Actual Infinity in Reality?

[RoadRunner79]

[polymath257]

[SteveII]

[RoadRunner79]

Back to the topic.... I think there is an question that wasn't answered, that should be.

  If it is said, that a line contains a continuum of points (however you choose to define them).  Despite the fact, that this supposed infinity ends at 1 which is contradictory to saying that it is infinite in number in itself. (note:  I'll use one as a destination in this writing, although it may be another length)   What is the point immediately prior to 1?  There is necessarily an instance, where you transition from "not 1" to "1" while traveling along this line. 

I don't think that those proposing an actual infinite can answer the question.   I believe that this question shows the bait and switch that is occurring (whether the presenter knows it or not).  I think it is also why I have found difficulty in these conversations in having someone define what the term "point" is.  (It much easier to play fast and loose, if you do not define your terms).  If the points along the line are in fact infinite, then there cannot be a transition from "not 1" to "1".  As the argument goes, no matter how small the number is between our last point and the destination, we can always make up another number which is yet smaller (nature of the decimal system).  And we can repeat this over and over again, never reaching 1.  The time doesn't matter; this will never end (which is correctly  the definition of infinite) .   This is what Zeno's dichotomy (runners) paradox shows .   And I don't think that this is being addressed.  To get from "not 1" to "1"  you have to end the infinity (thus not infinite).

If you follow the logic and the procedure that is used to get an infinity in this way, then you cannot logically reach the destination either (not if you are consistent).  Adding an infinity of points of time, does not change here that the process will never end (which is why time is inconsequential).  The fact, that it does end, and that motion is possible, shows that this idea of a infinity in any given line and any given motion, shows that this idea is not logical (or at least the way it is argued is not logical).  

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Why do you assume there is a point just before 1? In fact, I can prove that, in fact, there is no point that is the 'last point before 1'.

The proof is a simple one done via proof by contradiction. Suppose x is any point before 1, so x<1. Let y be the average of x and 1. Then x<y<1, showing x is NOT the last point. hence, no last point can exist.

We cannot 'reach the destination' IN THAT WAY. We still reach the destination, but not via stopping at each of those infinitely many points.

I'm not saying that you have to stop at the point. Passing  through is fine.  

Your demonstration did not offer a rebuttal, but further re-enforces my point.  This is exactly what I was saying. 

It's your own point that you are saying there is a contradiction with, even though you keep saying that there is no contradictions.  This is why myself and Steve are saying that you will never reach the destination of 1. 

This is using the same reasoning that you used to say that there is an infinite number of points, and showing that by that reasoning you cannot reach the destination.


Now if you would want to get into what you are calling an assumption that there is a point prior to our destination (A) where it is (!A); we can work through that.  However any way you get around this, I believe is going to cause you to abandon your model which is giving you an infinite number of "points".

[pocaracas]

dude...
1/3 = 0.3333333....
2/3 = 0.6666666....
3/3 = 0.9999999.... = 1

dude, 

1/3 does NOT equal 0.3333333...
2/3 does NOT equal 0.6666666...
3/3 does NOT equal  0.9999999...

you are just staying the same thing. It is wrong for the same reason. There is an infinite amount of 3's, 6's and 9's to the right. So it will NEVER equal the fraction or the whole number.

Well... when you say "never", you are implying the usage of time to perform the calculation, or the writing of the result, perhaps?...
But the ellipsis is there to signify that the representation of that number in the decimal system goes on "forever", it is infinite, never ending.

When I learned the algorithm for performing a division, it went something like this:
- 1 divided by 3
- how many times does 3 fit in 1? zero, actually...
- first digit: 0
- add decimal place on the 1 and the result, making it 1.0/3 and 0., so far.
- how many times does the 3 fit in the new value of 10 (we ignore the decimal point for simplicity - I could multiply the 3 by the decimal location we're considering on the answer, but it's too much of a hassle) ? 3
- 3*3 = 9, the remained to reach 10 is 1.
- while the remainder is not zero, repeat the above for increasing decimal location.
- You will note that you always get a 3 on the result, with a 1 in the remainder. This tells you that the 3s go on to infinity.
- the result is usually written in the form 0.(3)

This is just a decimal representation of a number in the decimal system.
Of course, as has been shown, some other system will give you a different representation.
1/3 is yet another representation, one that relies on the operation of division and is a bit more convenient for us humans than an infinite string of digits.


If we add 0.(3) to 0.(3), we get 0.(6). Add again and we get 0.(9), which, in our more convenient writing format, is 3/3, which is also identified as 1.


Which leads us to

If it is said, that a line contains a continuum of points (however you choose to define them). Despite the fact, that this supposed infinity ends at 1 which is contradictory to saying that it is infinite in number in itself. (note: I'll use one as a destination in this writing, although it may be another length) What is the point immediately prior to 1? There is necessarily an instance, where you transition from "not 1" to "1" while traveling along this line.

Here we should introduce the concept of the infinitesimal. Let's call it dx. dx is simply the limit, as x goes to zero, of x, or, in notation: lim(x->0) x.

The value directly before 1 would be: 1 - lim(x->0) x, or 1 - dx.

This is the only way I know how to represent this "number".
The infinitesimal is often used in derivative and integral algebra, where the derivative of a function of x is represented as df/dx, usually meaning derivative of f with respect to x, but the df/dx nomenclature is actually more powerful than just a simple representation of the derivative, as it allows some infinitesimal algebra when you want to integrate a derivative:
∫df/dx dx = f (+ a constant, but let's not go there)

Care for a simple example?

f(x) = x^2 = x*x

df/dx = f'(x) = lim(t->0) [f(x+t) - f(x)] / t
As you can see, this will always lead to an indeterminate 0/0, which is annoying, but it can be lifted, in most cases.
For our case:
df/dx = f'(x) = lim(t->0) [ (x+t)^2 - x^2 ] / t
= lim(t->0) [x^2 + t^2 + 2xt - x^2 ] / t
= lim(t->0) (t^2 + 2xt)/t
= lim(t->0) (t + 2x)
= 0 + 2x = 2x

So, the derivative of x^2 is 2x.
You can do it fox x^n and the result will be nx^(n-1)

Integration is the inverse operation and is always done, to the best of my knowledge, as a reduction of the function to be integrated into known derivatives.
Knowing that the derivative of x^2 is 2x, we can easily say that the integral of 2x is x^2.
Integration can also be though of as a successive sum of the function.

∫ 2x dx = Σ 2x dx, but remember that dx is an infinitesimal, so a normal sum isn't feasible.... however, computer numerical implementations do this all the time.

---

So from Hilbert's Hote we get:
infinity + infinity = infinity
infinity + infinity = infinity/2
infinity - 1 = infinity
infinity / 2 = infinity
infinity - infinity = 3

These are contradictory statements resulting from simple arithmetic operations (from 2).

I'd like to stress that those infinities there are not numbers. They are not bound by normal algebra.
I can give you examples, again, using limits.

infinity + infinity = infinity
lim(x->0) 1/x + 1/x^2 = ∞ + ∞ = ∞ >> I'm sure this one is obvious, huh?

infinity + infinity = infinity/2
lim(x->0) 1/(2x) + 1/(2x^2) = 1/2 { lim(x->0) 1/x + 1/x^2 } = 1/2 ∞ >> and, as we saw above, each of the initial parcels goes to infinity

infinity - 1 = infinity
lim(x->0) 1/x - 1 = ∞ - 1 = ∞
lim(x->0) 1/x - 1 = lim(x->0) 1/x - x/x
= lim(x->0) (1 - x) / x >> and now, because as the top x becomes close to zero, it becomes meaningless:
= lim(x->0) 1 / x = ∞

infinity / 2 = infinity
lim(x->0) 1/(2x) = 1/2 lim(x->0) 1/x = ∞/2 >> in this case, as x goes to zero, the result gets larger faster than 1/x alone!


infinity - infinity = 3
This one is trickier...
lim(x->∞) { (2x^2 - 5) / (x+3) } - lim(x->∞) { 2x } = ∞ - ∞, right?
we can do some algebra, though....
lim(x->∞) { (2x^2 - 5) / (x+3) } - lim(x->∞) { 2x } = lim(x->∞) { (2x^2 - 5) / (x+3) - 2x }
= lim(x->∞) {(2x^2 - 5) / (x+3) - 2x(x+3)/(x+3)}
= lim(x->∞) {[2x^2 - 5 - 2x(x+3)] / (x+3)}
= lim(x->∞) {[2x^2 - 5 - 2x^2 - 6x] / (x+3)}
= lim(x->∞) {(- 5 - 6x) / (x+3)}
= lim(x->∞) { - (6x + 5) / (x+3) } >> As x goes to infinity, the parts that don't have an x become meaningless, so this equals
= lim(x->∞) { - 6x / x }
= lim(x->∞) { - 6 }
= - 6

ok, it's not 3, but it's enough to show how to lift an indetermination of the type ∞ - ∞.
Sometimes, the result is ∞, sometimes it's -∞... in this case, it was a nice finite number.


I hope you had some fun with all this algebra. I know I did, but my boss didn't!

[polymath257]

Why do you assume there is a point just before 1? In fact, I can prove that, in fact, there is no point that is the 'last point before 1'.

The proof is a simple one done via proof by contradiction. Suppose x is any point before 1, so x<1. Let y be the average of x and 1. Then x<y<1, showing x is NOT the last point. hence, no last point can exist.

We cannot 'reach the destination' IN THAT WAY. We still reach the destination, but not via stopping at each of those infinitely many points.

I'm not saying that you have to stop at the point. Passing  through is fine.  

Your demonstration did not offer a rebuttal, but further re-enforces my point.  This is exactly what I was saying. 

It's your own point that you are saying there is a contradiction with, even though you keep saying that there is no contradictions.  This is why myself and Steve are saying that you will never reach the destination of 1. 

This is using the same reasoning that you used to say that there is an infinite number of points, and showing that by that reasoning you cannot reach the destination.


Now if you would want to get into what you are calling an assumption that there is a point prior to our destination (A) where it is (!A); we can work through that.  However any way you get around this, I believe is going to cause you to abandon your model which is giving you an infinite number of "points".

The contradiction is to the assumption that there is a last number before 1. That was a proof that there is not.

This is called a proof by contradiction. You assume something, get a contradiction, and thereby show it is false. In this case, the assumption was that there is a last number before 1. There isn't. But there is no reason to think there would be.

Your article was just hilarious in the complete lack of understanding of math contained in it. It assumes that infinite quantities act the same as finite. They don't. But there is no *logical* contradiction to aleph_0 +aleph_0 =aleph_0, as claimed.

At no point do we ever get a contradiction: a proof of both some statement and its negation.

It is just true.

[Amarok]

I thought Steve quit 

[RoadRunner79]

[polymath257]

The contradiction is to the assumption that there is a last number before 1. That was a proof that there is not.

This is called a proof by contradiction. You assume something, get a contradiction, and thereby show it is false. In this case, the assumption was that there is a last number before 1. There isn't. But there is no reason to think there would be.

If you are saying that you never reach the destination (A), then I would agree.  It was my understanding that you where saying that A was reached, as opposed to Zeno's conclusion that motion was impossible or you have your contradiction.

Here is my reasons for there needing to be an last point (note reasons, not an assumption). 

So we take a line between two points (0 and our destination) 
We'll define A as any point >= the destination, making any point less than the destination !A

We start at point 0 (!A)  and progress through the line of points (either finite or infinite) where the current point is always greater than the previous point.
1.If you reach the destination (A) then the last point you passed where !A is your final point.

I believe that your equation where you progress through where X<Y<A  is analogous to Zeno's paradox.   Where Y is the average between X and A;  and you keep repeating this sequence where then Y becomes X and you solve for Y again.

This involves the first principles of the Law of identity, the law of non-contradiction, and the law of excluded middle applied as we move along our line.  It assumes that we are progressing along the line, and that we do reach the destination.   We start off !A and progress through the point along the line until we reach A.

Now what myself and Steve have been saying, and that a number of people seem to not want to do the math on, is that given this approach; you will never reach A.  And I would agree, given that you never reach A, there will not be a last point prior to A (it's a nonsensical question in that case).  The contradiction would be in saying that both A and !A are true.  And what makes your claim of infinity also prevents you from ever ending at point A.

I think that we can take for granted, we are presented this in the middle.  Can I place another number in there... Sure.   However the problem arises when you start doing so on the ends, and realize that you either cannot finish or cannot start your journey with such thinking, and lead Zeno to conclude that motion is impossible.  Zeno according to the tale was a strict logician and even though it was demonstrated that motion was in fact possible, demanded that he be shown logically where he was wrong.

I hope that we are in agreement, that motion is possible. To me and others, Zeno's problem is not in that math, nor in his logic.  His problem was with his starting assumption of a continuum or having to cross an infinite.

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I think we reach the point A, but not by stopping at each point in that sequence. To stop would entail spending more than an instant at each point, and that would not allow us to reach the end, 1.

In that process, at no stage is the final point reached. But that isn't the complete description of how we get from 0 to 1. There is more there than just that process (which would finish in a finite amount of time).

In 1, you are assuming there is a last point. The proof above shows there is not. yes, we start at !A and proceed to A. But there is not a last point before A. Where is the contradiction?

There is no A and !A being true. We do reach that last point by going through each point in the sequence AND ALSO 1. If we limit ourselves to that sequence, we do not get to A, but we are not limited to being on that sequence. We do go through every point of the sequence *and* 1 also.

[RoadRunner79]