# One to one correspondence between transcendental and uncomputable numbers

I know that both sets are uncountable infinite but the transcendentals are not a subset of the uncomputables. I don’t know if there exist uncomputable numbers that are not transcendental. But my question is whether the two sets have the same cardinality.

• $2^{\aleph_0}$. – Lord Shark the Unknown Apr 20 at 6:53
• An algebraic number is always computable, hence an uncomputable number must be transcendental. – Peter Apr 20 at 6:54
• Since both all real numbers and all transcendental numbers have the same cardinality, and the uncomputable numbers are in between, they also have it, see Schröder–Bernstein theorem. – Conifold Apr 20 at 7:08
• @Conifold This argument is not valid. The set of the transcendental numbers is NOT a subset of the set of the uncomputable numbers. There are transcendental computable numbers. – Peter Apr 20 at 7:14
• @Peter: But the opposite argument is valid. Any non-computable real is transcendental. And since there are only countably many computable numbers... – Asaf Karagila Apr 20 at 7:33