Re: Is continuum completely filled up?

On Feb 12, 1:10 pm, "toshiaki" <fara...@xxxxxxxxx> wrote:
"Randy Poe" <poespam-t...@xxxxxxxxx> wrote in message

news:1171210459.553315.248210@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

When we deal with infinite numbers, some of them are not different,

No, this is not true. If you have two reals x and y which are
not equal, then they differ in some finite position. The reason
for this is that "differ" means that |x-y| is a real number,
which means it has a nonzero digit in some finite position.

You say that if x is not equal to pai,

By the way, the spelling is "pi".

In Japan the spelling "pi" likely cause us to make different
pronounciation.

|pai - x| =\= 0.

Yes.

But when we examine
a number 3.1415...., we cannot decide whether this is pai or not,

If it is not pi, it differs in some finite position. We
can decide by finding that digit, either through logic
or calculation.

The candidates of this finite position exist infinitely. Decimals are not
clearly decided, until its all digits are shown.

This is nonsense. If x is a real number, all of its digits
are fixed.

They have no label.

The existence of the n-th digit does not depend on
knowing what that n-th digit is.

because
how much we may examine, there remain infinite unexamined.

Yes, but the place where it differs is a finite position.
You don't need to examine all the digits, only the first
one where it differs.

All digits are in finite position, if I don't misunderstand.

All digits are in finite positions.

Algorisms or explicit description is only way to describe clear objects.

Incorrect. Pi is an example known for thousands of years.
Its description is clear and unambiguous.

I think that its definition is clear and unambiguous, isn't it?

Yes. Its definition is clear and unambiguous. This gives
us a way to determine if any number is equal to pi or
unequal to pi, even without knowing all the digits of pi.

- Randy

.

Relevant Pages

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