## Theorem Thursday: Liouville’s Approximation Theorem And How To Make Your Own Transcendental Number

As I get into the second month of Theorem Thursdays I have, I think, the whole roster of weeks sketched out. Today, I want to dive into some real analysis, and the study of numbers. It’s the sort of thing you normally get only if you’re willing to be a mathematics major. I’ll try to be readable by people who aren’t. If you carry through to the end and follow directions you’ll have your very own mathematical construct, too, so enjoy.

# Liouville’s Approximation Theorem

It all comes back to polynomials. Of course it does. Polynomials aren’t literally everything in mathematics. They just come close. Among the things we can do with polynomials is divide up the real numbers into different sets. The tool we use is polynomials with integer coefficients. Integers are the positive and the negative whole numbers, stuff like ‘4’ and ‘5’ and ‘-12’ and ‘0’.

A polynomial is the sum of a bunch of products of coefficients multiplied by a variable raised to a power. We can use anything for the variable’s name. So we use ‘x’. Sometimes ‘t’. If we want complex-valued polynomials we use ‘z’. Some people trying to make a point will use ‘y’ or ‘s’ but they’re just showing off. Coefficients are just numbers. If we know the numbers, great. If we don’t know the numbers, or we want to write something that doesn’t commit us to any particular numbers, we use letters from the start of the alphabet. So we use ‘a’, maybe ‘b’ if we must. If we need a lot of numbers, we use subscripts: a_{0}, a_{1}, a_{2}, and so on, up to some a_{n} for some big whole number n. To talk about one of these without committing ourselves to a specific example we use a subscript of i or j or k: a_{j}, a_{k}. It’s possible that a_{j} and a_{k} equal each other, but they don’t have to, unless j and k are the same whole number. They might also be zero, but they don’t have to be. They can be any numbers. Or, for this essay, they can be any integers. So we’d write a generic polynomial f(x) as:

(Some people put the coefficients in the other order, that is, and so on. That’s not *wrong*. The name we give a number doesn’t matter. But it makes it harder to remember what coefficient matches up with, say, x^{14}.)

A zero, or root, is a value for the variable (‘x’, or ‘t’, or what have you) which makes the polynomial equal to zero. It’s possible that ‘0’ is a zero, but don’t count on it. A polynomial of degree n — meaning the highest power to which x is raised is n — can have up to n different real-valued roots. All we’re going to care about is one.

Rational numbers are what we get by dividing one whole number by another. They’re numbers like 1/2 and 5/3 and 6. They’re numbers like -2.5 and 1.0625 and negative a billion. Almost none of the real numbers are rational numbers; they’re exceptional freaks. But they are all the numbers we actually *compute* with, once we start working out digits. Thus we remember that to live is to live paradoxically.

And every rational number is a root of a first-degree polynomial. That is, there’s some polynomial f(x) = a_0 + a_1 x that’s made zero for your polynomial. It’s easy to tell you what it is, too. Pick your rational number. You can write that as the integer p divided by the integer q. Now look at the polynomial f(x) = p – q x. Astounded yet?

That trick will work for any rational number. It won’t work for any irrational number. There’s no first-degree polynomial with integer coefficients that has the square root of two as a root. There *are* polynomials that do, though. There’s f(x) = 2 – x^{2}. You can find the square root of two as the zero of a second-degree polynomial. You can’t find it as the zero of any lower-degree polynomials. So we say that this is an algebraic number of the second degree.

This goes on higher. Look at the cube root of 2. That’s another irrational number, so no first-degree polynomials have it as a root. And there’s no second-degree polynomials that have it as a root, not if we stick to integer coefficients. Ah, but f(x) = 2 – x^{3}? That’s got it. So the cube root of two is an algebraic number of degree three.

We can go on like this, although I admit examples for higher-order algebraic numbers start getting hard to justify. Most of the numbers people have heard of are either rational or are order-two algebraic numbers. I can tell you truly that the eighth root of two is an eighth-degree algebraic number. But I bet you don’t feel enlightened. At best you feel like I’m setting up for something. The number r(5), the smallest radius a disc can have so that five of them will completely cover a disc of radius 1, is eighth-degree and that’s interesting. But you never imagined the number before and don’t have any idea how big that is, other than “I guess that has to be smaller than 1”. (It’s just a touch less than 0.61.) I sound like I’m wasting your time, although you might start doing little puzzles trying to make smaller coins cover larger ones. Do have fun.

Liouville’s Approximation Theorem is about approximating algebraic numbers with rational ones. Almost everything we ever do is with rational numbers. That’s all right because we can make the difference between the number we want, even if it’s r(5), and the numbers we can compute with, rational numbers, as tiny as we need. We trust that the errors we make from this approximation will stay small. And then we discover chaos science. Nothing is perfect.

For example, suppose we need to estimate π. Everyone knows we can approximate this with the rational number 22/7. That’s about 3.142857, which is all right but nothing great. Some people know we can approximate it as 333/106. (I didn’t until I started writing this paragraph and did some research.) That’s about 3.141509, which is better. Then there’s 355/113, which is not as famous as 22/7 but is a celebrity compared to 333/106. That’s about 3.141529. Then we get into some numbers only mathematics hipsters know: 103993/33102 and 104348/33215 and so on. Fine.

The Liouville Approximation Theorem is about sequences that converge on an irrational number. So we have our first approximation x_{1}, that’s the integer p_{1} divided by the integer q_{1}. So, 22 and 7. Then there’s the next approximation x_{2}, that’s the integer p_{2} divided by the integer q_{2}. So, 333 and 106. Then there’s the next approximation yet, x_{3}, that’s the integer p_{3} divided by the integer q_{3}. As we look at more and more approximations, x_{j}‘s, we get closer and closer to the actual irrational number we want, in this case π. Also, the denominators, the q_{j}‘s, keep getting bigger.

The theorem speaks of having an algebraic number, call it x, of some degree n greater than 1. Then we have this limit on how good an approximation can be. The difference between the number x that we want, and our best approximation p / q, has to be larger than the number (1/q)^{n + 1}. The approximation might be higher than x. It might be lower than x. But it will be off by *at least* the n-plus-first power of 1/q.

Polynomials let us separate the real numbers into infinitely many tiers of numbers. They also let us say how well the most accessible tier of numbers, rational numbers, can approximate these more exotic things.

One of the things we learn by looking at numbers through this polynomial screen is that there are transcendental numbers. These are numbers that can’t be the root of *any* polynomial with integer coefficients. π is one of them. e is another. Nearly all numbers *are* transcendental. But the proof that any particular number is one is hard. Joseph Liouville showed that transcendental numbers must exist by using continued fractions. But this approximation theorem tells us how to make our own transcendental numbers. This won’t be any number you or anyone else has ever heard of, unless you pick a special case. But it will be yours.

You will need:

- a
_{1}, an integer from 1 to 9, such as ‘1’, ‘9’, or ‘5’. - a
_{2}, another integer from 1 to 9. It may be the same as a_{1}if you like, but it doesn’t have to be. - a
_{3}, yet another integer from 1 to 9. It may be the same as a_{1}or a_{2}or, if it so happens, both. - a
_{4}, one more integer from 1 to 9 and you know what? Let’s summarize things a bit. - A whopping great big gob of integers a
_{j}, every one of them from 1 to 9, for every possible integer ‘j’ so technically this is infinitely many of them. - Comfort with the notation n!, which is the factorial of n. For whole numbers that’s the product of every whole number from 1 to n, so, 2! is 1 times 2, or 2. 3! is 1 times 2 times 3, or 6. 4! is 1 times 2 times 3 times 4, or 24. And so on.
- Not to be thrown by me writing -n!. By that I mean work out n! and then multiply that by -1. So -2! is -2. -3! is -6. -4! is -24. And so on.

Now, assemble them into your very own transcendental number z, by this formula:

If you’ve done it right, this will look something like:

Ah, but, how do you know this is transcendental? We can prove it is. The proof is by contradiction, which is how a lot of great proofs are done. We show nonsense follows if the thing isn’t true, so the thing must be true. (There are mathematicians that don’t care for proof-by-contradiction. They insist on proof by charging straight ahead and showing a thing is true directly. That’s a matter of taste. I think every mathematician feels that way sometimes, to some extent or on some issues. The proof-by-contradiction is easier, at least in this case.)

Suppose that your z here is not transcendental. Then it’s got to be an algebraic number of degree n, for some finite number n. That’s what it means not to be transcendental. I don’t know what n is; I don’t care. There is some n and that’s enough.

Now, let’s let z_{m} be a rational number approximating z. We find this approximation by taking the first m! digits after the decimal point. So, z_{1} would be just the number 0.a_{1}. z_{2} is the number 0.a_{1}a_{2}. z_{3} is the number 0.a_{1}a_{2}000a_{3}. I don’t know what m you like, but that’s all right. We’ll pick a nice big m.

So what’s the difference between z and z_{m}? Well, it can’t be larger than 10 times 10^{-(m + 1)!}. This is for the same reason that π minus 3.14 can’t be any bigger than 0.01.

Now suppose we have the best possible rational approximation, p/q, of your number z. Its first m! digits are going to be p / 10^{m!}. This will be z_{m} And by the Liouville Approximation Theorem, then, the difference between z and z_{m} has to be at least as big as (1/10^{m!})^{(n + 1)}.

So we know the difference between z and z_{m} has to be larger than one number. And it has to be smaller than another. Let me write those out.

We don’t need the z – z_{m} anymore. That thing on the rightmost side we can write what I’ll swear is a little easier to use. What we have left is:

And this will be true whenever the number m! (n + 1) is greater than (m + 1)! – 1 for big enough numbers m.

But there’s the thing. This isn’t true whenever m is greater than n. So the difference between your alleged transcendental number and its best-possible rational approximation has to be simultaneously bigger than a number and smaller than that same number without being equal to it. Supposing your number is anything but transcendental produces nonsense. Therefore, congratulations! You have a transcendental number.

If you chose all 1’s for your a_{j}‘s, then you have what is sometimes called *the* Liouville Constant. If you didn’t, you may have a transcendental number nobody’s ever noticed before. You can name it after someone if you like. That’s as meaningful as naming a star for someone and cheaper. But you can style it as weaving someone’s name into the universal truth of mathematics. Enjoy!

I’m glad to finally give you a mathematics essay that lets you make something you can keep.

## Andrew Wearden 3:29 pm

onThursday, 30 June, 2016 Permalink |Admittedly, I do have an undergrad math degree, but I thought you did a good job explaining this. Out of curiosity, is there a reason you can’t use the integer ‘0’ when creating a transcendental number?

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## Joseph Nebus 6:45 am

onSunday, 3 July, 2016 Permalink |Thank you. I’m glad you followed.

If I’m not missing a trick there’s no reason you can’t slip a couple of zeroes in to the transcendental number. But there is a problem if you have nothing but zeroes after some point. If, say, everything from on were zero, then you’d have a rational number, which is as un-transcendental as it gets. So it’s easier to build a number without electing zeroes rather than work out a rule that allows zeroes only in non-dangerous configurations.

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## My Mathematics Reading For The 13th of June | nebusresearch 4:01 pm

onTuesday, 13 June, 2017 Permalink |[…] know whether is transcendental or not. It’s hard proving numbers are transcendental. If you go out trying to build a transcendental number it’s easy, but otherwise, you have to hope you know your number is the exponential of an algebraic […]

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