Good book recommendation on number systems.

[Ackbach]

What would be a good book that starts with the Peano axioms about the natural numbers and constructs, rigorously, the natural numbers, whole numbers, integers, rationals, reals, and finally complexes? I would prefer it if the book did not leave out major steps, but instead went in a logical order.

Ideally, it would do this and only this, unless it is your opinion that there should be one or two more intermediate steps or final steps (quaternions?). I am interested in a book that doesn't divert from this path much, if at all.

I don't have a terrific idea of the prerequisites involved; I'm guessing some set theory and maybe mathematical logic? Analysis? Topology? The more self-contained the book, the better, to a point.

Thoughts?

Thanks very much for your time!

 

[Petek]

The classic book on this topic, which almost exactly meets your requirements, is Landau's Foundations of Analysis. The writing is very dry, however.

 

[Ackbach]

The classic book on this topic, which almost exactly meets your requirements, is Landau's Foundations of Analysis. The writing is very dry, however.

That looks perfect! Thanks so much!

 

[Deveno]

It would be helpful if you indicated *why* you desire such a book.

For example, you say:

... that starts with the Peano axioms about the natural numbers and constructs, rigorously, the natural numbers...

If one constructs the natural numbers, say via the von Neumann construction, one can then prove many (but not all, at least not without assuming some *other* things, such as the axiom of infinity) of the Peano axioms.

On the other hand, the Peano axioms (at least the second-order variety) determine, up to isomorphism, the natural numbers, and thus there is no need to "construct" them.

An abbreviated "modern construction":

Start with a *natural number object* $N$: a recursively defined, recursively enumerable set that is *stable* (here, this means it supports recursive definitions with parameters). By construction, induction is valid for this set.

Show that it is possible to define two binary operations $+$ and $\ast$ such that:

1. $(N,+,0)$ is a commutative monoid
2. $(N,\ast,1)$ is a commutative monoid
3. The function $L_x: N \to N$ given by $L_x(n) = x\ast n$ is monoid-homomorphism of $(N,+)$ into itself.

The above are often listed separately as associativity, commutativity and identity axioms (1 & 2); (3) is better known as the distributive law.

Since our natural number object comes with an injective function $s:N \to N\setminus\{0\}$, the two operations are usually defined (recursively) as:

$n + 0 = n$
$n + s(m) = s(n + m)$

$n \ast 0 = 0$
$n \ast s(m) = n + (n \ast m)$

One then constructs the Grothendieck groupification of $N$, with equivalence multiplier $1$ (since $N$ is cancellative). Essentially, this entails "adjoining the negative numbers", but I mention it here because it is analogous to the later construction of the rationals (in fact, the multiplicative group of the non-zero rationals *is* the Grothendieck groupification of the monoid of non-zero integers).

Then one forms the field of quotients of the integral domain $\Bbb Z$, both these constructions are "categorical" in the sense that they represent the "minimal" way we can extend our near-ring to a ring, and our ring to a field (it turns out that the cancellation property of natural numbers:

$a\ast b = a\ast c \implies b = c$ is *very important*).

Next, one takes the Cauchy completion of the rational numbers. This is arguably the "hardest part", and the "construction via Dedekind cuts" is probably the most accessible with a minimum of fuss.

There are certain subtleties in dealing with negative real numbers in the Dedekind construction, which often leads authors (I believe Landau does this) to define the "positive cone" first (something we can only do in the presence of an order-something $\Bbb Q$ inherits from $\Bbb Z$ who in turn inherits it from $\Bbb N$).

Finally, the complexes are formed by adjoining a square root of $-1$ (it doesn't matter which one, a fact which essentially says complex-conjugation is a field automorphism that fixes the reals), or equivalently, forming the quotient ring $\Bbb R[x]/\langle x^2 + 1\rangle$

***************

Landau's book is a classic, and if you're doing this for your own personal edification, one cannot do much better (it is a bit dated). As to prerequisites? Well, that's a bit tricky-although completeness is, essentially a topological property (it has to do with properties of subsets of our structure), one doesn't *need* to know topology to construct the real numbers, although topology helps us understand *why* we desire it as a property. Also, if one is basically just going to "accept" that one has a decent idea of what natural numbers are (and this, surprisingly, is a matter of some contention), then a lot of the set theory can be dispensed with (a basic understanding of "limited extensionality" that is, that for a property $P$ and a set $S$, that $\{x \in S: P(x)\}$ is a well-defined set, along with the basic notions of union, intersection, subset, cartesian product and complement should be enough to get by). A knowledge of basic algebraic structures would be useful, the whole point of creating the rationals is to create an (infinite) field generated by $1$.

I note in passing that many of the constructions actually used above involve *equivalence relations*, which one would do well to become quite comfortable with.

 

[Ackbach]

I see that I used the word "construct" a bit too loosely. All I'm interested in is a rigorous, axiomatic-deductive method of "arriving at" all the number systems analysts use on a regular basis (mainly the rationals, for computers, and the reals and complexes, for calculus). That is, I'm interested in knowing "what's behind all this?" And by "all this", I mean what's often assumed in high school algebra or precalculus books: commutative and associative laws, identities, inverses, distributive laws, etc.

Moreover, I keep running across terms like "Dedekind cut", and I really don't have a good notion of what that is, but it's often referenced at important points.

I am very definitely not interested in a Russell and Whitehead level of proof. I'm quite willing to accept a few more axioms in order to simplify proofs a bit.

Finally, I ran across this idea that you could go from naturals to integers to rationals to reals to complexes in the book The Art of Mathematics, and the author's write-up made me curious about it.

 

[Deveno]

Well, the question:

"What *is* a real number?" is a deep one.

The facile answer is "an element of a complete Archimedean ordered field", which rather begs the question.

The "field" part is easy, fields can be defined axiomatically, and contain the usual rules we employ to do arithmetic on a more or less daily basis. The "ordered" part is a bit more subtle, but can also be defined more or less axiomatically. You want the trichotomy rule, and the compatibility conditions:

$a < b \implies a + c < b+ c$
$a < b$ and $c > 0$ implies $ac < bc$.

The Archimedean bit is harder to explain-basically it means we don't allow "infinite numbers" or "infinitesimal numbers". The easiest to understand intuitively statement of this is:

Let $F$ be an ordered field, and $x \in F$ be any element. Then $F$ is an Archimedean ordered field if there is a positive integer $n$ such that $n > x$.

This property is used all the time in calculus, in the following way: suppose we have some $\epsilon > 0$, and we want to find some small number $s < \epsilon$. We can, by the Archimedean property, find a positive integer $N$ such that:

$N > \dfrac{1}{\epsilon}$.

Then $\dfrac{1}{N} < \epsilon$, as desired.

Sometimes, this is stated in the following way, for real numbers:

Every real number lies between two rational numbers.

The "complete" bit is hardest to explain of all-and it can be stated in a variety of different ways:

Any bounded-above non-empty set of real numbers has a least upper bound
Any Cauchy series converges
Any closed and bounded set of real numbers is compact
The closure of any set of real numbers contains all of its limit points
Every sequence of nested closed intervals has a non-empty intersection
Any Dedekind cut of the real numbers is generated by a (unique) real number

The second diagonal argument of Cantor showed that there were uncountably many real numbers-the first diagonal argument shows that the set of algebraic numbers, that is solutions to polynomials with rational coefficients, is countable. The proof that there even exist transcendental numbers at all, is rather involved (I am referring to the Lindemann-Weierstrass theorem), but an easy consequence is that if $\alpha \in \Bbb C$ is the solution to a rational polynomial, then:

$e^{\alpha}$

is transcendental. This idea can be adapted to show that $e$ (take $\alpha = 1$) is transcendental, and more importantly, that $\pi$ is transcendental:

If $\alpha = \pi$ were algebraic, then $i\pi$ would likewise be algebraic ($i$ is the solution to the rational polynomial $x^2 + 1$, and algebraic numbers form a field-although I will not prove that here), and so:

$e^{i\pi} = -1$

would be transcendental, contradiction.

The point being, it was known that "non-algebraic reals" were "out there", so it was theoretically possible to construct rational sequences that "converged" to these non-algebraic values, so even the extension of the rationals to their *algebraic* closure had "holes" in it.

So, the answer (at least as best as I can formulate it) to the question:

"What is a real number?"

is: a real number represents an ideal measurement that is able to be approximated to any desired degree of accuracy by rational numbers (of arbitrarily high denominator value).

Now I've glossed over a LOT of "down and dirty details" (and there ARE a lot of them, and they are "involved"). But the moral of the story is:

We start with the rationals, and find all the limit points and lump them all together.

Of course, to even begin this program we have to have the rationals "well in hand". If all we have is integers to begin with, this is no problem:

Take the integers, $\Bbb Z$. Make another copy, and throw away $0$. Now we have the cartesian product:

$\Bbb Z \times (\Bbb Z -\{0\}) = \Bbb Z \times (\Bbb Z)^{\ast}$.

So now we have PAIRS of integers $(a,b)$ where $b \neq 0$. We define an equivalence relation (this equivalence relation is due to Eudoxus, who studied ratios) on this set:

$(a,b) \sim (c,d) \iff ad = bc$ ("cross-multiplication").

I will not prove this is an equivalence relation, you may do this for yourself. I will state one property which is easily proven:

$(ka,kb) \sim (a,b)$ ("cancel top and bottom").

If we denote the equivalence class of $(a,b)$ by $[(a,b)]$, it can be shown that there is a unique pair $(a',b')$ for which:

1. $\gcd(a',b') = 1$
2. $b' > 0$

We call $(a',b') \in [(a,b)]$ the fraction $a/b$ in *reduced form*.

The set $(\Bbb Z \times (\Bbb Z)^{\ast})/\sim$ can be made into a ring, by setting:

$[(a.b)] + [(c,d)] = [(ad+bc,bd)]$ (one must verify that this addition does not depend on the particular pairs $(a,b)$ and $(c,d)$ but only on the equivalence classes), and:

$[(a,b)]\ast[(c,d)] = [(ac,bd)]$ (the same goes for the product).

We can embed the ring $\Bbb Z$ in this new ring by sending $k \mapsto [(k,1)]$ (this embedding is 1-1, but not onto, and preserves the addition and multiplication of integers), thus we are "justified" in calling $\Bbb Q$ an *extension* of $\Bbb Z$.

If, however, we only have the NATURALS "at hand", we can perform a similar slight of hand:

We take PAIRS of natural numbers: $(m,n)$, and define the following equivalence:

$(m,n) \sim (m',n') \iff m+n' = m' + n$

(here we think of $m$ as "the positive part", and $n$ the "negative part" (black ink/red ink), that is: $(m,n)$ represents $m-n$).

We can define addition and multiplication like so, on $(\Bbb N \times \Bbb N)/\sim$:

$[(m,n)] + [(m',n')] = [(m+m',n+n')]$
$[(m,n)]\ast[(m',n')] = [(mm' + nn', mn' + m'n)]$

Note that $[(k+m,k+n)] = [(m,n)]$, and thus that $[(m,m)] = [(0,0)]$.

We can similarly embed the naturals in this system by sending $m \mapsto [(m,0)]$.

So now we are "back to the naturals" which for brevity, I will take as given.