Posted by: Alexandre Borovik | May 26, 2008

Why is arithmetic difficult, II

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

7 + \square = 12,

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

but struggled with

\square +6 =11,

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 set N or \mathbb{N}. [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 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

m + 1 =S(m); \quad m+S(n) = S(m+n).

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: S(n) = n'.

Theorem 1: If x\ne y then x' \ne y'.

Proof: Otherwise, we would have x' = y' and hence, by Axiom 4, x=y.

Theorem 2: x' \ne x.

Proof: Let \mathcal{M} be the set of all x for which this holds true.

I) By Axiom 1 and Axiom 3, 1' \ne 1, therefore 1 belongs to \mathcal{M}.

II) If x belongs to \mathcal{M}, then x' \ne x, and hence by Theorem 1, (x')' \ne x', so that x' belongs to \mathcal{M}. By Axiom 5, \mathcal{M} therefore contains all the natural numbers, i.e. we have for each x that x' \ne x.

Theorem 3: If x \ne 1, then there exists one (hence, by Axiom 4, exactly one) u such that x=u'.

Proof: Let \mathcal{M} be the set consisting of the number 1 and of all those x for which there exists such a u. (For any such x, we have of necessity that x \ne 1 by Axiom 3.)

I) 1 belongs to \mathcal{M}.

II) If x belongs to \mathcal{M}, then, with u denoting the number x, we have x '=u', so that x' belongs to \mathcal{M}.

By Axiom 5, \mathcal{M} therefore contains all the natural numbers; thus for each x\ne 1 there exists a u such that x=u'.

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

1) x + 1 = x' for every x,

2) x + y'  = (x + y)' for every x and every y.

x + y is called the sum of x and y, or the number obtained by
addition of y to x.

Proof: A) First we will show that for each fixed x there is at most one possibility of defining x + y for all y in such a way that x+ 1 = x' and x + y'  = (x + y)'   for every y.
Let a_y and b_y be defined for all y and be such that

a_1 = x', \quad     b_1 = x', a_{y'} = (a_y)', \quad b_{y'} = (b_y)' for every y.
Let \mathcal{M} be the set of all y for which

a_y = b_y .

I) a_1 = x' = b_1; hence 1 belongs to \mathcal{M}.

II) If y belongs to \mathcal{M}, then a_y = b_y, hence by Axiom 2, (a_y)' =  (b_y)', therefore

a_{y'} =  (a_y)' = (b_y)' = b_{y'},

so that y' belongs to \mathcal{M}.

Hence \mathcal{M} is the set of all natural numbers; i.e. for every y we have a_y=b_y.

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

x + 1 = x' and x + y' = (x + y)' for every y.

Let \mathcal{M} be the set of all x for which this is possible (in exactly one way, by A) ) .

I ) For x=1, the number x+ y =y' is as required, since

x + 1=1'=x',

x + y' = (y')' = (x + y)'.

Hence 1 belongs to \mathcal{M}.

II) Let x belong to \mathcal{M}, so that there exists an x + y for all y.
Then the number x'+y=(x+y)' is the required number for x', since
x' + 1 = (x + 1)' = (x')'

and

x'+ y' = (x+y')' = ((x+y)')' =(x' + y)'.

Hence x' belongs to \mathcal{M}. Therefore \mathcal{M} contains all x.

Theorem 5 (Associative Law of Addition):

(x+y)+z=x+(y+ z).

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

Theorem 6 (Commutative Law of Addition):

x+y=y+ x.

Proof: Fix y , and let \mathcal{M} be the set of all x for which the assertion holds.

I) We have y+1=y' and furthermore, by the construction in the proof of Theorem 4,
1 + y -=y', so that

1+ y =y+ 1

and 1 belongs to \mathcal{M}.

II) If x belongs to \mathcal{M}, then x+y=y+x, therefore

(x + y)' = (y + x)' = y + x'.

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

x' + y = (x + y)'

hence

x'+ y=y+x',

so that x' belongs to \mathcal{M}. The assertion therefore holds for all x.

***

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,

2 \times 2 = 4,

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

2 = 1 + 1,

4 = (((1 + 1) + 1) + 1),

and then prove the theorem.

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