I expand an earlier post with the same title.

My colleague EHK told me about a difficulty she experienced in her first encounter with arithmetic, aged 6. She could easily solve “put a number in the box” problems of the type

,

buy counting how many 1′s she had to add to 7 in order to get 12

but struggled with

,

because she did not know where to start. Worse, she felt that she could not communicate her difficulty to adults.

A brief look at Peano axioms for formal arithmetic provides some insight in EHK’s difficulties. I quote Wikipedia, with slight changes:

The Peano axioms define the properties of

natural numbers, usually represented as a setNor [I skip axioms for equality relation — AB.]The [...] axioms define the properties of the natural numbers. The constant 1 is assumed to be a natural number, and the naturals are assumed to be closed under a “successor” function S(n).

Axiom 1. 1 is a natural number.

Axiom 2. For every natural numbern,S(n) is a natural number.Axioms 1 and 2 define a unary representation of the natural numbers: the number 2 isS(1), and, in general, any natural numbernisS^{n-1}(1). The next two axioms define the properties of this representation.Axiom 3. For every natural number

nother than 1,S(n) ≠ 1. That is, there is no natural number whose successor is 1.

Axiom 4. For all natural numbersmandn, ifS(m) =S(n), thenm=n. That is,Sis an injection.The final axiom, sometimes called theaxiom of induction, is a method of reasoning about all natural numbers; it is the only second order axiom.Axiom 5. If

Kis a set such that:

1is inK, and- for every natural number
n, ifnis inK, thenS(n) is inK,then

Kcontains every natural number.

Thus, Peano arithmetic is a formalisation of that very counting by one that little EHK did, and addition is defined in a precisely the same way as EHK learned to do it: by a recursion

Commutativity of addition is a non-trivial (although still accessible to a beginner) theorem. To force you to feel some sympathy to poor little EHK, I reproduce its proof from Edmund Landau’s famous book *Grundlangen der Analysis*, 1929. I am using notation from Landau’s book: .

**Theorem 1:** If then .

Proof: Otherwise, we would have and hence, by Axiom 4, .

**Theorem 2:** .

Proof: Let be the set of all for which this holds true.

I) By Axiom 1 and Axiom 3, , therefore belongs to .

II) If belongs to , then , and hence by Theorem 1, , so that belongs to . By Axiom 5, therefore contains all the natural numbers, i.e. we have for each that .

**Theorem 3:** If , then there exists one (hence, by Axiom 4, exactly one) such that .

Proof: Let be the set consisting of the number and of all those for which there exists such a . (For any such , we have of necessity that by Axiom 3.)

I) belongs to .

II) If belongs to , then, with denoting the number , we have , so that belongs to .

By Axiom 5, therefore contains all the natural numbers; thus for each there exists a such that .

**Theorem 4, and at the same time Definition 1:** To every pair of numbers , we may assign in exactly one way a natural number, called ( to be read “plus”), such that

1) for every ,

2) for every and every .

is called the sum of and or the number obtained by

addition of to .

Proof: A) First we will show that for each fixed there is at most one possibility of defining for all in such a way that and for every .

Let and be defined for all and be such that

, for every .

Let be the set of all for which

.

I) ; hence belongs to .

II) If belongs to , then , hence by Axiom 2, , therefore

,

so that belongs to .

Hence is the set of all natural numbers; i.e. for every we have .

B) Now we will show that for each it is actually possible to define for all in such a way that

and for every .

Let be the set of all for which this is possible (in exactly one way, by A) ) .

I ) For , the number is as required, since

,

.

Hence belongs to .

II) Let belong to , so that there exists an for all .

Then the number is the required number for , since

and

.

Hence belongs to . Therefore contains all .

Theorem 5 (Associative Law of Addition):

.

Proof: skipped, since not needed for proving commutativity of addition.

**Theorem 6 (Commutative Law of Addition):**

.

Proof: Fix , and let be the set of all for which the assertion holds.

I) We have and furthermore, by the construction in the proof of Theorem 4,

, so that

and belongs to .

II) If belongs to , then , therefore

.

By the construction in the proof of Theorem 4, we have

hence

,

so that belongs to . The assertion therefore holds for all .

*******

Landau’s book is characterised by a specific austere beauty of entirely formal axiomatic development, dry, cut to the bone, streamlined. It is claimed that logical austerity and precision were Landau’s characteristic personal traits. [Asked for a testimony to the effect that Emmy Noether was a great woman mathematician, he famously said: “I can testify that she is a great mathematician, but that she is a woman, I cannot swear.”]

Grundlagen der Analysis opens with two prefaces, one intended for the student and the other for the teacher. The preface for the student begins thus:

1. Please don’t read the preface for the teacher.

2. I will ask of you only the ability to read English and to think logically-no high school mathematics, and certainly no higher mathematics. [...]

3. Please forget everything you have learned in school; for you haven’t learned it.

Please keep in mind at all times the corresponding portions of your school curriculum; for you haven’t actually forgotten them.

4. The multiplication table will not occur in this book, not even the theorem,

,

but I would recommend, as an exercise for Chap. I, 4, that you define

,

,

and then prove the theorem.

[...] Why is arithmetic difficult, II « Mathematics under the Microscope [...]

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things to look at (June 10th - June 11th) | stimulant - changing things around. . .on June 12, 2008at 3:01 am