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You get to a contradiction if you assume a/b = sqrt(2) where a,b integers using just algebra no? No bijection between rationals and real, so uncountable. Not su
by kelipso 4y ago
You get to a contradiction if you assume a/b = sqrt(2) where a,b integers using just algebra no? No bijection between rationals and real, so uncountable. Not sure where other assumptions come in but don't remember Cantors diagonal argument.
- floxy 4y agoYou may, or may not enjoy chapter 5: https://arxiv.org/pdf/math/0404335.pdf https://arxiv.org/pdf/math/0404335.pdf ..."Why should I believe in a real number if I can’t calculate it, if I can’t prove what its bits are, and if I can’t even refer to it? And each of these things happens with probability one! The real line from 0 to 1 looks more and more like a Swiss cheese, more and more like a stunningly black high-mountain sky studded with pin-pricks of light."