On the Implications of Ababou’s Constant Max Zhu 2018-05-09
Contents 1 Intro duction 1.1 Motivation . . . . . . . . . . . . . . . . . . 1.2 The Natural Numbers . . . . . . . . . . . . 1.2 1.2.1 Con Constr structi uction on of Natu atural ral Numbe umberrs . . 1.2 1.2.2 Arit Arithm hmeetic tic With With Na Natu turral Num Numbers bers . 1.2.3 The Natural Number bers Have No End 1.3 The number bers have an end (?!) . . . . . . . . 1.3.1 Ababou’s Constant . . . . . . . . . .
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2 Attempts to Clear Contradictions 2.1 Attempt 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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It is necessary to make a scientific revolution to correct the course of mathematics based on its theory of Numbers have an end. -Mohamed Ababou-
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1 1.1 1.1
Intr In trod oduc ucti tion on Moti Motiv vatio ation n
The majority of people with some interest in mathematics know that the natural numbers are unbounded, which is to say that the natural numbers have no “end”. It follows that rational numbers and real numbers also have no end. For this reason, Mohamed Ababou’s claim that the “numbers have an end” can be quickly quickly rejected rejected with little thought thought given. given. It does not help Ababou’s case that the majority of his arguments are incoherent, and his diagrams unintelligible. In fact, there are entire communities dedicated to the mockery of ideas such as those proposed by Ababou. In this paper I will first demonstrate that the numbers do not have an end using a set of fundamental axioms, then I will entertain Ababou’s idea that the numbers have an end to see what implications this would have. Specifically, this paper will revolve around these two questions: Q1. What contradictions, if any, arise from the assertion that numbers have an end? Q2. Can this assertion be developed into a consistent system?
1.2 1.2
The The Na Natu tura rall Num Number berss
Consider the set of natural numbers, N. It is common knowledg knowledgee that there is no largest natural number. number. Before Before proving this statement, statement, I will describe describe the Peano axioms which define the set N, as well as definitions of a few well known concepts. 1.2.1 1.2.1
Constru Constructi ction on of Natu Natural ral Num Numbers bers
Definition 1.2.1 (Peano Axioms) The Peano axioms are as follows: (i) 0 is a natural number. (ii) There is a function S such that for all natural numbers n, S(n) is a natural number, (iii) For all natural numbers m and n, m = n iff S(m) = S(n), and (iv) There is no natural number n such that S(n) = 0. (v) Let P(n) be a predicate such that P(0) is true and P(n) implies P(S(n)). Then, P(n) is true for every natural number n. Remark. Axiom (v) justifies justifies the use of mathematical mathematical induction. induction. Remark. The symbo symbols ls 1, 2, 3, ... are simpl simply y shortha shorthand nd for S(0), S(0), S(S(0)), S(S(0)), S(S(S(0))), S(S(S(0))), ...
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1.2.2 1.2.2
Arithm Arithmeti etic c With With Natura Naturall Numbers Numbers
Arithmetic is one of the first things we learn to do with numbers, so let us define some common arithmetical concepts. Definition 1.2.2 (Equality) Let m, n, z be natural numbers. Then, the equality relation (=) has the following properties: (Reflexive) m = m is true. (Symmetric) If m = n, then n = m. (Transitive) If m = n and n = z, then m = z. ( N is closed under equality) If m = n and m is a natural number, then n is a natural number as well. Definition 1.2.3 (Addition) Let m, n be natural numbers. Then, addition (+) defined as follows: + 0 = m m + S (n) = S (m + n) m
Definition 1.2.4 (Subtraction) Let m, n, z be natural natural numbers. numbers. Then, Then, subtraction (-) is defined as follows: m
− n = z ⇐⇒
m = n
+z
Definition Definition 1.2.5 (Inequalitie (Inequalities) s) Let m, n be natural numbers. Then, m ≤ n iff there is some natural number z such that m + z = n. Also, m ≥ n iff n ≤ m. Remark. If m ≤ n and m ≥ n, then m = n. I will not prove prove this this familiar familiar property of inequalities here. Now, we can prove the statement that Ababou rejects. 1.2.3 1.2.3
The Natur Natural al Numbe Numbers rs Have Have No End End
Here, “end” means some natural number greater than all other natural numbers. Lemma 1.2.6 For every natural number n, n is not equal to S(n). Proof. By induction. induction. Let Let n be a natural natural number. number. Since Since S(n) = 0 is false by (iv), therefore S(0) does not equal 0 for the base case. Suppose n does not equal S(n) for some natural number n. Then, S(n) does not equal S(S(n)) by (iii). Therefore by induction, the statement holds.
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Theorem 1.2.7 There is no natural number n such that m ∈ N =⇒
n
≥ m.
Proof. For contradict contradiction ion suppose suppose there there exists a natural natural number number n such that m ∈ N =⇒ n ≥ m. Now, n
+ 1 = n + S (0) (0) = S (n + 0) = S (n)
This follows from the definition definition of addition. addition. Thus, n ≤ S(n) by definition. Also, n ≥ ≥ S(n) by hypothesis, so n = S(n). This clearly contradicts our lemma, so n cannot exist. Remark. It easily follows follows that that there is no largest rational rational or real number number.. I will not provide a construction of these types of numbers in this section.
1.3 1.3
The The num number berss hav have an end end (?!) (?!)
In spite of numerous proofs to the contrary, Mohamed Ababou continually refuses to accept the fact that numbers do not end, often speaking through broken English and incomprehensible images. However, it may be interesting to examine the implications implications of taking Ababou’s Ababou’s claims at face value. That will be the goal of the rest of this paper. 1.3.1 1.3.1
Ababou’s Ababou’s Consta Constant nt
Definition 1.3.1 Define AB to be a natural number such that m ∈ AB ≥ m. Axiom 1.3.2 (Ababou’s axiom)
N
=⇒
AB exists.
Since Ababou is so persistent in his claim that the numbers have an end, it makes sense to define Ababou’s constant, or AB, to be the end of numbers. Note that the definition merely states that such a constant exists, and makes no claim on its value.
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Attempt ttemptss to to Clea Clearr Con Contra tradic dictio tions ns
From now on, we will do away with most of the axioms used in the introduction, and attempt to create a new system which follows Ababou’s axiom.
2.1 2.1
Attem ttemp pt 0
First, we must deal with the fact that Ababou’s axiom blatantly contradicts theorem theorem 1.2.6. To make our system consisten consistent, t, we must make changes changes to the axioms axioms we used. used. Let us look at each axiom axiom in turn, turn, see what what purpose purpose they they served in the old system, and make the necessary alterations. (i) 0 is a natural natural number. number. This axiom is necessary to guarantee a nonempty set. We should, however, modify it to include Ababou’s constant. (ii) There is a function function S such that for all natural numbers numbers n, S(n) is a natural natural number, This This axiom axiom is used used to repres represen entt the notion notion of “succe “successo ssor” r” or “next” “next”.. We should keep it as is. (iii) For all natural natural numbers numbers m and n, m = n iff S(m) = S(n) Now we run into a major problem. This axiom was was used to prove prove Lemma 1.2.5, 1.2.5, so we need to change change it. Let us try placing placing the restriction restriction that m and n cannot equal Ababou’s constant. (iv) There There is no natural number number n such that S(n) = 0. This axiom states that 0 is the “first” natural number. There are no problems here, as long as 0 is not equal to Ababou’s constant. (v) Let P(n) be a predicate such such that P(0) is true and P(n) P(n) implies P(S(n)). P(S(n)). Then, P(n) is true for every natural number n. (vi) S(AB) =
AB
Axiom (vi) is a new addition, which states that AB is indeed the end of the naturals, naturals, and that one can’t go beyond it through the successor successor function. By axiom (ii) S(AB) must be a natural number, and it makes the most sense to define it as AB. Here are our new axioms.
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Definition 2.1.1 (New Peano Axioms) The New Peano axioms are as follows: (i) 0 and AB are distinct natural numbers. (ii) There is a function S such that for all natural numbers n, S(n) is a natural number, (iii) For all natural numbers m and n such that neither m nor n are equal to AB, m = n iff S(m) = S(n), (iv) There is no natural number n such that S(n) = 0, and (v) S( AB AB) = AB (vi) Let P(n) be a predicate such that P(0) is true and P(n) implies P(S(n)). Then, P(n) is true for every natural number n. Let us see how addition works with Ababou’s constant. Theorem 2.1.2
AB
+ n = AB for all natural numbers n.
Proof. By induction. AB + 0 = AB by definition for the base case. Suppose AB + n = AB for some natural number n. Then, AB + S (n)
As required.
= S (AB + n) = S (AB) = AB
So far so good. Now let us see how Ababou’s constant works with subtraction. Theorem 2.1.3
AB
-
AB =
n is true for all natural numbers n.
Proof. Let n be be a natural natural number. number. By theorem theorem 2.1.2, AB + n = AB. Therefore by definition of subtraction, AB - AB = n. Corollary Corollary 2.1.3.1 1 2.1.3.1 1 = 2. Proof. By theorem, theorem,
AB
-
AB =
1 = 2.
Corollary Corollary 2.1.3.2 I 2.1.3.2 I am Mohamed Ababou. Proof. Consider Consider the the set set {Me, Mohamed Ababou }. Clearly, this set contains 2 elements. Since 1 = 2, this set contains 1 element, so me and Mohomed Ababou must be equal.
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