# Prove the existence of uncountably many transcendental numbers

• Feb 3rd 2009, 10:44 AM
ninano1205
Prove the existence of uncountably many transcendental numbers
It says we need to use the theorem
"The union of a finite or countable number of countable sets A1,A2,... is itself countable." &
"The set of real numbers in the closed unit interval [0,1] is uncountable."
• Feb 3rd 2009, 10:58 AM
ThePerfectHacker
Here is a hint. All algebraic numbers are roots of non-zero polynomials with rational coefficient. There are countable many non-zero polynomials with rational coefficients and each of these polynomials has finitely many zeros. This means that the cardinality of the algebraic numbers is countable. Can you see what to do next?
• Feb 3rd 2009, 11:31 AM
ninano1205
Sounds right?
The set of real numbers is a union of Algebraic numbers and transcendental numbers and uncountable. The set of Algebraic numbers is countable. Therefore the set of transcendental numbers should be uncountable, otherwise the set of real numbers would be countable.
• Feb 3rd 2009, 03:33 PM
ThePerfectHacker
Quote:

Originally Posted by ninano1205
The set of real numbers is a union of Algebraic numbers and transcendental numbers and uncountable. The set of Algebraic numbers is countable. Therefore the set of transcendental numbers should be uncountable, otherwise the set of real numbers would be countable.

Exactly. (Clapping)