1. (I came up with this idea after thinking about the proof that e is irrational.)
For a positive integer n, let T_n denote the set of all real numbers which are the root of a polynomial with integer coefficients, with degree at most n and whose coefficients do not exceed n in absolute value.
Note that every real algebraic number must be an element of T_n for some n, and that T_n is a subset of T_m whenever m
> n.
Thus, T_1 consists of the roots of the polynomials x, x - 1, x + 1, -x + 1, -x - 1, and therefore is equal to the set {-1, 0, 1}.
Now, we define a sequence of numbers U_n, recursively as follows:
U_1 = 0.
For all n
> 1, U_n is defined as the smallest element of T_n that is larger than U_(n-1), but smaller than the arithmetic mean of U_(n-1) and the next smallest element of T_(n-1). If there is no such number, U_n is defined to be U_(n-1).
For example, suppose the set T_2 were equal to {-3, -2, 0, 0.3, 0.5, 2}. Now, U_1 was 0, and the next smallest element of T_1 was 1, so we need the smallest element of T_2 larger than 0 but smaller than 0.5. Here we get 0.3. If no such number existed then U_2 would be 0 again (so removing 0.3 would give us U_2 = 0).
U_n is a non-decreasing sequence with an upper bound (indeed 0.5 is an upper bound), so as n goes to infinity the sequence converges, and let L be what it converges to.
L must be a transcendental number, no? If it were algebraic, it would be an element of T_n for some positive integer n, but we defined the sequence so that it would have to lie between two consecutive elements of T_n (and closer to the smaller than the larger).

If my explanation is confusing, please tell me where you're lost and I hope I can explain a better way.

So a few questions:
1) Why didn't mathematicians think of this when determining if transcendental numbers exist, or how to find one?
2) Is there a name for this kind of technique?

2.

3. Your first sentence is the first statement of the proof that algebraic numbers are countable. Each T_n is finite so the union of all of them is countable. What's left over among the reals is uncountable and (by definition) transcendental.

I did not rea the rest of your post - MEGO.

4. That certainly proves that transcendental numbers exist, but it doesn't give any examples of one. I constructed one by finding a number that lies between two consecutive elements of T_n for each positive integer n. I made a convergent sequence U_n such that for every positive integer m, all the U_n's after a certain point will lie between two consecutive elements of T_m.

5. I haven't looked at it complete detail, but it seems your method may have a loophole, at some point U_n may not exist.

6. Originally Posted by mathman
I haven't looked at it complete detail, but it seems your method may have a loophole, at some point U_n may not exist.
The algebraic numbers are a dense set, so there will always be some algebraic number lying between two consecutive values of T_n.
If there doesn't happen to be one inside T_(n+1), then U_(n+1) is set to be the same as U_n, and you continue until you find a T_m that does work, and this defined U_m.

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