that if a largest natural number existed, then there was a larger natural number than it. Tilt!Pippen wrote:How about that?Bill Spight wrote:I know I kind of went around the bush there, but I wanted to be clear that we did not identify the largest natural number, we simply talked about it as though it existed.
You prove that L cannot be the largest number, because L+1 is its successor.
No. If L does not exist, then neither does L+1. (Or for that matter, L-1, or 2*L, or any other number expressed in terms of L.) A non-natural number does not have a successor which is a natural number. I do not have to prove anything about such supposed numbers.I say: Fine, but L+1 is a natural number too.
The proof does not claim to be for all natural numbers, only for the possible largest natural number. Which, it turns out, does not exist.whatever your proof states, it is not for all natural numbers and therefore your proof is incomplete.
Hmm...I think L does exist, because by assumption it was assumed to be a natural number, but it didn't have the property to have not a successor. THAT's what you proved. So you proved that any number that can be pulled into L is a) a natural number and b) with a successor, therefore can't be the largest one. But since L+1 is not covered under L then the question arises if L+1 is the natural number that hasn't a successor.Bill Spight wrote:No , I proved that L does not exist.
No. You've got it backwards. Go back to what I said. I did not start by positing an L, and then assuming that L is a natural number, and then assuming that it is the largest natural number, and then proving that the third assumption was wrong, so that we end up with a natural number, L, that is not the largest. Where does that get us?Pippen wrote:Hmm...I think L does exist, because by assumption it was assumed to be a natural number, but it didn't have the property to have not a successor.Bill Spight wrote:No , I proved that L does not exist.
No, I did not pull any number into L. L is not necessary to the proof. It is just a convenient way to refer to the largest natural number.THAT's what you proved. So you proved that any number that can be pulled into L
We don't kill 3 because it follows from the definition of natural number. So if we kill it we are no longer talking about a hypothetical natural number.Here is my line of thinking:
1. We assume a natural number L that has the property to have no successor.
2. We find through Peano 2 that any n € N has a successor n+1 € N.
3. Through mp we find that L must have a successor L+1.
4. Lines 1 and 3 contradict each other, so either one must be false (by the way I have no idea why we do not kill 3 here, maybe someone can tell which one of two contradictory statements can be erased), and we agree?!? that it's line 1, therefore: L has not the property to have no successor.
5. But this proof in 1-4 just tells us about L and not about L+1. L+1 is not covered by L (otherwise L=L+1) therefore a very formal proof would have to prove now that L+1 has not the property to have no successor and so forth.
Does the largest natural number+1 make sense to you?Not Pippen wrote:4. Lines 1 and 3 contradict each other, so either one must be false (by the way I have no idea why we do not kill 3 here, maybe someone can tell which one of two contradictory statements can be erased), and we agree?!? that it's line 1, therefore: the largest natural number has not the property to have no successor.
5. But this proof in 1-4 just tells us about the largest natural number and not about the largest natural number+1. The largest natural number+1 is not covered by the largest natural number (otherwise the largest natural number=the largest natural number+1) therefore a very formal proof would have to prove now that the largest natural number+1 has not the property to have no successor and so forth.
Stop right there.Pippen wrote:
1. We assume a natural number L that has the property to have no successor.
Ok, I understand that and agree with it. But why then do they construct this reductio proof if it follows already DIRECTLY from Peano 2 that there are infinite many nat. numbers? Peano 2: ∀n: (n ∈ N -> n' ∈ N). With Peano 2 we can directly show that for all n follows that if n is a natural number it must have a successor, so (because of the allquantor or the fact that n stands basically for any object of N resp.) we can conclude: no natural number cannot have no successor. Why the indirect way at all?Mike Novack wrote:Stop right there.Pippen wrote:
1. We assume a natural number L that has the property to have no successor.
The "natural numbers" were defined as "things" which have certain properties. Those properties include:
Every natural number has a successor.
So if some "thing" doesn't have a successor then it can't be a natural number.