Infinity

KALSTER · 09-04-2008

Quote:

Originally Posted by Little Bang

Name something in the observed universe that can be proven to be infinite.

Well, something can only be proven infinite by using mathematics. In a big bang universe though, nothing is physically infinite AFAIK.

Tormod · 09-04-2008

Quote:

Originally Posted by KALSTER

In a big bang universe though, nothing is physically infinite AFAIK.

The universe might well be infinite even in a big bang universe. We currently have no way to prove - or even test - what's beyond the visible universe.

Qfwfq · 09-06-2008

Quote:

Originally Posted by KALSTER

But can we deal with that without all the cardinality business?

With topology. Not that I see what's wrong with cardinality...

Quote:

Originally Posted by Erasmus00

Actually its pretty well accepted that $0.\overline{9999}$ is an equivalent decimal expansion to 1.

More than pretty well accepted, I'd say it's pretty well proven.

Quote:

Originally Posted by Buffy

Sure, but it the proofs on both sides are so unsatisfying, and at the same time fun to manipulate!

But the proof that it's equal to 1 is by no means unsatisfying Buffy! Remember the geometric series?

$0.\overline{9}=0.9\sum_{n=0}^{\infty}(0.1)^n=0.9\frac{1}{1-0.1}=0.9\frac{1}{0.9}=1$ (and that means exactly!!!)

Quote:

Originally Posted by Tormod

The amount of people who believe strange things?

According to Einstein though it would be human stupidity...

KerryK · 09-07-2008

Quote:

Originally Posted by Qfwfq

With topology. Not that I see what's wrong with cardinality...

More than pretty well accepted, I'd say it's pretty well proven.

But the proof that it's equal to 1 is by no means unsatisfying Buffy! Remember the geometric series?

$0.\overline{9}=0.9\sum_{n=0}^{\infty}(0.1)^n=0.9\frac{1}{1-0.1}=0.9\frac{1}{0.9}=1$ (and that means exactly!!!)

Here's another proof, although I don't quite know how to make it look right (maybe Tormod can help out my igorance).

Let x = 0.999999' (where I'm letting the ' mark mean "followed by an infinite number of the same".

Consider 10x = 9.999999' (right?)

Now, subtract an x from each side:

10x - x = 9.99999' - x

But this means that 9x = 9

Hence, x = 1

How's that?

Kerry.

Tormod · 09-07-2008

Kerry, you need to read up on using the LaTeX tag here at Hypography.

http://hypography.com/forums/physics...th-v2-0-a.html

KerryK · 09-07-2008

I knew you'd direct me correctly. I'll do that. Thank you!

Tormod · 09-07-2008

Quote:

Originally Posted by KerryK

Now, subtract an x from each side:

10x - x = 9.99999' - x

But this means that 9x = 9

Wouldn't that be $9x=9.999999'-x?$

Buffy · 09-07-2008

Quote:

Originally Posted by Qfwfq

But the proof that it's equal to 1 is by no means unsatisfying Buffy! Remember the geometric series?

$0.\overline{9}=0.9\sum_{n=0}^{\infty}(0.1)^n=0.9\frac{1}{1-0.1}=0.9\frac{1}{0.9}=1$ (and that means exactly!!!)

Sure its unsatisfying because:

$\sum_{n=0}^{\infty}(0.1)^n=\frac{1}{1-0.1}$

is definitional! It's just leaving the $lim$ stuff out!

Oh you know I'm just playing here dear...

Quote:

Originally Posted by KALSTER

If you have an infinite data set where $0.\overline{9999}$ is supposed to describe the probability of a certain property to be exhibited by each element of the data set, does that mean that among any number of elements in the infinite data set you would care to consider that all of them would exhibit this particular property?"

(concerning the above) If you were to limit the data set to say $10^{100}$ elements...

Going down the route of analyzing this by bringing in probability is probably going to be less edifying, but there are a few points to be made here.

If you are really going to be doing what is called "Discrete Probability"--that is use a limited set of countable elements--then the definition of any particular probability is given by:

$p = \frac {number\ of\ elements\ possessing\ property\ x} {number\ of elements\ in\ the\ dataset}$

Thus if you are going to use a finite data set, $0.\overline{9999}$ really is no longer relevant unless you use the substitution that equates that with $1$ . What you would really have is (simplifying down to a smaller dataset):

$p = \frac {n-1} {n} = \frac {999999} {1000000} = 0.999999$

because there's no "infinite representation" in a discrete probability

Conversely if the dataset were infinite, then the substitution $0.\overline{9999}=1$ would say that every element of the infinite set would have the property. But this also presents one of the counter proofs! That is, by saying that the probability is $0.\overline{9999}$ in an infinite set RATHER THAN saying that the probability is $1.0$ you are definitionally saying that exactly one out of the infinite set does not have the property!

This is basically what you seem to be trying to get at in saying:

Quote:

Originally Posted by KALSTER

...But if the probability for this property to be exhibited per cycle were to be < 1, with any number of 9's after the comma, and you could leave the data set for as long as you wished and the number of elements that did not exhibit the property would start going up.

but the "number of elements that did not exhibit the property" would not "go up" because the "inverse probability" implied by $0.\overline{9999}$ is (colloquially) $0.00...infinite\ number\ of\ zeros....0001$ which unfortunately, plugging into the formula for probability is $\frac 1 \infty$ . Mathematically speaking (and this is a corollary to the original argument) these are equivalent to zero, but the 1 in the numerator still *implies* "there is one element without the property x."

Note that this is *really* more of an example of the problem of *applying* mathematics to not only "the real world" but to other conceptual--as opposed to concrete--scientific disciplines (don't you dare tell a mathematician that statistics is mathematics or he'll stick his chalk in your eye!).

What gets left out is the narrative between the bullets, which would tell us who's going to do what and how we're going to achieve the generic goals on the list,

Buffy

KerryK · 09-08-2008

Quote:

Originally Posted by Tormod

Wouldn't that be $9x=9.999999'-x?$

Tormod,

No, it wouldn't. If you take 9.99999' and subtract the 0.99999' from it, you're left with 9. No tricks. But x = 0.99999'.

So, 9.999999' - x = 9.

And 10x - x = (10-1)x = 9x.

So, 9x = 9 and x = 1.

(Sorry, I've not had the time to check out LaTex.)

Kerry

Tormod · 09-08-2008

I don't get the logic behind that.

10x=9.99999'

Subtract x:

10x-x=9.99999'-x

Would yield:

9x = 9.99999' - x.

Where am I going wrong? Do you divide by nine?

-T

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