3 ms·
It doesn't just prove that there's infinitely many irrationals -- that is trivially easy to prove. For instance we know pi is irrational, therefore 2pi, 3pi, 4p
by hugh3 16y ago
It doesn't just prove that there's infinitely many irrationals -- that is trivially easy to prove. For instance we know pi is irrational, therefore 2pi, 3pi, 4pi, 5pi,... etc are all irrational as well.
Cantor's proof proves that there's no one-to-one correspondence between reals and integers (whereas there is a one-to-one correspondence between integers and rationals), and therefore the infinite set of reals is "bigger" than the infinite set of rationals.
That there is an infinite number of irrationals should follow as a trivial consequence of this. (Left as an exercise for the reader...)
- ihodes 16y agoI understand this: I was attempting to show, intuitively, how his theory could demonstrate that there is more than one irrational number. Apologies if I wasn't clear: this was something that popped into my head to help clear this up!
- jplewicke 16y agoWhat's also somewhat interesting is there's a non-trivial countable set of irrational numbers: those formed by taking the roots of polynomials with raitonal coefficients. http://en.wikipedia.org/wiki/Algebraic_number http://en.wikipedia.org/wiki/Algebraic_number has the low-down, but it's definitely neat that square roots, etc. are still a countable set.