I've found this exercise on Boolos' Logic and Computability:
A real number $x$ is called algebraic if it is a solution to some equation of the form:
$$c_{\small d}x^{\small d}+c_{\small d-1}x^{\small d-1}+c_{\small d-2}x^{\small d-2}+\text{ }…\text{ }+c_{\small 2}x^{\small 2}+c_{\small 1}x+c_{\small 0}=0$$
Where the $c_i$ are rational numbers and $c_d\neq 0$. For instance, for any rational number $r$, the number $r$ itself is algebraic, since it is the solution to $x-r=0$; and the square root $\sqrt{r}$ or $r$ is algebraic, since it is a solution to $x^2-r=0$.
(b) A real number that is not algebraic is called transcendental. Prove that transcendental numbers exist.
This exercise is nice, but I don't know if I know enough to do it. I thought it was simple to do because it's located on an early chapter of the mentioned book, but searching for it on Wikipedia I've seen that it seems to be difficult to do it – So, the fact that it's located on an early chapter of Boolos' book could indicate that perhaps there is a simple way that is show in the book and I'm unaware of such way, is that it?
I'll use the tags logic and computability because of the title of the book.
Edit: Oh, I forgot one thing, I don't know if it's a good idea or if it's possible, but is there a way of proving it pretending I don't know that such numbers do exist with antecedence? I guess it's easier if you know they exist beforehand although I don't know if it's possible to do that way.
Best Answer
You presumably know the real numbers are uncountable. It is not hard to show the algebraic numbers are countable. This immediately implies your desired result.