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 […] 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 number n, S(n) is a natural number.Axioms 1 and 2 define a unary representation of the natural numbers: the number 2 is S(1), and, in general, any natural number n is Sn-1(1). The next two axioms define the properties of this representation.
Axiom 3. For every natural number n other than 1, S(n) ≠ 1. That is, there is no natural number whose successor is 1.
Axiom 4. For all natural numbers m and n, if S(m) = S(n), then m = n. That is, S is an injection.The final axiom, sometimes called the axiom of induction, is a method of reasoning about all natural numbers; it is the only second order axiom.
Axiom 5. If K is a set such that:
- 1 is in K, and
- for every natural number n, if n is in K, then S(n) is in K,
then K contains 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
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
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.