Calculus is about net change, so we'd be looking for the net change from 0 to 1. You can put that on a number line call it F(b) - F(a) and you have the Fundamental Theorem of Calculus.

And again that doesn't have anything to do with solving Zeno's paradox.

We I get done with work, I can do up a visual I think which may help.

No graph is going to help. If asked to count the reals from 0 to 1, which is the first number you count after 0? There isn't one. You can't even start. That's the logic of Zeno's paradox, and no amount of calculus is going to help us here.

Treating a supertask just as a geometric series is misplaced.

I am not concerned with the parts as the parts are infinite, that is not the point, I am interested in the sum of the parts which is finite.

I am interested in the sum of the parts which is finite. — Jeremiah

Which doesn't have anything to do with solving Zeno's paradox.

Yes it does, this distance from Achilles to the Turtle is finite, even after accounting for the movement of the Turtle. The only thing Zeno's paradox does is create an infinite number of points on a finite number line.

Yes it does, this distance from Achilles to the Turtle is finite, even after accounting for the movement of the Turtle. The only thing Zeno's paradox does is create an infinite number of points on a finite number line. — Jeremiah

The number line between 0 and 1 is finite, and yet it is still impossible to count the real numbers between them.

Here's the typical solution to Zeno's paradox:

...if I can divide up a finite distance into an infinite number of small distances, then adding all those distances together should just give me back the finite distance I started with. (An infinite sum such as the one above is known in mathematics as an infinite series, and when such a sum adds up to a finite number we say that the series is summable.)

Now the resolution to Zeno’s Paradox is easy. Obviously, it will take me some fixed time to cross half the distance to the other side of the room, say 2 seconds. How long will it take to cross half the remaining distance? Half as long—only 1 second. Covering half of the remaining distance (an eighth of the total) will take only half a second. And so one. And once I have covered all the infinitely many sub-distances and added up all the time it took to traverse them? Only 4 seconds, and here I am, on the other side of the room after all.

But now let's apply this reasoning to my example of counting the reals:

It will take me some fixed time to count half the numbers between 0 and 1, say 2 seconds. How long will it take to count half the remaining numbers? Half as long—only 1 second. Counting half of the remaining numbers (an eighth of the total) will take only half a second. And so on. And once I have counted all the infinitely many reals and added up all the time it took to count them? Only 4 seconds, and there I am, having counted them after all.

There are two issues with this. The first is that it begs the question by asserting as a premise that counting the reals is possible (assume it takes 2 seconds to reach 0.5) and the second is that this premise must be false given that there isn't a first number to count to after 0.

The paradox is less to do with reaching the end and more to do with starting at all.

Just for clarity's sake: the problem you're pointing up is that the reals are uncountable. You could look at the rationals and say, there is no first one after zero, but this doesn't actually matter, because the rationals can be re-arranged into a list. You can just pick what to count as the first, the second, etc. You cannot do this with the reals.

Yes, rationals. That's the right term.

but this doesn't actually matter, because the rationals can be re-arranged into a list — Srap Tasmaner

Except to keep this analogous to movement the counting has to be in ascending order. We don't jump to the half-way point and then back to some earlier point.

Except to keep this analogous to movement the counting has to be ordered. We don't jump to the half-way point and then back to some earlier point. — Michael

I'd have to brush up on this to answer properly, but my instinct is that that's an interpretation problem, essentially a matter of labeling. There's the standard interpretation, associated with the number line, of what order numbers are in, but they don't have to be. That may not look like much of an answer.

And infinite tasks of any kind are still beyond the capabilities of finite beings. The difference between the rationals and the reals is that even if you had infinite time or could count infinitely fast, you still couldn't count the reals.

Of course if space is granular, then our task is finite, yes?

But yet it is not impossible to sum them, which is what we are concerned with.

I'd have to brush up on this to answer properly, but my instinct is that that's an interpretation problem, essentially a matter of labeling. There's the standard interpretation, associated with the number line, of what order numbers are in, but they don't have to be. That may not look like much of an answer. — Srap Tasmaner

Then to make it simpler, imagine a machine moving from A to B, where B is 1 metre from A. At set intervals the machine records the distance in metres it has travelled such that it can be mapped to the geometric series:

$0m$ $... \frac{1}{8}m, \frac{1}{4}m, \frac{1}{2}m, 1m$

What is the first position on its recording after the start point, i.e. the first count after 0? There can't be one. It's impossible to even begin such a count, and so it is impossible to even begin such a movement.

But yet it is not impossible to sum them, which is what we are concerned with. — Jeremiah

We're concerned with solving Zeno's paradox. Summing a geometric series doesn't solve Zeno's paradox.

Of course if space is granular, then our task is finite, yes? — Srap Tasmaner

Not necessarily space but movement. I don't see a problem with continuous space but with movement occurring by "jumping" from one position to the next without passing through some half-way point. It seems the only way for movement to make sense at all. It might be unintuitive, but that's better than the incoherence that continuous movement seems to be.

This is getting confusing, so big thanks to @Jeremiah!

What you're pointing out now, I think, is that the rationals (or, I guess $\{ \frac{1}{2^n}\}$) are not well-ordered under $\lt$, and that's true.

The reason we care is because we're talking about movement, and movement looks like a matter of going from one place to the next, where "next" is already defined in a particular way.

Which gives rise to another paradox? Given an infinite amount of time I could hop to all the rationals between my starting point and any destination (inclusive), but I cannot do them in order from closest to where I start to farthest (i.e., at my destination). So whatever that is, it doesn't look much like movement in the usual sense.

Getting back to our finite world, any finite subset of $\{ \frac{1}{2^n}\}$ is well-ordered under $\lt$, so that's what we're looking for I guess.

I am unclear on whether progress is being made, which is pretty freaking ironic.

Given an infinite amount of time I could hop to all the rationals between my starting point and any destination (inclusive) — Srap Tasmaner

Could you, though? Is there a way to prove it without begging the question, as this "solution" does, and taking as a premise that it takes $n$ seconds to count some percentage of them?

As with any supertask, I'd say that completing it is impossible. Unlike the ordered variation, where it's impossible to start, in this case it's impossible to finish. It would be a variation of Thomson's lamp where for each number counted you toggle the switch. Is the lamp on or off when you finish?

You are just talking about a continuous number line, which is actually a requirement for the FTC. The curve must be continuous from a to b.

You are just talking about a continuous number line, which is actually a requirement for the FTC. The curve must be continuous from a to b. — Jeremiah

But isn't it simply the case that you are confusing the continuum with reality?

Don't tortoises and demi-gods obey the laws of physics rather than the rules of certain branches of mathematics?

Map/territory confusion.
The paradoxical claim is:

This would actually make all motion impossible — Jeremiah

The realm is ultimately physics. If the mathematical models (maps) cause paradoxes so much the worse for their application in this instance.

I don't see what this has to do with mathematics. — Michael

This doesn't seem consistent with your continuing focus on the maths.

Map/territory confusion. — Baden

I don't think so. It seems much more like an unfamiliarity with the laws that govern reality, and the mistaken assumption that these laws admit the (decoherent) continuum.

The realm is physics. If the mathematical models (maps) cause paradoxes so much the worse for their application in this instance. — Baden

The statement of what exists in reality, how it behaves, and why, can be expressed in any language. Mathematics lends itself to a particularly efficient expression of the laws of reality, in a form amenable to testing. It's not a map, it's an assertion.

If it took me 1 second for each hop, it would take me countably many seconds to do all the rationals, the same number of seconds it would take to hop to all the natural numbers in order. The only point here is that you cannot hop to all the reals in countably many seconds.

A supertask has countably many steps completed in a finite amount of time. Seems like we could get go "infinitely fast" and get a finite amount of time, instead of taking an infinite amount of time at a finite speed. I'm not clear whether the definition of "supertask" precludes going infinitely fast in this sense. Maybe "infinitely fast" doesn't even make sense the way "infinite amount of time" does. Maybe that's part of the point?

If it took me 1 second for each hop, it would take me countably many seconds to do all the rationals, the same number of seconds it would take to hop to all the natural numbers in order. — Srap Tasmaner

Countably, sure, but still infinite. You never actually finish.

A supertask has countably many steps completed in a finite amount of time. — Srap Tasmaner

Thomson's lamp paradox shows that this leads to a contradiction.

For the programmers, assume a computer that could run "infinitely fast" counts each integer and checks to see if it is even. What is the value of $$state$ at the end? True or false?

At the end of what? When it reaches the largest integer?

You were talking about it being possible to finish counting (if "infinitely fast") a countably infinite set weren't you? The above was an example to show that that doesn't make sense.

That is the different between a convergent series and a divergent series. We know the distance between Achilles and the turtle converges to a finite number, so you are kind of arguing a moot point.

Also it should be pointed out that infinity is just an amount too big to measure. It could be discrete; it does not necessarily have to be continuous; such as, there are an infinite number of stars in the heavens.

Physics is ultimately in the realm of mathematics. The entire backbone of science is mathematics, as well as the rest of the skeleton. Basic human comprehension is rooted in mathematics, everything is in the realm of mathematics. That's the very reason I study it.

Maybe the concept of a supertask is what is being misplaced here. 0 is your starting point, now all I need is the end point, let's pick 1. Net change on a straight line is 1.