On Fri, Nov 16, 2012 at 6:54 PM, Allan Wechsler <acwacw@gmail.com> wrote:
People often don't like to talk about n-adic numbers when n isn't prime, because the resulting structure fails to be a field. I think that's the problem, anyway -- I don't remember too clearly.
It's worse than that; there's no good way to define multiplication in the n-adic numbers for n composite. You can define a metric on the rationals by | a - b | = 1/n^k, where k is the largest power of n dividing a - b. The n-adics are then formed by taking the topological completion of this metric space. If multiplication is continuous, you can then extend multiplication to the n-adics by continuity. But continuity of multiplication depends on the fact that | a | * | b | = | (a * b) | which only holds for n prime. Andy
On Fri, Nov 16, 2012 at 6:48 PM, Dan Asimov <dasimov@earthlink.net> wrote:
I'm not very familiar with the n-adic numbers, but I think that
...999.0... = -1
is true in the 10-adic numbers, where the reasoning Marc uses can be justified rigorously.
(Alternatively, just add 1 to ...999.0... to get 0.)
--Dan
On 2012-11-16, at 6:32 AM, Andy Latto wrote:
On Fri, Nov 16, 2012 at 12:52 AM, Marc LeBrun <mlb@well.com> wrote:
Since this was originally an historical question perhaps as an arithmetical post-modernist punk I should hold my piece, but perhaps those that hold that
...0.999... = 1
will also equably agree that
...999.0... = -1?
After all, are we not told (presented with precisely the same corroboration as given for 0.999...=1), that the geometrical series with term ratio r=10:
9 10^0 + 9 10^1 + 9 10^2 + ...
must be equal to 9/(1-r) = 9/(1-10) = 9/-9 = -1?
This isn't post-modern, it's pre-classical. This is exactly the way Euler reasoned, and he came up with the same sort of conclusions. But the reasons I, and a modern mathematician, say that .99999.... = 1 not just because of the proof of the formula for the sum of a geometric series (which, if you examine it closely, shows only that *if* the series has a limit, *then* the limit is 1), but because of the epsilon-definition of a limit, and a proof I can provide, using that definition, that this series indeed does sum to 1. If you try to provide a similar epsilon-delta proof that the series ...9999.0 sums to -1, you will fail, and I can provide an epsilon-delta proof that this series does *not* have a limit.
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