Countable subset under irrational

Question

Argument: There exists a denumerable subset of the set of irrational numbers

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My argument is that if you add a rational number to an irrational number it will still be an irrational number but you can count to it

ex: 1/n+sqrt(2) is this a vaild example?

Answer 1

Yes, it is valid.

Alternatively, $\{\sqrt2 n: n \in \mathbb{Z}^+\}$

Answer 2

Yes, rational+irrational = irrational. This can be proven by noting that if there is a rational number r and irrational i such that s=r+i is rational, then s-r = i, which would mean that there are two rational numbers whose difference is irrational.

Answer 3

Here’s a countable subset of the irrationals, with the additional property that its elements are independent over the rationals: Enumerate the primes, and take their square roots.