Actual Infinity in Reality?

[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.

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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!