The cardinality of the set of transcendental numbers is larger than that of the irrationals.

Question

In a video, one guy said that there are more transcendental numbers than irrational numbers. I wonder why that would be. The only knowledge I have about infinities is that $|\mathbb{N}|=|\mathbb{Z}|=|\mathbb{Q}|<|\mathbb{R}|$. Can someone show me a proof?

Answer 1

One has to be careful with what "more" can mean in this situation. If $\mathbb T$ is the set of transcendent (real) numbers, then it is true that $$\mathbb T \subsetneq \mathbb R \backslash \mathbb Q \,.$$

So, in this sense there are more irrationals than rationals.

But, since the set of algebraic numbers is countable, it is easy to conclude that $$|\mathbb T|=|\mathbb R|=|\mathbb R \backslash \mathbb Q|$$ So, speaking in terms of cardinality, there are as many transcendent numbers as real numbers.

Answer 2

As a supplement to the comments, I think it can illustrative to reason why the transcendental numbers are uncountable.

Indeed, the transcendental numbers are defined as the complement of the algebraic numbers. And a number is defined as algebraic if it is the solution of a polynomial with integer coefficients.

But every polynomial can be thought of as a string of integer coefficients. For example, we could represent the polynomial $x^2+2$ as the string "(1,0,2)".

Furthermore, since the number of solutions is finite, as per the fundamental theorem of algebra, then we can enumerate the solutions of a polynomial. For example, we could represent $\sqrt 2$ as the second solution of $x^2+2$, or "(1,0,2),2".

This representation is not unique, but it is still surjective: every algebraic number can be represented in this way.

But every one of our custom code words is a finite list of integers! So the cardinality of the algebraic numbers is less than the cardinality of the finite lists of integers.

Since the set of finite lists of integers is countable, then the set of algebraic numbers must be countable too.

But if we substract a countable subset from an uncountable one, the cardinality of the uncountable set is unfaced.

So $|R| = |R\setminus Alg| = |Trans|$.