Are there more transcendental numbers or irrational numbers that are not transcendental?

Question

This is not a question of counting (obviously), but more of a question of bigger vs. smaller infinities. I really don't know where to even start with this one whatsoever. Any help? Or is it unsolvable?

Answer 1

The non-transcendental numbers (otherwise known as the algebraic numbers – Wikipedia link) comprise a countably infinite set, whereas the transcendental numbers are uncountably infinite.

(Why are there only countably many algebraic numbers? Because we can group them according to what polynomial in $\mathbb{Q}[x]$ they are a root of, and any such polynomial has finitely many roots, and there are only countably many such polynomials.)

The point is: in colloquial terms, there are more transcendental numbers than algebraic numbers.

Therefore, there are certainly more transcendental numbers than there are algebraic numbers that also are not rational.

Answer 2

Hint: the set of algebraic numbers is Countable.

For proving this, you can show that the polynomials with integer coefficients form a countable set, and that the set of their roots is also countable. The latter is just the Set of Algebraic Numbers (over $\mathbb{C}$). As the Algebraic Numbers over $\mathbb{R}$ form a subset of the former set, it must also be Countable.

Answer 3

The set of algebraic numbers $\mathbb A$ is countable, so $\mathbb A\cap (\mathbb R\setminus \mathbb Q)$ is also countable. On the other hand, the set of transcendental numbers $\mathbb R\setminus\mathbb A$ must be uncountable, so $$|\mathbb R\setminus\mathbb A|>|\mathbb A\cap(\mathbb R\setminus\mathbb Q)|.$$

Answer 4

Note that the non-transcendental irrational numbers form a subset of $\Bbb A = \overline {\Bbb Q} \subset \Bbb C$, the set of complex algebraic numbers (i.e. the algebraic closure of $\Bbb Q$).

From a purely set-theoretic perspective, let us show that $\Bbb A$ is countable. For $a_0, \dots, a_n \in \Bbb Q$, let $R(a_0, \dots, a_n)$ be the set of the roots of the polynomial $a_0 + \dots + a_n x^n$ - a finite set with at most $n$ distinct elements. Note that since every element of $\Bbb A$ is the root of a such polynomial, then

$$\Bbb A = \bigcup \limits _{n \ge 1} \bigcup \limits _{(a_0, \dots, a_n) \in \Bbb Q ^{n+1}} R(a_0, \dots, a_n) .$$

Since $\Bbb Q$ is countable, so will be $\Bbb Q ^{n+1} \space \forall n \ge 1$, so the inner union is a countable union of finite sets and therefore countable. Now, the outer union will be a countable union of countable sets, so again countable. The transcendental numbers will therefore be uncountable, therefore much more.

From a measure-theoretic point of view, countable sets are null sets for the Lebesgue measure, therefore again the set of the transcendental ones will be much larger.

Finally, from a topological point of view, it is slightly more technical to show that both types of numbers that you ask about are dense in $\Bbb C$, so from this point of view, they are equally many.

These statements remain true even when you consider only real algebraic and transcendental numbers.

Answer 5

1. All transcendental numbers are irrational.

2. The square root of 2 is irrational, but not transcendental.

Therefore, the set of irrational numbers is larger than the set of transcendental numbers -- in real numbers.