Proof of a fact about real numbers - transcendental and algebraic

[dextercioby]

TL;DR

I quote a passage from Wikipedia and ask a possibly dumb question about real numbers.

Here's a quote from here:

Properties​

A transcendental number is a (possibly complex) number that is not the root of any integer polynomial. Every real transcendental number must also be irrational since every rational number is the root of some integer polynomial of degree one.[17] The set of transcendental numbers is uncountably infinite. Since the polynomials with rational coefficients are countable, and since each such polynomial has a finite number of zeroes, the algebraic numbers must also be countable. However, Cantor's diagonal argument proves that the real numbers (and, therefore also the complex numbers) are uncountable. Since the real numbers are the union of algebraic and transcendental numbers, it is impossible for both subsets to be countable. This makes the transcendental numbers uncountable.

Here's my question. If transcendental numbers are all numbers that are not algebraic, how do we know that the disjoint union of the sets of algebraic numbers and transcendental numbers is exactly $\mathbb R$ and not an uncountable proper subset of it? (Here, I assume that $\mathbb R$ is defined as the completion of $\mathbb Q$, or simply put, the disjoint union of the set of rational and irrational numbers).

 

[fresh_42]

I don't understand your question. $\mathbb{R}$ is the disjoint union of all numbers that are blue, and the set of numbers that are not blue. There are no other possibilities. Now, if transcendental is defined as not blue, and algebraic as blue, where should be numbers besides these two possibilities? That would only make sense if we define transcendental other than not algebraic.

 

[FactChecker]

For any universal set, $U$, and a subset of it, $S$, it is true that $S \cup \neg S = U$.

 

[dextercioby]

Well, yes, if you say a transcendental number is a real number which is not algebraic, then the question is dumb. Sorry, I thought I had a case here.

 

[dextercioby]

Sorry, let me try to revive this. The set of irrational numbers is uncountable, the set of all transcendentals is uncountable and is a proper subset of all irrational numbers. This begs the question: is the set of all irrational numbers which are not transcendental also uncountable?

 

[PeroK]

Sorry, let me try to revive this. The set of irrational numbers is uncountable, the set of all transcendentals is uncountable and is a proper subset of all irrational numbers. This begs the question: is the set of all irrational numbers which are not transcendental also uncountable?

The set of algebraic numbers is countable.

Note that the set of computable numbers is also countable and includes all algebraic numbers and a subset of the transcendental numbers.

 

[jbriggs444]

This begs the question: is the set of all irrational numbers which are not transcendental also uncountable?

So the set includes just "not transcendental" numbers. So just a subset of the algebraic real numbers, right?

Clearly countable then.

 

[dextercioby]

So the set includes just "not transcendental" numbers. So just a subset of the algebraic real numbers, right?

Clearly countable then.

Now that you wrote it and made it so obvious, I had to go make a picture. Not a Venn diagram, but an Excel one. So I reckon now it should be visibly obvious.