Why can no one in sci.math understand my simple point?
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From: WM on 15 Jun 2010 07:39 On 15 Jun., 12:39, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > That's *all* that matters, for Cantor's theorem. The claim > is that for every list of reals, there is another real > that does not appear on the list. The claim is only proved for every finite subset of the list. Then it is extrapolated to an infinite list without more reason than to extrapolate other laws from the finite to the infinite. For every finite subset of Hercules' list we see that it contains as many of digits of pi as can be contained. Regards, WM
From: WM on 15 Jun 2010 07:40 On 15 Jun., 10:03, Derek Holt <ma...(a)warwick.ac.uk> wrote: > A little more precisely, there does not exist a computable bijection > from the natural numbers to the set of all computable reals. > > But the set of computable reals is of course countable so there does > exist a list of all computable reals - but not a computable list. In what form does an uncountable real exist? Regards, WM
From: WM on 15 Jun 2010 07:42 On 15 Jun., 08:58, fishfry <BLOCKSPAMfish...(a)your-mailbox.com> wrote: > In article <87ocucFrn...(a)mid.individual.net>, > > "|-|ercules" <radgray...(a)yahoo.com> wrote: > > Consider the list of increasing lengths of finite prefixes of pi > > > 3 > > 31 > > 314 > > 3141 > > .... > > > Everyone agrees that: > > this list contains every digit of pi (1) > > No, I don't agree, so "Everyone agrees that ..." is false. > > The list consists of a collection of integers. Item n on the list are > the first n digits of pi, starting from 3 and ignoring the decimal > point. So the 1000th item on the list is 31... pi to 1000 places. > > There is no one element of the list that contains pi in its entirety. > And the reason is because each 'n' represents a FINITE NUMBER. Like 6, > or 100043, or a zillion eleven. And on that line we find a zillion > eleven digits of pi. But no more! > > No one item on the list contains pi in its entirety. > > Do you understand that? > > What is true is that: if you ask me for, say, pi to a trillion digits, > I'll say, "No problem, here it is, it's the trillionth item on the > list." > > But if you ask me for ALL the digits of pi, I have to say, "Sorry, > that's not on the list." And if it is constructed within one line in the same way as done above in several lines? What is then? Can a countable number of digits of pi be constructed? Regards, WM
From: WM on 15 Jun 2010 07:43 On 15 Jun., 10:28, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > > But no single member of the list contains all digits of pi. > > Pi doesn't appear anywhere on the list.- And wheres does it appear ,as an infinite sequence not obtaiend from a finite formula? Regards, WM
From: WM on 15 Jun 2010 07:50
On 15 Jun., 12:26, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > (B) There exists a real number r, > Forall computable reals r', > there exists a natural number n > such that r' and r disagree at the nth decimal place. In what form does r exist, unless it is computable too? But if r is computable, then this theorem shows that the computable numbers are uncountable. Contradiction. And if r is not computable, then it is impossible to prove disagreement with any r'. Regards, WM |