"jbriggs444" <firstname.lastname@example.org> wrote in message news:email@example.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.