Why does Rudin say "the rational number system is inadequate as a field"?

Question

In the INTRODUCTION of chapter 1 of Baby Rudin, he says

The rational number system is inadequate for many purposes, both as a field and as an ordered set.

Addition and multiplication of rational numbers are commutative and associative, and multiplication is distributive over addition. Both 'zero' and 'one' exist.

Plus, as I recall, rational numbers are the smallest subfield of $\mathbb{C}$. So exactly what does Rudin mean by “inadequate as a field”?

Answer 1

Rudin isn't questioning $\mathbb{Q}$'s status as a field — he's questioning it's suitability for doing algebra and analysis.

For the purposes of algebra, $\mathbb{Q}$ is not a very nice field to work with: for example, many (most?) polynomials don't even have a root, let alone factor completely, and dealing properly with this deficiency is the subject of the frontiers of mathematical research!

(for comparison, while not every polynomial over $\mathbb{R}$ has a root, it has a very simple relationship with the complex numbers $\mathbb{C}$, a field over which every nonconstant polynomial has a root)

Similarly, for the purposes of analysis, $\mathbb{Q}$ is not a very nice ordered set; for example, they form a totally disconnected space, making them nearly useless for capturing even basic familiar geometric notions such as continuity.

Answer 2

The author means that $\mathbb Q$ as an ordered field is incomplete, i.e. not every Cauchy sequence in $\mathbb Q$ converges in $\mathbb Q$, or equivalently not every nonempty subset of $\mathbb Q$ that is bounded above has a least upper bound in $\mathbb Q$. This makes $\mathbb Q$ "inadequate" for many purposes in analysis, where the least-upper-bound property is required. For example, when $a$ is a positive rational number and $n$ a positive integer, you want the equation $x^n=a$ to have a solution; this does not always happen in $\mathbb Q$.

$\mathbb Q$ can be made complete by enlarging it to the set $\mathbb R$ of real numbers. There are two main ways by which this can be achieved: either by considering equivalence classes of Cauchy sequences in $\mathbb Q$, or by means of Dedekind cuts. It can be shown that $\mathbb R$ as a complete ordered field is unique: any two complete ordered fields are isomorphic.

Answer 3

Your remark “Addition and multiplication of rational numbers are commutative and associative, and multiplication is distributive over addition. Both 'zero' and 'one' exist” is inappropriate. Of course all of this is true because otherwise $\mathbb Q$ would not be a field.

However, as far as Analysis is concerned, $\mathbb Q$ is inadequate because:

• in $\mathbb Q$, a Cauchy sequence is not necessarily convergent;
• a monotonic and bounded sequence of rational numbers doesn't have to converge to a rational number;
• a bounded set of rational numbers doesn't need to have a supremum or an infimum in $\mathbb Q$;
• a continuous function defined on a closed and bounded interval of $\mathbb Q$ may well be unbounded;
• a continuous function from $\mathbb Q$ into $\mathbb Q$ may not have the intermediate value property.

And so on.

Answer 4

The real numbers are a subfield of the complex numbers. Not every sequence of rational numbers converges to a rational number: rational numbers aren't a complete set.

Answer 5

To add to the other (excellent) answers, the rationals are inadequate even for elementary mathematics. Starting with only a vague school-level idea of what numbers are, when we draw the graph of (say) $y=x^2-2$, it's visually obvious that the graph crosses the $x$ axis. With school arithmetic, we can even calculate to high accuracy just where the crossing points are. But the points do not exist if we are restricted to rational numbers. Similarly, under this restriction, no sense can be made of even simple geometric ratios, such as between the diagonal and side of a square or the circumference and diameter of a circle (albeit proof of the latter impossibility needing much more advanced mathematics).

Answer 6

Another way to look at it is there are a lot numbers of interest or importance in calculus that are not rational, namely irrational numbers such as $\pi$, $e$, $\sqrt{2}$... In fact, the set of irrational numbers is a far bigger subset of the real numbers than is the set of rationals.

Answer 7

Once I was surprised to see that many dozen pages in the Chapter on Riemann Stieltjes Integrals given in Apostole's Mathematical Analysis are valid even in the rational number system. So there can be a good part of analysis for rational numbers also.