291 reputation
110
bio website
location
age
visits member for 1 year, 10 months
seen Dec 11 '13 at 23:13

Jul
10
awarded  Yearling
Jul
2
awarded  Curious
Jan
18
awarded  Popular Question
Mar
15
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
Mar
14
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.
Mar
14
awarded  Critic
Mar
14
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.
Mar
14
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
Mar
14
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.
Mar
14
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
Mar
14
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?
Mar
14
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?
Mar
14
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.
Mar
14
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?
Mar
14
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.
Mar
14
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.
Mar
13
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.
Mar
13
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.
Mar
13
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.
Mar
13
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.