> So the argument is that if you assume you can always pick an interval that doesn't contain a specified algebraic you won't land on an algebraic. No, the argu

> So the argument is that if you assume you can always pick an interval that doesn't contain a specified algebraic you won't land on an algebraic. No, the argument is that there is a sequence that converges to some real number, but that real number is not algebraic, hence there exist non-algebraic real numbers. I do agree that the article doesn't clearly describe how to construct said sequence - perhaps the second page of [1] will be clearer. Basically we always have an interval (a_n, b_n) where the a_n and b_n are algebraic, and we keep narrowing the interval by going down the list of algebraic numbers until we find two numbers x and y such that a_n < x < y < b_n; then (x, y) is the new interval. The sequence of the a_n converge but not to an algebraic number, because if it did, that number would eventually become an endpoint of the sequence, which is a contradiction. [1]: http://www.cwu.edu/~harperj/472%20Winter%202012/Cantor-rev2.pdf http://www.cwu.edu/~harperj/472%20Winter%202012/Cantor-rev2....