## Now there’s completeness and then there’s completeness…

This post achieves a fortuitous segue from the last post into my serious of articles on the beauty of Galois theory.

In the previous post I introduced Dedekind cuts as a means of constructing the real number line, and I said that this perspective is responsible for the completeness of the real numbers $\mathbb{R}$.

Now, that was completeness in the topological sense.  There is another, very different notion of algebraic completeness.

A number system is called algebraically complete if every polynomial equation in one variable with coefficients from that number system can be solved in that number system.

So is $\mathbb{R}$ algebraically complete?  Certainly it is not, because there is no solution to $x^2+1=0$ in the real numbers.  We would have to add a number (by custom called $i$) defined by the fact that it is a solution to that equation.  If we combine that with $\mathbb{R}$, then we get the complex numbers $\mathbb{C}$.

It turns out that $\mathbb{C}$ is algebraically complete.  This is a very deep theorem, and it says that we’ll never run into the same kind of trouble we did with $x^2+1=0$ ever again, not even with highly complex polynomials.  The is the celebrated Fundamental Theorem of Algebra.

Because $\mathbb{C}$ is algebraically complete and because any number system containing $\mathbb{R}$ which is algebraically complete has to contain $\mathbb{C}$, we say $\mathbb{C}$ is the algebraic completion of $\mathbb{R}$.

Let’s ask a different question.  Is $\mathbb{C}$ the algebraic completion of the rational numbers $\mathbb{Q}$?  We already know that the real numbers can be defined in a way motivated by topology.  If all real numbers were roots of polynomials with rational coefficients, then we could also say that the real numbers have an existence motivated by algebra.

So are all complex numbers roots of polynomials with rational coefficients?  Are all real numbers?

It turns out the answer is a resounding no.  There is a number system, called the algebraic number system, which is formed by beginning with $\mathbb{Q}$ and adding all the roots of polynomials with rational coefficients.  (This system contains real and nonreal numbers, but if you don’t like imaginaries you can finesse the point by only adding roots for polynomials which assume both positive and negative values.)

Numbers in this system are called algebraic, and most real numbers are not algebraic.  That’s just by counting.  There are only countably many algebraic numbers, but there are uncountably many real numbers.

The above observation is elementary, and guarantees that nonalgebraic numbers (called transcendental because they transcend being defined in terms of their polynomial properties; the number $x=\sqrt5$ is defined by the fact that $x^2=5$, and transcendental numbers cannot be defined in that way) are highly abundant.  But to actually get your hands on a transcendental number, and to know it is transcendental, this is a very hard thing.  $\pi$ and $e$ are transcendental, but these are extraordinarily hard facts to verify.

Any number you can write with integers, addition, subtraction, multiplication, division, and all types of radicals (square roots, cube roots, etc.) is evidently algebraic (convince yourself of that).  On the other hand, it’s not true that every algebraic number can be written in that way.  This is deep and unexpected, and the tale of this fact is part of the epic tale of the quintic (5th degree) equation.  (It turns out there is no equation analogous to the quadratic equation for 5th degree or higher polynomials, and the reason why not is very interesting).