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

$7 + \square = 12$,

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:

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.

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