The theorem states. Let A be a set and Let c be a non empty countable set disjoint from A. If A has a denumerable subset then A union C is equinumerous to A and conversely.

I was thinking of letting A be the rational numbers, and letting C be the irrational numbers that way it's disjoint, and then the subset of A would be integers, but then so the union of integers and irrational numbers would be equinumerous to rational numbers, but that doesn't help with the equinumerous of irrational and real numbers. Not sure what to do.


Let $A$ be the irrational number and $C$ be the rational number, then $A$ and $C$ are disjoint.

Check that $A$ has a denumerable subset and you can conclude that $A \cup C = \mathbb{R}$ is equinumerous to $A$, the set of irrational numbers.

What remains for you to do is to check that $A$ has a denumerable subset, can you complete that?

  • $\begingroup$ Perfect ok so a subset of irrational numbers that is denumerable? hmm could I just say the set of irrational numbers less than sqrt(2)? $\endgroup$ – fwefwf Apr 23 '18 at 5:10
  • $\begingroup$ hmmm, the set of irrational numbers less than $\sqrt2$ is not denumerable. Hint: $\{ 1, 2, \ldots, \}$ is denumerable, try to link them with $\sqrt2$ to make it irrational. $\endgroup$ – Siong Thye Goh Apr 23 '18 at 5:12
  • $\begingroup$ I can't think of a possible subset of irrational numbers? what could it possibly be? could you just say the set of all numbers equal to sqrt(2) that way it's just one value and that is obviously denumarable? $\endgroup$ – fwefwf Apr 23 '18 at 5:13
  • $\begingroup$ Just multiply $\mathbb{N}$ with $\sqrt2$ elementwise. you can also do addition, subtraction, division. $\endgroup$ – Siong Thye Goh Apr 23 '18 at 5:16

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