Prove that the set of irrational numbers is uncountable?

Answer 1

What do you know so far?

If you know that the set of real numbers is uncountable and that the set of rational numbers is countable, then you may conclude that the set of irrational numbers is uncountable since otheriwse the set of real numbers would be the union of two countable sets!

If you don't know that the set of real numbers is uncountable, it must be proved. Similarly for the set of rational numbers being countable and the union of two countable sets being countable.

Answer 2

The set of all irrational numbers is uncountable (since the rationals are countable and the reals are uncountable). The set of algebraic irrationals, that is, the non-transcendental irrationals, is countable. Using the absolute value to measure distances, the irrational numbers become a metric space which is not complete. However, this metric space is homeomorphic to the complete metric space of all sequences of positive integers; the homeomorphism is given by the infinite continued fraction expansion. This shows that the Baire category theorem applies to the space of irrational numbers. Whereas the set of all reals with its usual topology is connected, this Baire space, topologized in the same way as the reals, namely with the order topology, is totally disconnected: there is no path from any irrational to any other along the irrational line.

Answer 3

That's essentially a non existence proof. A set is countable if there's a bijection between the whole numbers (N) and the set.

The proof involves showing that for any function, f, from N->irrationals, you can construct and irrational that's not in the image, hence f is not a surjection and hence not a bijection. It involves representing an irrational as an infinite decimal and then finding another irrational between 2 elements. It's not a trivial manner. I think that it was Cantor who did this.

I'd recommend Wikipedia:
http://en.wikipedia.org/wiki/Uncountable