Why is arithmetic difficult, II

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 S^n-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, 4, that you define

$2 = 1 + 1$ ,

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

and then prove the theorem.

About these ads

Posted in Uncategorized

Responses

[…] Why is arithmetic difficult, II « Mathematics under the Microscope […]
By: things to look at (June 10th - June 11th) | stimulant - changing things around. . . on June 12, 2008
at 3:01 am

Reply

M	T	W	T	F	S	S
« Apr				Jun »
			1	2	3	4
5	6	7	8	9	10	11
12	13	14	15	16	17	18
19	20	21	22	23	24	25
26	27	28	29	30	31

Mathematics under the Microscope