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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.

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    $\begingroup$ $2^{\aleph_0}$. $\endgroup$ – Lord Shark the Unknown Apr 20 at 6:53
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    $\begingroup$ An algebraic number is always computable, hence an uncomputable number must be transcendental. $\endgroup$ – Peter Apr 20 at 6:54
  • $\begingroup$ 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. $\endgroup$ – Conifold Apr 20 at 7:08
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    $\begingroup$ @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. $\endgroup$ – Peter Apr 20 at 7:14
  • $\begingroup$ @Peter: But the opposite argument is valid. Any non-computable real is transcendental. And since there are only countably many computable numbers... $\endgroup$ – Asaf Karagila Apr 20 at 7:33
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Here

https://en.wikipedia.org/wiki/Computable_number

under "properties" it is stated the the set of computable numbers is countable.

Hence, the set of uncountable numbers must be uncountable infinite and hence have the same cardinality as the set of transcendental numbers which is uncountable infinite as well.

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