|
|
Re: Bijection Between Complex Numbers and Real Numbers
Posted:
Feb 24, 2012 7:12 AM
|
|
"jbriggs444" <jbriggs444@gmail.com> wrote in message
news:6d5c038b-55ec-4b34-88e4-ab624115740f@9g2000vbq.googlegroups.com...
On Feb 24, 1:00 am, "Peter Webb" <r.peter.webb...@gmail.com> wrote:
> The explicit bijection between digit strings and real numbers gets
> you completely past the dual-representation problem.
> __________________________________________________
> I don't see how.
Right there. That is you being a stupid idiot.
If you have a bijection between the digit strings and the
real numbers then, by definition of bijection, you do
not have a single real number represented by two distinct
digit strings.
That takes you past the dual-representation problem.
__________________________________
Certainly, but it does not allow for a trivial digit interleaving to form a
bijection between R and R^2, which is the step I asked to be shown and which
you snipped.
> If you think I have said something incorrect in this thread, you
> should cut-and-paste it in context and we can discuss. Its almost
> certainly
> some lack of mathematical knowledge on your part.
Challenge met above. The lack is yours.
_____________________________________
I did in fact post a correct bijection based on digit interleaving elsewhere
in this thread. We have seen nothing from you. If you think that using the
mapping R -> P(N) to form a bijection between R and R^2 using digit
interleaving is trivial, show us how it is done. You will note that with the
digit interleaving bijection I provided earlier I provided a working
example, if you claim that a more trivial digit interleaving bijection
exists please show us how it works. Put up or shut up.
|
|
