Infinity of the set of transcendental numbers. I am looking for a proof of the infiniteness of the set of all (real) transcendental numbers. The first idea is to say that there exists, at least, one transcendental number (for example $e$). Now I am trying to prove that if $x$ is transcendental, then $x+1$ is also transcendental. Any idea or suggestion is welcome. (I do not even know wheter what I am trying to prove is even true).
Do you think this is a good idea to show  the infiniteness of the set of all transcendental numbers? Is there a better solution? Thank you in advance.
 A: There is in fact a far simpler approach.
The algebraic numbers are countably infinite, because they arise only as roots of polynomials with integral coefficients, which themselves are countably infinite. But the real numbers are uncountably infinite. Therefore, there must exist uncountably many transcendental numbers.
A: hint: if if x+1 is not trascendental, then x must be not trascendental. think how can you show it. and of course showing that there is infinite number of trascendental number can be shown if you find only one such number and prove that it is trascendental. of course, you can prove without existence with cardinal numbers.
A: This is easy, by way of contraposity: If $x+1$ is not transcendental, then it is algebraic, so it is the root of some polynomial $f(t)$ with real coefficients. Then $x$ is a root of $f(t+1)$, so it is also algebraic.
There is a more common solution: There are countably many rational numbers, so for each degree there are countably many polynomials with rational coefficients, so there are countably many roots to these polynomials. Thus there are only countably many algebraic numbers, and therefore there must be $|\Bbb R|$ many transcendental numbers.
