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Liouville's proof of existence of transcendental numbers

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By Yatir Halevi on Wednesday, September 18, 2002 - 07:46 pm:

How did Liouville 'construct' a transcendental number?


Yatir

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By Michael Doré on Thursday, September 19, 2002 - 01:00 am:

He discovered the following result, called Liouville's theorem (or at least this is one result called Liouville's theorem - there's a different one in complex analysis).

Suppose a is algebraic of degree n (that is it is the root of a nth degree irreducible polynomial with integer coefficients). Then you can find c > 0 such that for all rational p/q (with q > 0) we have |a-p/q| > c/qⁿ.

To prove this, let P be an irreducible polynomial of degree n with integer coefficients such that P(a) = 0. P has bounded gradient in [a-1,a+1] so there exists c > 0 such that for any rational p/q in [a-1,a+1] we have:

|a - p/q| > c |P(a) - P(p/q)|

(Note that we don't need to worry about rationals outside [a-1,a+1] since these have |a - p/q| >= 1 so can't possibly affect the theorem.)

Now just note that |qⁿ P(p/q)| is a non-negative integer and it is not zero (as P is irreducible) so it is at least 1. And P(a) = 0. Hence:

|a - p/q| > c/qⁿ

as required.

This theorem allows us to construct many transcendental numbers. The key is to find a number which is so well approximated by rationals that the conclusion of Liouville's theorem cannot possibly hold for any n, thus proving the number is transcendental. For example it is an easy exercise now to show that 1/2^0! + 1/2^1! + 1/2^2! + ... is transcendental. Historically this was the first number to be proved as such.

Most transcendental numbers aren't well enough approximated by rationals to allow Liouville's theorem to prove they're transcendental. Still I rather like the above proof of Liouville's theorem as it combines analysis and algebra in a nice way.

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By Yatir Halevi on Thursday, September 19, 2002 - 05:28 am:

Michael, could you please explain how to prove that the number you gave as an example is transcendental?

Yatir

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By Michael Doré on Thursday, September 19, 2002 - 11:24 am:

Consider more generally the sum x = S^¥ _i=11/a_i where a_i are strictly increasing positive integers with a_i|a_i+1. Suppose this is algebraic of degree n. We will use Liouville to derive a necessary condition on the a_is.

We can find c > 0 such that |x - p/q| > c/qⁿ for any rational p/q. Take p/q = S^m _i=11/a_i. Then q = a_m. We get:

S^¥ _i=m+11/a_i > c/a_mⁿ

and this must hold for all m.

Let's estimate the thing on the left hand side. We have a_i >= 2^i-m-1a_m+1 for all i >= m+1. So:

S^¥ _i=m+11/(2^i-m-1a_m+1) > c/a_mⁿ

Using the formula for a geometric progression this gives us:

2/a_m+1 > c/a_mⁿ

for all n.

In other words, if x is algebraic of degree n then there exists A > 0 such that a_m+1 < A a_mⁿ for all m.

So to ensure that x is not algebraic we just need to make sure this condition is violated for every positive integer n. For instance if (for every n) a_m+1/a_mⁿ tends to infinity as m tends to infinity, then the condition is certainly violated. If you set a_m = 2^m! then a_m+1/a_mⁿ = 2^{(m+1)! - n*m!} which tends to infinity as m tends to infinity (for any fixed n).

Hence S^¥ _i=01/2^i! is transcendental. You can replace 2 with any other base, such as 10, if you want. Thus the number 1.110001000000000000000001000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000001... is transcendental.

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