Theorem 19: there is a transcendental number

I went to a very nice talk by a colleague the other day.  I can’t tell you about an hour’s worth of talk in one blog post, but I can tell you about a few of the highlights.

I’ve previously written about the fact that the square root of 2 is irrational.  That is, it can’t be written as the ratio of two whole numbers (as a fraction, if you like).  In some ways, that makes it seem like an unpleasant number: it’s not as `clean’ as something like 1/3 or 17/15.  But it’s not totally disgusting!  We can get some control on it, because we know that it satisfies the equation x^2 = 2, or, putting that another way, x^2 - 2 = 0.  That’s quite a nice equation, actually.  (Not quite as simple as 3x - 1 = 0 or 15 x - 17 = 0, perhaps, but still pretty nice.)  It’s a polynomial equation (it involves powers of x), and the coefficients (the numbers in front of the powers of x) are integers.  We say that it is an algebraic number: it satisfies a polynomial equation with integer coefficients.  (We could replace “integer coefficients” by “rational coefficients” here without changing anything — you might like to convince yourself of this.)

We could think about some other numbers and ask whether they’re algebraic too.  Is \sqrt[3]{15} algebraic?  Yes, because it satisfies the equation x^3 - 15 = 0, and that is a polynomial equation with integer coefficients.  Is the golden ratio, \frac{1+\sqrt{5}}{2}, algebraic?  I’ll let you find a polynomial equation that it satisfies.  Is \pi algebraic?  Is e (the base of the natural logarithm) algebraic?  Are there any numbers that aren’t algebraic?

Just before we move on, I’d like to make sure you’ve noticed that every rational number is algebraic.  (Check this!)  So if there is a number that isn’t algebraic, it must be irrational.  We call numbers that aren’t algebraic (if they exist) transcendental.  We’d like to know whether there is a transcendental number, and whether \pi and e are transcendental.  It turns out that there is a transcendental number, and that’s this week’s theorem.

Theorem There is a transcendental number.

It’s not quite clear how we can possibly show that a number is transcendental.  As with showing that a number is irrational, we’re being asked to prove a negative statement (there is no polynomial equation satisfied by the number/the number cannot be expressed as a ratio), so maybe we should use contradiction, although we have no idea how hard it might be.

Amusingly, it might actually be easier to show that there are lots of (infinitely many) transcendental numbers than it is to show that a particular number is transcendental!  Why is this?  The proof uses ideas from last week’s post: countability and uncountability.  As I mentioned, but didn’t prove, there, the real numbers are uncountable: they cannot be written in a list.  One can prove that the algebraic numbers are countable, without too much difficulty.  So there must be lots of (uncountably many) real numbers that aren’t algebraic, that is, that are transcendental.

That’s a pretty slick proof (even if you write out the proofs that I omitted), but you may feel that it’s unsatisfactory.  After all, we still don’t have an explicit example of a transcendental number, despite knowing that almost all real numbers are transcendental!

Perhaps the most famous early example of a transcendental number is Liouville‘s number, which is defined to be L = \sum_{n=1}^{\infty} 10^{-n!}.  (Here n! denotes n factorial.)  So its decimal expansion starts 0.110001000000000000000001000000….

Why is this transcendental? 

Liouville’s insight was that irrational algebraic numbers cannot be approximated very well by rationals.  Precisely, if \alpha is an irrational algebraic number, and if n is the degree of \alpha (so \alpha satisfies a polynomial of degree n with integer coefficients but not one of lower degree), then there is some constant C so that |\alpha - \frac{p}{q}| > \frac{C}{q^n} for any integers p and q>0.  I think I’m going to skip the proof for now (although it’s not very difficult).  I might revise that decision later.

Why does this help?  Well, if we can show that Liouville’s number L can be approximated extremely well by a rational (and is irrational), then we’ll be in business.  But in fact L has been constructed precisely to have this property.  We can get a very good approximation to L by chopping off the sum at some point: the number \sum_{n=1}^N 10^{-n!} is rational and extremely close to L.  Again, I think I’ll skip the details for now.

So L is transcendental.

We’ve answered one of our questions from earlier: there is a transcendental number (in fact, there are uncountably many), and we have given an example.  But we haven’t said anything about whether \pi and e are transcendental.  It turns out that they are, but it took a few years from Liouville exhibiting his number to other people finding proofs.  Hermite gave the first proof that e is transcendental, and a few years later Lindemann proved that \pi is transcendental.  These proofs are a bit harder than proving that Liouville’s number is transcendental, but not impossibly so!

Further reading

There’s the usual selection of Wikipedia pages: Liouville’s number, transcendental numbers (including a list of numbers known to be transcendental), etc.  One classic book on the subject is Transcendental number theory, by Alan Baker (who himself made large contributions to the field).  There’s also a shorter description (including the things I’ve mentioned above, together with a proof of the transcendence of e) in Baker’s Concise Introduction to the Theory of Numbers.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: