Re: OT: Once again Inez delves into the murky waters of philosophy.

J. J. Lodder wrote:
Rupert Morrish <rupert@xxxxxxxxxxx> wrote:

J. J. Lodder wrote:
Rupert Morrish <rupert@xxxxxxxxxxx> wrote:

J. J. Lodder wrote:
John Wilkins <j.wilkins1@xxxxxxxxx> wrote:

J. J. Lodder <nospam@xxxxxxxxxxxxxxxx> wrote:

John Wilkins <j.wilkins1@xxxxxxxxx> wrote:

J. J. Lodder <nospam@xxxxxxxxxxxxxxxx> wrote:

John Wilkins <j.wilkins1@xxxxxxxxx> wrote:
The standard term for what logicians and mathematicians often do
to prove a claim is "reductio ad absurdum": to show that X is
true, prove that not-X will lead you to a contradiction. This is
a bit like what
Inez is after.
Brouwer objects,

Yeah, well I don't trust intuitions.
Ah, you believe in god instead,

Not at all. I'm an agnostic about mathematical objects.
You didn't recognize a reference.
In one of their public confrontations at some conference
there was Brouwer was lecturing, and explaining things
like not all reals having a decimal expansion.
(which was considered most shocking at the time)
I'm still most shocked. Can you give an example of a real number that
does not have a decimal expansion?
One of the ways to arrange it is to construct a real number r
such that for succeeding approximations to it
1 - 10^{-n} < r < 1 + 10^{-n} holds,
with the greater or smaller depending on the outcome
of some Brouwerian question.
So you can know the number r to arbitrary precision,
(with n equal to the number of known decimals of pi for example)
but you still don't know the first decimal,
for you don't know if it starts with
0.9999999.... or 1.0000000.....
and there is no way to find out,

Best,

Jan

I'm not sure that answers the question. If you knew the value of r, you'd know the decimal expansion.

But you don't.
You only know it to arbitrary precision.
1 - 10^{-n} < r < 1 + 10^{-n}
for as far ijn n as you can/want to calculate.

I still have the feeling that this example is demonstrating something other than what you claim. The fact that you can't determine the decimal expansion of a number you haven't specified doesn't mean that there is an actual real number that doesn't have a decimal expansion. Until there's only 1 real number in the range (1 - 10^{-n}, 1 + 10^{-n}).

This looks more like one of those series where you can prove convergence, but the series oscillates around the convergent value.


More to the point, we do know the digits of r - 1 (though not the sign), so I'm not sure what the utility
of this example is.

You don't know even the first one.
The number is either 0,9999999999.........88888888...(repeating 8888-s),
or
1.0000000..................1111111...(repeating 1-s)
or unknown which of the two.

Most definitions of decimal expansion I've seen specifically state that
an infinite series of 9s is the same as 1 (in the appropriate place). r
would appear to be 1.
That's correct: decimal expansions are not unique.
0.999... equals 1.000....

However: This number does not equal 0.99999... or 1.00000...
If we were to know it's value
it would be either greater or smaller than one.

Best,

Jan
(BTW, subtracting 1 one sees that there are real numbers
which are not positive or negative)

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