A simple question.

[LastPoet]

@Adrian: That is a consequence of the theorem that the Set of Real Numbers isn't a Numerable Set right? The demostration of that is awesome And the fact that the rational number set & the irrational number set are dense.

From the literature I have read 2^(infinity) is used to represent the cardinality of sets that have (like R) the power of the continuum.

@ Zenith: The infinite sums to wich we can give a numerical result are called convergent series

[Zenith]

@ Zenith: The infinite sums to wich we can give a numerical result are called convergent series

Yeah, I think I remember that. But what do you want to say with it?

all agree that 0.(1) is a number, so why won't 0.0000...0001 be a number too? both have an infinite number of decimals...

[LastPoet]

Because it wouldn't be a finite spot where you could stop to put the last '1'. You cannot put Real numbers in sequence, like you can with natural or even rational numbers, because of the completeness principle. In the sequence 1, 1/10, 1/100, 1/1000,...,1/(10^n) For all the positive arbitrary real number x, no matter how small you fixate it, I'll find an order P, that for all n > p => Abs(1/(10^n)) < x. The limit for that sequence is 0. A number written in 0,00...0001 is finite.

[Zenith]

You say that 0.00...0001 is finite.
How can you call 0.00...0001 and 0.11...1111 infinite??
the former is the representation of product(i=1,infinite) of 1/(10^i) = 1/10 * 1/100 * 1/1000 * ...
and the later is the representation of sum (i = 1, infinity) of 1/10^i = 1/10 + 1/100 + 1/1000 + ... = 0.(1)

The former might be written 0.(0)1 but I guess this is not allowed to be written. Anyway, you can't find a number smaller than it because there are an infinite number of decimals (which mean, an infinite number of "0" decimals, all before that "1").

[Tiberius]

You say that 0.00...0001 is finite.
How can you call 0.00...0001 and 0.11...1111 infinite??
the former is the representation of product(i=1,infinite) of 1/(10^i) = 1/10 * 1/100 * 1/1000 * ...
and the later is the representation of sum (i = 1, infinity) of 1/10^i = 1/10 + 1/100 + 1/1000 + ... = 0.(1)

0.00...0001 is finitely long because it ends with a 1 (i.e. there is nothing after the last 1). 0.111... is infinitely long because it doesn't have an end by definition.

As I've said before, using an infinite geometric series doesn't use infinity in the actual calculation. The series tends to infinity, and the answer it calculates is derived via a specific geometric formula instead: https://secure.wikimedia.org/wikipedia/e...ric_series

For your sum example:

1/10 + 1/100 + 1/1000 ...

Where a = 1/10, and r = 1/10, the sum of the infinite series is a / (1 - r), which is:

(1/10) / (1 - 1/10)

= 0.1 / 0.9

= 0.111...

For the product example:

1/10 * 1/100 * 1/1000, since the r value is less than 1, the product is said to diverge to zero as the calculation progresses. Whether the actual product of the calculation is in fact 0 is still debated by mathematicians, but it doesn't matter in this case because if it isn't, we are unable to write the actual product in mathematical terms.

Anyway, you can't find a number smaller than it because there are an infinite number of decimals (which mean, an infinite number of "0" decimals, all before that "1").

There can't be an infinite number of "0" digits before a "1". In order for there to be a "1" on the end, there needs to be an end. If there are an infinite number of "0" digits then there cannot possibly be an end for the "1" to go on. So either your string of "0" digits has and end for the "1" to go on (therefore making the number finitely long), or there isn't such a number that you speak of.

[theVOID]

You say that 0.00...0001 is finite.
How can you call 0.00...0001 and 0.11...1111 infinite??

They aren't infinite, all you have there is an undefined number of places before the 1, you've still got an end to the series thus it's not infinite.

[Zenith]

0.00...0001 is finitely long because it ends with a 1 (i.e. there is nothing after the last 1). 0.111... is infinitely long because it doesn't have an end by definition.

By the same logic I can say that 0.(1) = 0.11...111 is finite because it ends with "1". In other words, my example hasn't an end by definition, either. If this "0.00...001" does not look as you'd like, then use this representation:

"1/10 * 1/100 * 1/1000 ..." - because this is clearly not an "invalid" thing.

As I've said before, using an infinite geometric series doesn't use infinity in the actual calculation.

I don't care if infinite geometric series use or not the infinity in the actual calculation. I also don't care how this thing should be called. But, if you insist that "1/10 * 1/100 * 1/1000 ..." must not be used, perhaps you can show where it's written that this thing is forbidden to be used.

[theVOID]

It DOES have an end, you simply have an undefined number of places between the start and end of the sequence. Since the sequence has an end it is, by definition, finite.

[Zenith]

ok, if it is finite, show me a number greater than 0 but less than this number.

[theVOID]

You haven't properly defined the number, so that's not possible.

You have a number with a start and an end but no defined length, because it has a beginning and an end it is necessarily finite.