Para-consistent logic

[SmoothOper]

Fundamentally, as a computer scientist, I find it really annoying when Mathematicians play fast and loose with sets, then you try to implement their calculations and algorithms, and they don't work because they used the axiom of choice. Think compiler error.

Having proved that you are not a mathematician, you now prove that you are not a computer scientist.

I have this book on my table. Mathematical Go, do you think the Axiom of Choice is compatible with proofs in go, I know what if we could form an equivalency between certain positions in go, yeah yeah then we could order them yeah and there would be like quarter points

[RBerenguel]

Fundamentally, as a computer scientist, I find it really annoying when Mathematicians play fast and loose with sets, then you try to implement their calculations and algorithms, and they don't work because they used the axiom of choice. Think compiler error.

Having proved that you are not a mathematician, you now prove that you are not a computer scientist.

I have this book on my table. Mathematical Go, do you think the Axiom of Choice is compatible with proofs in go, I know what if we could form an equivalency between certain positions in go, yeah yeah then we could order them yeah and there would be like quarter points

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.

[Bill Spight]

For one thing, integers are not a subset of rationals, strictly speaking. A rational is an ordered pair of integers. True, there are rationals that are equal to integers, but that is not exactly the same thing.

Have you read Benacerraf's "What Numbers Could Not Be"? I don't have settled views here, but I have enough sympathy towards the mathematical structuralist view that I get a funny feeling when I hear that rationals are really just ordered pairs of integers.

Related: do you have an opinion about what the reals really are?

No, I have not read Benacerraf. Nor do I say that rationals are just ordered pairs of integers. They have certain defining properties. I am not at all up on structuralism. What I have been saying is what I learned as a kid. AFAIK, based upon what I just saw on Wikipedia about structuralism, the natural number 1, the integer 1, the rational number 1, the real number 1, the complex number 1, and the game (Conway number) 1, are all different, from a structuralist point of view, because they are part of different structures. Dunno. You could enlighten me on that.

I do not have an opinion about what real numbers really are. I like Dedekind cuts, I do not like numbers born on day omega. {shrug} Buckminster Fuller thought that real numbers do not occur in nature. When I first heard that, I was skeptical, but now I tend to agree. If there are actual space-time continua, why quantum mechanics?

Con permiso, let me sketch out some of these different number systems, as I learned about them.

In the beginning was counting, such as notching sticks to denote numbers of cows, sheep, or days, etc. (Not what I learned as a kid, but from a cognitive point of view, counting numbers may be seen as part of sensory-motor schemata, which are basic preverbal schemata.) Today there is some question of whether to include 0 in the natural numbers, but historically, we know that people counted before they had a concept of zero.

There is no problem adding natural numbers, as the sum of two natural numbers is also a natural number. (Natural numbers are closed under addition.) However, there is a problem with subtraction. 5 - 2 = 3, but what is 2 - 5?

Well, 0 and negative numbers were discovered/invented. They were not natural numbers, but integers. So what are integers? Integers may be constructed from natural numbers, as pairs of integers. (Structure! )

First, let's define equality for these integers. (A,B) = (C,D) iff A+D = B+C.

Now addition: (A,B) + (C,D) = (A+C,B+D)

Now subtraction: (A,B) - (C,D) = (A+D,B+C)

Note that we can define subtraction for integers in terms of addition for natural numbers. Pretty cool, eh?

Well, we do not write integers this way. How can we write them in the usual fashion? Using -> to indicate "write as", here is how.

(A,A) -> 0
(A+1,A) -> +1
(A,A+1) -> -1
Etc.

Bingo! (Note that I have written +1 for the integer, to distinguish it from the natural number 1. Conventionally we can drop the +. )

Well, the integers are not closed under division. We can extend the integers to the rationals and define them as pairs of integers, as long as the second integer is not 0. When we do so, we can define division of rationals in terms of the multiplication of integers. (See my earlier post about that.) Our write rules are these.

(A,A) -> 1
(A,B) -> A/B

The rationals are not closed under exponentiation. (The ancient Greeks discovered irrational "numbers", such as the square root of 2, but I am not sure that they regarded them as numbers.) We can extend them to the real numbers, but not so simply as we extended the natural numbers and the integers. Most real numbers have no finite representations.

There is no solution to the equation, x^2 + 1 = 0, in real numbers, but we can extend the reals to the complex numbers. A complex number is an ordered pair of real numbers. (Ordered pairs again! )

There are other kinds of numbers, surreals, hyperreals, etc., each belonging to different structures.

Above I referred to different kinds of 1s. Is there an archetype, 1? Maybe so. See Mick Jagger in the movie, "Performance", as an embodiment of the First Arcanum of the tarot.

[Bill Spight]

Fundamentally, as a computer scientist, I find it really annoying when Mathematicians play fast and loose with sets, then you try to implement their calculations and algorithms, and they don't work because they used the axiom of choice. Think compiler error.

Having proved that you are not a mathematician, you now prove that you are not a computer scientist.

I have this book on my table. Mathematical Go, do you think the Axiom of Choice is compatible with proofs in go, I know what if we could form an equivalency between certain positions in go, yeah yeah then we could order them yeah and there would be like quarter points

I hope that you have enjoyed the book.

[hyperpape]

I shall have to say more when I am not on a phone. But the Benacerraf paper is just 12 pages, and delightful (minus a bit about lions in the middle). http://isites.harvard.edu/fs/docs/icb.t ... cerraf.pdf

[RBerenguel]

I shall have to say more when I am not on a phone. But the Benacerraf paper is just 12 pages, and delightful (minus a bit about lions in the middle). http://isites.harvard.edu/fs/docs/icb.t ... cerraf.pdf

Added to my "to-read PDFs" Seems pretty light, but even some lines of logic formalisms scare me, I've always had a hard time with it (stupidly enough formal logic was not part of our degree.)

[Bill Spight]

I shall have to say more when I am not on a phone. But the Benacerraf paper is just 12 pages, and delightful (minus a bit about lions in the middle). http://isites.harvard.edu/fs/docs/icb.t ... cerraf.pdf

Thanks for the reference to the Benacerraf paper. It seems that he and I pretty much agree that, as he states, "numbers are not objects at all, because in giving properties . . . of numbers you merely characterize an abstract structure - and . . . the 'elements' of the structure have no properties other than those relating them to other 'elements' of the same structure." As I indicated, I allow for the possibility of numerical archetypes, but they are not merely mathematical in nature.

[Pippen]

No, L was the largest, by assumption. In that case there is no natural number, L+1 (or otherwise), which could also be the largest.

The problem is that this assumption was wrong, so L wasn't the largest one and that leaves the question open if not some other number is the largest one.

My problem is this: What the hell is a variable?

A variable is a placeholder for an (one!) arbitrary object of a set. If that's the definition I'd object that it would mean that a variable like x could just stand for one object at a time and that means it wouldn't stand for the other objects at that time. E.g. L would stand just for one assumed largest number and that means even if we prove that L isn't the largest one it'd follow nothing for all the numbers that weren't represented by L.

Therefore I'd define: A variable is a placeholder for an (one!) arbitrary object of a set, where we assume/clarify additionally that all objects of the set can be put in this placeholder. Then L would have represented basically all natural numbers and then we could say: Every L has a L+1 and if L stands somehow for all numbers than there cannot be any maximum number.

[Tryss]

The problem is that this assumption was wrong, so L wasn't the largest one and that leaves the question open if not some other number is the largest one.

My problem is this: What the hell is a variable?

A variable is a placeholder for an (one!) arbitrary object of a set. If that's the definition I'd object that it would mean that a variable like x could just stand for one object at a time and that means it wouldn't stand for the other objects at that time. E.g. L would stand just for one assumed largest number and that means even if we prove that L isn't the largest one it'd follow nothing for all the numbers that weren't represented by L.

Therefore I'd define: A variable is a placeholder for an (one!) arbitrary object of a set, where we assume/clarify additionally that all objects of the set can be put in this placeholder. Then L would have represented basically all natural numbers and then we could say: Every L has a L+1 and if L stands somehow for all numbers than there cannot be any maximum number.

But L is not a placeholder, it's a label, a name you attach to the number defined by "the largest number". If you prefer you can remove the variable altogether and write the proof without it :

The successor of "the largest number" is larger than "the largest number", so it's contradictory and "the largest number" doesn't exist

Now where you can object is that "can I use an object I just defined by a propriety and not constructed?"

[Bill Spight]

No, L was the largest, by assumption. In that case there is no natural number, L+1 (or otherwise), which could also be the largest.

The problem is that this assumption was wrong, so L wasn't the largest one and that leaves the question open if not some other number is the largest one.

My problem is this: What the hell is a variable?

A variable is a placeholder for an (one!) arbitrary object of a set. If that's the definition I'd object that it would mean that a variable like x could just stand for one object at a time and that means it wouldn't stand for the other objects at that time. E.g. L would stand just for one assumed largest number and that means even if we prove that L isn't the largest one it'd follow nothing for all the numbers that weren't represented by L.

Therefore I'd define: A variable is a placeholder for an (one!) arbitrary object of a set, where we assume/clarify additionally that all objects of the set can be put in this placeholder. Then L would have represented basically all natural numbers and then we could say: Every L has a L+1 and if L stands somehow for all numbers than there cannot be any maximum number.

Here is what I wrote:

Suppose that there is a largest natural number. Call it L. Then there is a natural number, L + 1. Call it M. Then M > L, which means that L is not the largest natural number, and our supposition is false.

As Tryss points out, L and M are labels. ("Call it L. . . . Call it M.") Logical variables. Pronouns, if you will.

I could have written, "Suppose that there is a largest natural number. Then it has a successor. . . ." Instead of the pronoun, "it", I used "L", that's all.

[Mike Novack]

No, L was the largest, by assumption. In that case there is no natural number, L+1 (or otherwise), which could also be the largest.

The problem is that this assumption was wrong, so L wasn't the largest one and that leaves the question open if not some other number is the largest one.
.

No it doesn't, because that "induction" applied to all members of the set (of the "things" that are natural numbers) is one of the axioms. If this axiom has not been necessary to capture the essence of what the natural numbers "are" it wouldn't have been included.

I'll restate that axiom: For all members of the set N, if S (any statement) being true for n means it is true for the successor of n and it is true for the member of the set that is not the successor of any other, then it is true for all members of the set.

So, you prove (using the axioms) that if L is not the largest member of the set then successor of L is not the largest member <you show that successor(successor (L)) is in the set> and that 0 is not the largest member of the set. Once you have that you use the induction axiom to say true for all members of the set (not the largest).

[Bill Spight]

What the hell is a variable?

A variable is a placeholder for an (one!) arbitrary object of a set. If that's the definition I'd object that it would mean that a variable like x could just stand for one object at a time and that means it wouldn't stand for the other objects at that time.

Well, a logical variable is bound (stands for something, without existential import) only once. That's it. It cannot stand for something one time and for something else another time. Once bound, it remains bound.

So if L stands for the largest element in a set that can have only one, and we prove that there is an element that is larger than L, we do not have to try again. In fact, we do not get to try again. L is already bound to the largest element, whatever it may be. The largest element is L. If we try to bind another variable, M, to the largest element, then M = L. If we have already shown that M > L, we cannot do that.

[hyperpape]

Let's back up: do you accept proof by contradiction, Pippen?

[Mike Novack]

Let's back up: do you accept proof by contradiction, Pippen?

That's an important question.

But you don't need it for this one. You just need to prove these two and then use the induction axiom.

1) If L is not the largest member of the set then successor(L) is not the largest member of the set.
2) The one member of the set that is not the successor of any member is not the largest.

[Pippen]

If you prefer you can remove the variable altogether and write the proof without it :

The successor of "the largest number" is larger than "the largest number", so it's contradictory and "the largest number" doesn't exist

Now where you can object is that "can I use an object I just defined by a propriety and not constructed?"

I would counter like this (just that you know my line of thought, I know it must be wrong or I'd be more clever than all the math since 2500 years which is very very unlikely^^):

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....

Yes, I accept proof of contradiction. The problem is - as above shown - that I think it leads to an infinite proof-sequence. Only if a variable x stands basically for all natural numbers then you could show directly that no number has no successor which leads to the proposistion: "No natural number can be the largest one, because all have an successor" (=infinity).

I just do not know the most basic stuff: How does a variable work? 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?