A short and elegant proof to demonstrate that the number of trascendental numbers is infinite ?

Answer 1

Once you have one transcendental number, say x, then the collection {x+y:y is rational} is an infinite collection of transcendental numbers. To get x, it is fairly easy to show that
sum {1/10^(n!): n=1 to infinity}
is transcendental.

Alternatively, the collection of all real numbers is uncountable and the collection of algebraic numbers is countable. Thus, the collection of transcendental numbers is actually uncountable.

Alternatively, show that every number of the form
sum{ a_n /10^(n!):n=1 to infinity}
where each a_n=1 or 2
is transcendental. This is a whole Cantor set of transcendental numbers.

Answer 2

Cantor showed by his diagonal argument that the set of real numbers is uncountably infinite. On the other hand, the set of polynomials with integer coefficients is countably infinite, and every such polynomial has a finite number of zeroes, meaning that the solutions of integer coefficient polynomials (which defines non-transcendental real numbers) is countably infinite. Removing the countable set of algebraic numbers from the uncountable set of real numbers leaves the set of transcendental numbers, which must therefore be not just infinite, but uncountably so.

Answer 3

n(trascendental)=infinity