18 Jan 2019

Griss (1.1) “Negationless Intuitionistic Mathematics, I” Section 1.1, “Construction of the Natural Numbers”, summary

 

by Corry Shores

 

[Search Blog Here. Index-tags are found on the bottom of the left column.]

 

[Central Entry Directory]

[Mathematics, Calculus, Geometry, Entry Directory]

[Logic and Semantics, entry directory]

[Griss, entry directory]

[Griss, “Negationless Intuitionistic Mathematics, I”, entry directory]

 

 

[The following is summary. I am not a mathematician, so please consult the original text instead of trusting my summarizations, which are surely mistaken or inelegantly articulated. Bracketed comments are my own. Proofreading is incomplete, so please forgive my mistakes.]

 

 

 

 

Summary of

 

George François Cornelis Griss

(G.F.C. Griss)

 

“Negationless Intuitionistic Mathematics, I”

 

1

“The Natural Number”

 

1.1

“Construction of the Natural Numbers”

 

 

 

 

 

Brief summary:

(1.1.1) We will construct the natural numbers using negationless intuitionistic mathematical principles (see section 0). We first simply imagine an object, call it “1”. It remains the same. Thus it is the same as 1. The symbolic formulation for this is: 1 = 1. (1.1.2) We next imagine another object that we call 2, which is also selfsame, meaning that, in symbolic formulation, 2 = 2; and, these two objects are distinguishable from one another, or in symbolic formulation, 1 ≠ 2, 2 ≠ 1. (1.1.3) Objects 1 and 2 (see sections 1.1.1 and 1.1.2) form a set. So 1 and 2 are members of the set {1, 2}. (For now, the set is simply these two.) If an object were to belong to this set, that object would be either 1 or 2. If that object is distinguishable from 1, then it is 2. If that object is distinguishable from 2, then it is 1. (1.1.4) We next imagine another object and set element. We call it 3. It remains selfsame, so in symbolic formulation, 3 = 3. Also, 3 is distinguishable from 1 and 2, so in symbolic formulation, 1 ≠ 3, 3 ≠ 1, 2 ≠ 3, 3 ≠ 2. (1.1.5) Objects 1, 2, and 3 (see sections 1.1.1, 1.1.2, and 1.1.4) form the set {1, 2, 3}. (The set is limited to these three.) Any object belonging to this set would  be either 1, 2, or 3.  So, “if it is distinguishable from 3, it is an element of {1, 2}.” (1.1.6) We can also imagine there being any additional number to the set that is selfsame and distinguishable from the rest of the members: “If, in this way, we have proceeded to {1, 2, …, n}, we can, again, imagine an element n′, remaining the same, n′ = n′, and distinguishable from each element p of {1, 2, ... , n}, in formula n′p, pn′.” (1.1.7) The set member n′ in addition to the set {1, 2, …, n} (see section 1.1.6) form the set {1, 2, …, n′}. Any number belonging to {1, 2, …, n′} either is a member of {1, 2, ... , n} or it is n′ itself. We can determine which in the following way. “If it is distinguishable from each element of {1, 2, ... , n}, it is n′; if it is distinguishable from n′, it is an element of {1, 2, ... , n}.” (1.1.8) We can obtain a finite set {1, 2, …, m} if we cease our additions with the mth element. Or we can obtain the countably infinite set {1, 2, …} by proceeding with the additions unlimitedly. (1.1.9) If we want large sets and we choose a new symbol for each one, then the symbolization can become difficult. (Either a large number of distinct simple symbols will need to be continuously invented, or redundancy methods, like simply combining strokes or even using numerative systems like decimal, will sooner or later create symbols that become unmanageably long.)

 

 

 

 

 

 

Contents

 

1.1.1

[A Selfsame Object as 1 = 1]

 

1.1.2

[Another Distinct Selfsame Object, 2]

 

1.1.3

[Set {1, 2}]

 

1.1.4

[Yet Another Distinct Selfsame Object, 3]

 

1.1.5

[Set {1, 2, 3}]

 

1.1.6

[Adding Additional Members Indefinitely]

 

1.1.7

[Determining Members of the Set of Numbers {1, 2, …, n′}]

 

1.1.8

[Finite and Infinite Sets]

 

1.1.9

[Difficulties with Representing All the Distinct Members of Large Sets]

 

 

 

 

 

Summary

 

1.1.1

[A Selfsame Object as 1 = 1]

 

[We will construct the natural numbers using negationless intuitionistic mathematical principles (see section 0). We first simply imagine an object, call it “1”. It remains the same. Thus it is the same as 1. The symbolic formulation for this is: 1 = 1.]

 

[ditto]

Imagine an object, e.g. 1. It remains the same 4), 1 is the same as 1, in formula 1 = 1.

(1131)

4) Here, again, I do not enter into philosophical questions. Cf. my philosophical sketch “Idealistische filosofie” (Van Loghum Slaterus, Arnhem, 1946).

(1131)

[contents]

 

 

 

 

 

 

1.1.2

[Another Distinct Selfsame Object, 2]

 

[We next imagine another object that we call 2, which is also selfsame, meaning that, in symbolic formulation, 2 = 2; and, these two objects are distinguishable from one another, or in symbolic formulation, 1 ≠ 2, 2 ≠ 1.]

 

[ditto]

Imagine another object, remaining the same, and distinguishable 4) from 1. e.g. 2; 2 = 2; 1 and 2 are distinguishable (from one another), in formula 1 ≠ 2, 2 ≠ 1.

(1131)

4) Here, again, I do not enter into philosophical questions. Cf. my philosophical sketch “Idealistische filosofie” (Van Loghum Slaterus, Arnhem, 1946).

(1131)

[contents]

 

 

 

 

 

 

1.1.3

[Set {1, 2}]

 

[Objects 1 and 2 (see sections 1.1.1 and 1.1.2) form a set. So 1 and 2 are members of the set {1, 2}. (For now, the set is simply these two.) If an object were to belong to this set, that object would be either 1 or 2. If that object is distinguishable from 1, then it is 2. If that object is distinguishable from 2, then it is 1.]

 

[ditto]

They form the set {1, 2}; 1 and 2 belong to the set. If conversely an object belongs to this set, it is 1 or 2. If it is distinguishable from 1, it is 2; if it is distinguishable from 2, it is 1.

(1131)

[contents]

 

 

 

 

 

 

1.1.4

[Yet Another Distinct Selfsame Object, 3]

 

[We next imagine another object and set element. We call it 3. It remains selfsame, so in symbolic formulation, 3 = 3. Also, 3 is distinguishable from 1 and 2, so in symbolic formulation, 1 ≠ 3, 3 ≠ 1, 2 ≠ 3, 3 ≠ 2.]

 

[ditto]

Imagine again another object (element), remaining the same and distinguishable from 1 and from 2, e.g. 3; 3 = 3; 1 and 3 are distinguishable, 2 and 3 too, in formula 1 ≠ 3, 3 ≠ 1, 2 ≠ 3, 3 ≠ 2.

(1131)

[contents]

 

 

 

 

 

 

1.1.5

[Set {1, 2, 3}]

 

[Objects 1, 2, and 3 (see sections 1.1.1, 1.1.2, and 1.1.4) form the set {1, 2, 3}. (The set is limited to these three.) Any object belonging to this set would  be either 1, 2, or 3.  So, “if it is distinguishable from 3, it is an element of {1, 2}.”]

 

[ditto]

They form the set {1, 2, 3}. If an element belongs to {1, 2, 3}, it belongs to 1, 2 or it is 3. If it is distinguishable from each element of {1, 2}, it is 3; if it is distinguishable from 3, it is an element of {1, 2}.

(1131)

[contents]

 

 

 

 

 

 

1.1.6

[Adding Additional Members Indefinitely]

 

[We can also imagine there being any additional number to the set that is selfsame and distinguishable from the rest of the members: “If, in this way, we have proceeded to {1, 2, …, n}, we can, again, imagine an element n′, remaining the same, n′ = n′, and distinguishable from each element p of {1, 2, ... , n}, in formula n′p, pn′.”]

 

[ditto]

If, in this way, we have proceeded to {1, 2, …, n}, we can, again, imagine an element n′, remaining the same, n′ = n′, and distinguishable from each element p of {1, 2, ... , n}, in formula n′p, pn′.

(1131)

[contents]

 

 

 

 

 

 

1.1.7

[Determining Members of the Set of Numbers {1, 2, …, n′}]

 

[The set member n′ in addition to the set {1, 2, …, n} (see section 1.1.6) form the set {1, 2, …, n′}. Any number belonging to {1, 2, …, n′} either is a member of {1, 2, ... , n} or it is n′ itself. We can determine which in the following way. “If it is distinguishable from each element of {1, 2, ... , n}, it is n′; if it is distinguishable from n′, it is an element of {1, 2, ... , n}.”]

 

[ditto]

They form the set {1, 2, …, n′}. If an element belongs to {1, 2, …, n′}, it belongs to {1, 2, ... , n} or it is n′. If it is distinguishable from each element of {1, 2, ... , n}, it is n′; if it is distinguishable from n′, it is an element of {1, 2, ... , n}.

(1131)

[contents]

 

 

 

 

 

 

1.1.8

[Finite and Infinite Sets]

 

[We can obtain a finite set {1, 2, …, m} if we cease our additions with the mth element. Or we can obtain the countably infinite set {1, 2, …} by proceeding with the additions unlimitedly.]

 

[ditto]

We can cease with the mth element and get a finite set {1, 2, …, m}; we can also proceed unlimitedly and get the countably infinite set {1, 2, …}.

(1131)

[contents]

 

 

 

 

 

 

 

1.1.9

[Difficulties with Representing All the Distinct Members of Large Sets]

 

[If we want large sets and we choose a new symbol for each one, then the symbolization can become difficult. (Either a large number of distinct simple symbols will need to be continuously invented, or redundancy methods, like simply combining strokes or even using numerative systems like decimal, will sooner or later create symbols that become unmanageably long.)]

 

[ditto]

If one continuously chooses an arbitrary symbol as a new object, this soon leads to difficulties. One can try to prevent this by proceeding systematically in the choice of new symbols. So you can take I, II, III, etc.; after proceeding to a certain symbol you get a new symbol by adding a I. This is not practical either. It is better to use a numerative system, though, in principle, the same difficulties arise.

(1131)

[contents]

 

 

 

 

 

 

 

 

Bibliography:

 

Griss, G.F.C. (1946). “Negationless Intuitionistic Mathematics, I,’’ Proceedings of the Koninklijke Nederlandse Akademie van Wetenschappen, 49, 1127–1133.

Journal PDF here:

http://www.dwc.knaw.nl/DL/publications/PU00014659.pdf

Article PDF here:

http://www.dwc.knaw.nl/DL/publications/PU00018278.pdf

Listing of Griss at this journal:

http://www.dwc.knaw.nl/toegangen/digital-library-knaw/?pagetype=publist&search_author=PE00000531

 

.

No comments:

Post a Comment