Para-consistent logic

[hyperpape]

I don't know in general. I find the reductio slightly easier to state.
Another conceivable reason is that you can study Peano arithmetic without the axiom of induction (PA-), so the reductio is more general. I believe the ordinals model PA-, for instance, though they do have a closely related form of induction, so it may not matter for them.

[Mike Novack]

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?

Not quite. Did you follow the proof the way I did it?

A bit "naive" (in the formal logic sense) to jump to the conclusion. The axiom of "induction" was included for a purpose. Necessay. The way you are thinking about getting to "all" is the way a positive direction proof might have been done prior to formalization of logic (19th Century) and is open to exactly the objection SmoothOper raised << that you can show it for any n you try, you are showing it only for those finite number of n's that you try ---- your jump to "that means all" can't be done when the logic is by formal symbols >>

That is perhaps when other here are demonstrating by the negative proof.

[Pippen]

@Mike: I don't think you need the axiom of induction for such a proof. Peano 2 is sufficient.

1. n ∈ N -> n' ∈ N; n stands for any possible object of N, precisely: n holds a place for one object out of N that is assumed to be takeable by any object of N. (This is the reason why you can universally generalize from n to ∀n here. Maybe someone knows universal generalization and can show how to come from "n ∈ N -> n' ∈ N" to ∀n, n': (n ∈ N -> n' ∈ N) in a formal way).

2. Since there can't be any natural number that is not represented by n it follows that because of Peano 2 and mp no natural number cannot have no successor.

[Mike Novack]

@Mike: I don't think you need the axiom of induction for such a proof. Peano 2 is sufficient.

1. n ∈ N -> n' ∈ N; n stands for any possible object of N, precisely: n holds a place for one object out of N that is assumed to be takeable by any object of N. (This is the reason why you can universally generalize from n to ∀n here. Maybe someone knows universal generalization and can show how to come from "n ∈ N -> n' ∈ N" to ∀n, n': (n ∈ N -> n' ∈ N) in a formal way).

2. Since there can't be any natural number that is not represented by n it follows that because of Peano 2 and mp no natural number cannot have no successor.

Not enough for you that Peano did? (thought it was necessary)

Did you understand SmoothOper's objection? Every time you select an n within N you can show that this n is not the largest. But no matter for how many n's you have shown this to be true, there still exists an n' for which you have not (yet) shown it to be true. Yes of course, you can now show it true for this n', but you don't have a finite proof (no last step).

[SmoothOper]

@Mike: I don't think you need the axiom of induction for such a proof. Peano 2 is sufficient.

1. n ∈ N -> n' ∈ N; n stands for any possible object of N, precisely: n holds a place for one object out of N that is assumed to be takeable by any object of N. (This is the reason why you can universally generalize from n to ∀n here. Maybe someone knows universal generalization and can show how to come from "n ∈ N -> n' ∈ N" to ∀n, n': (n ∈ N -> n' ∈ N) in a formal way).

2. Since there can't be any natural number that is not represented by n it follows that because of Peano 2 and mp no natural number cannot have no successor.

Not enough for you that Peano did? (thought it was necessary)

Did you understand SmoothOper's objection? Every time you select an n within N you can show that this n is not the largest. But no matter for how many n's you have shown this to be true, there still exists an n' for which you have not (yet) shown it to be true. Yes of course, you can now show it true for this n', but you don't have a finite proof (no last step).

I agree with that proof. It isn't necessary to generate every element, since he proved it for every element of the set each element behaving according to certain properties.

I have problems with proofs that have one object in the set and an infinite number of arguments. IE when the natural numbers are defined recursively.

[cyclops]

Does there exist a last post in this thread?

[skydyr]

Does there exist a last post in this thread?

Sorry, no.

[lemmata]

@Mike: I don't think you need the axiom of induction for such a proof. Peano 2 is sufficient.

1. n ∈ N -> n' ∈ N; n stands for any possible object of N, precisely: n holds a place for one object out of N that is assumed to be takeable by any object of N. (This is the reason why you can universally generalize from n to ∀n here. Maybe someone knows universal generalization and can show how to come from "n ∈ N -> n' ∈ N" to ∀n, n': (n ∈ N -> n' ∈ N) in a formal way).

2. Since there can't be any natural number that is not represented by n it follows that because of Peano 2 and mp no natural number cannot have no successor.

Not enough for you that Peano did? (thought it was necessary)

Did you understand SmoothOper's objection? Every time you select an n within N you can show that this n is not the largest. But no matter for how many n's you have shown this to be true, there still exists an n' for which you have not (yet) shown it to be true. Yes of course, you can now show it true for this n', but you don't have a finite proof (no last step).

To prove that there is no natural number without a successor, the axiom of induction is unnecessary. Axiom 2 says that every natural number has a successor. These two statements are logical negations of each other.

Order theory definition of maximal element: Let R be a binary relation on the set P. An element of P is said to be R-maximal if there does not exist a R-greater element of P.

The successor relation defines an order relation: For all pairs m,n in N, let mRn if and only if m succeeds n and say that "m is greater than n".

To get a total, reflexive, transitive, well-founded relation ≥ that extends R, you need the rest of the axioms.

The relevant facts: To show the non-existence of a natural number that is greater than or equal to (according to the order that is the transitive and reflexive closure of R) every natural number, you need the other axioms (which ensure that there is an appropriate extension of R to a nice total order). The non-existence of a natural number that has no greater natural number is logically equivalent to Axiom 2.

What this conversation reveals is that when we get down to the foundations of mathematics, even things like "largest" have to be defined carefully in order for two people not to talk past each other.

PS: I always found the well-ordering principle to be a much nicer way of stating the induction axiom. Equivalent, but one is more aesthetically pleasing than the other (subjectively speaking of course).

[SmoothOper]

Anyway has anyone else noticed how ugly, Bill Spight, Dr. Straw, and RB's arguments become when confronted with the fact that division isn't defined over integers? Not to practice apologetics, but it seems that academics have no sense of shame.

[RBerenguel]

Anyway has anyone else noticed how ugly, Bill Spight, Dr. Straw, and RB's arguments become when confronted with the fact that division isn't defined over integers? Not to practice apologetics, but it seems that academics have no sense of shame.

Weren't you banned? In any case, division is defined over the integers. It's called integer division, in that case.

[SmoothOper]

Anyway has anyone else noticed how ugly, Bill Spight, Dr. Straw, and RB's arguments become when confronted with the fact that division isn't defined over integers? Not to practice apologetics, but it seems that academics have no sense of shame.

Weren't you banned? In any case, division is defined over the integers. It's called integer division, in that case.

Ahem, can you rephrase that in a way that points out the flaws in your thinking?

For example you could say integer division is defined, but in such away that doesn't map to the rationals, therefore cantors proof is flawed. That would be a step in the right direction towards fixing your flawed character, though the character issue may never really resolve itself at thus rate, and at this point no amount of technical "mumbo jumbo" will allow you to save face, the argument is over.

[cyclops]

... In any case, division is defined over the integers. It's called integer division, in that case.

No need to fight nonsense with nonsense. Integer division is no division as a\b=c doesn't imply bc=a so indeed there is no true division operation defined over I-{0}.

[RBerenguel]

... In any case, division is defined over the integers. It's called integer division, in that case.

No need to fight nonsense with nonsense. Integer division is no division as a\b=c doesn't imply bc=a so indeed there is no true division operation defined over I-{0}.

Integer division is what we consider division over the integers. It just has sometimes remainders, because, well, it's not the proper inverse of product. It's no nonsense, and works in any other similar ring, and with it you can do lots of interesting things. If you want to close it and make it a proper inverse of products, you complete the ring to a field, that's it again, and with it you get even more neat properties and can do funny things like extending rings of polynomials to fields.

[cyclops]

Integer division is closed already ( i.e. defined on Z X Z0 ). What you close is some (proto-
) division defined on a specific subset of Z X Z0 ( consisting of the (proto-)dividable tuples ). a is protodivable by b if there is a number c such that bc==a. Thus you generate the rationals. That is what they taught me in my first year course "number theory" some forty+ years ago. Somewhere I must still have Walter Rudin's "Principles of Analysis", a very nice, be it dry book. It seems you mix up integer division with this proto division. But both are stricly no divisions on Z.
So what about SmoothOper? Strangely enough Cantor's proof has nothing to do with division or rationals so this whole discussion is futile.

[SmoothOper]

Integer division is closed already ( i.e. defined on Z X Z0 ). What you close is some (proto-
) division defined on a specific subset of Z X Z0 ( consisting of the (proto-)dividable tuples ). a is protodivable by b if there is a number c such that bc==a. Thus you generate the rationals. That is what they taught me in my first year course "number theory" some forty+ years ago. Somewhere I must still have Walter Rudin's "Principles of Analysis", a very nice, be it dry book. It seems you mix up integer division with this proto division. But both are stricly no divisions on Z.
So what about SmoothOper? Strangely enough Cantor's proof has nothing to do with division or rationals so this whole discussion is futile.

This is just what shamademians do when they've been caught red handed with a contradiction(claiming to be able to generate the rationals from integers without defining division on integers or rationals), they wheedle and nit pick, point at your credentials, insult you intelligence or knowledge, introduce a bunch of irrelevant verbiage, then say it's not politically popular, yada yada yada. Now, how do we want our crow?