AlexHeuman
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 Jul10 awarded Yearling Jul2 awarded Curious Jan18 awarded Popular Question Mar15 revised What was the first bit of mathematics that made you realize that math is beautiful? (For children's book) added 1323 characters in body Mar14 accepted If $x\lt y$ for arbitrary real x and y there exists a real r $r$ such that $x \lt r \lt y$ and hence infinitely many. Mar14 awarded Critic Mar14 comment If $x\lt y$ for arbitrary real x and y there exists a real r $r$ such that $x \lt r \lt y$ and hence infinitely many. Thanks, I see what you're saying. That's a nice way of doing it. Mar14 revised If $x\lt y$ for arbitrary real x and y there exists a real r $r$ such that $x \lt r \lt y$ and hence infinitely many. edited tags Mar14 comment If $x\lt y$ for arbitrary real x and y there exists a real r $r$ such that $x \lt r \lt y$ and hence infinitely many. @PhilipBenjMarcobyEragon I'm not sure I know what you mean. Mar14 revised If $x\lt y$ for arbitrary real x and y there exists a real r $r$ such that $x \lt r \lt y$ and hence infinitely many. deleted 1 characters in body; edited tags Mar14 comment If $x\lt y$ for arbitrary real x and y there exists a real r $r$ such that $x \lt r \lt y$ and hence infinitely many. @GitGud, thanks for the answer, but the section of the book I'm going through is just introducing the well-ordering principle and the Archimedian property of numbers. I think that my proof should be based on these ideas. Which is why I have presented my proof in such a way that I have. What I'm mostly trying to gain from this question is, is there an error with my proof? If so, what is it? Mar14 comment If $x\lt y$ for arbitrary real x and y there exists a real r $r$ such that $x \lt r \lt y$ and hence infinitely many. Also, is there an error with my proof? Mar14 comment If $x\lt y$ for arbitrary real x and y there exists a real r $r$ such that $x \lt r \lt y$ and hence infinitely many. I understand that, but with induction, you usually show for the case k and k+1, here you are just doing repeating processes. I could be wrong, but that just doesn't seam kosher to me. Mar14 comment If $x\lt y$ for arbitrary real x and y there exists a real r $r$ such that $x \lt r \lt y$ and hence infinitely many. I suppose that I can repeat the argument, but would there be a way to show that this works infinitely many times, or just as many times as I repeat the argument, and is there something wrong with my proof? Mar14 comment If $x\lt y$ for arbitrary real x and y there exists a real r $r$ such that $x \lt r \lt y$ and hence infinitely many. I can't, because I have to show that there are infinitely many r between x and y. Taking the average, does not show that. Mar14 asked If $x\lt y$ for arbitrary real x and y there exists a real r $r$ such that $x \lt r \lt y$ and hence infinitely many. Mar13 accepted If x is rational, $x\ne 0$, and $y$ irrational, prove $x+y, x-y, xy, x/y$ and $y/x$ are all irrational. Mar13 comment If x is rational, $x\ne 0$, and $y$ irrational, prove $x+y, x-y, xy, x/y$ and $y/x$ are all irrational. Okay, I get what you're saying now. Thanks for your help. Mar13 comment If x is rational, $x\ne 0$, and $y$ irrational, prove $x+y, x-y, xy, x/y$ and $y/x$ are all irrational. I don't seem to follow your reasoning. I think I'm missing something key here. Mar13 asked If x is rational, $x\ne 0$, and $y$ irrational, prove $x+y, x-y, xy, x/y$ and $y/x$ are all irrational.