Not that anyone cares but I just realized that this is not actually paradoxical and I can prove it mathematically! (I think) Bear with me since I’ve like just barely learned this stuff this week.
Proof Let S be the set of all steps needed to be taken. It can be written as S = {(distance to be traveled)(2-n): n in the Natural numbers}. Thus, S shares cardinality with the natural numbers and is countably infinite.
However, time is continuous. Thus, it has the cardinality of the continuum (real numbers) which means any time interval contains an uncountably infinite amount of moments. Let us denote an arbitrary time interval as T.
Because | T | > | S | there is no injection from T to S. Thus if each step has only 1 time value, there will be moments of time left over, and since the hand is not in two places at once we know each step must have its own time value, so this must be the case.
Therefore, when moving in steps like this, one will run out of infinite steps before they run out of moments in time to complete those steps. Hence, any finite distance can be traversed in this way over some bounded interval of time. QED.
Basically, you can traverse any distance in any time interval as long as physics allows you to move at a fast enough speed. Even if it doesn’t, there may be a limit to how fast you can traverse the distance, but it is still bounded. You can traverse any finite distance like this before existence runs out of time.
(I’m still learning. So if there’s an error in my proof please be gentle lol)
I think this idea works, and it’s consistent with the way the ‘paradox’ was presented. I just need to point out that movements aren’t performed in steps, the intermediate positions are also continuous, so there are undoubtedly infinite positions to map with undoubtedly infinite instants. To me, that’s the “true” solution.
As I see it, the only reason the paradox presents halving steps is because it’s the easiest way to demonstrate that there are infinitely many steps. It doesn’t bother to show that there’s uncountably infinite steps because countable is sufficient.
But as I said, your proof (or disproof?) works for the argument as it was presented, so it’s good!
I much prefer the physics approach: disproof by experiment! “Look at my gavel. According to your theory, it has infinitely many steps to go through before it reaches the pad, so it can never hit the pad.” *bang* “Motion denied.”
I haven’t read the actual proof you wrote, but I think the paradox was shown to, well, not be a paradox multiple times in history. It was thought of long ago, when our understanding of mathematics and physics was more limited, and we didn’t have the tools to prove the flaws exist.
It’s pretty cool that a problem created so long ago can still be relevant, and that random people find it interesting enough to solve in Lemmy comments though! Keep learning and rock on!
I thought it was first halfway proved. And then an extra 1/4th proved. And then an additional 1/8th proved. And then the proof went an extra 1/16th. So if you can wait an infinite time, it will be proved.
Not that anyone cares but I just realized that this is not actually paradoxical and I can prove it mathematically! (I think) Bear with me since I’ve like just barely learned this stuff this week.
Proof Let S be the set of all steps needed to be taken. It can be written as S = {(distance to be traveled)(2-n): n in the Natural numbers}. Thus, S shares cardinality with the natural numbers and is countably infinite.
However, time is continuous. Thus, it has the cardinality of the continuum (real numbers) which means any time interval contains an uncountably infinite amount of moments. Let us denote an arbitrary time interval as T.
Because | T | > | S | there is no injection from T to S. Thus if each step has only 1 time value, there will be moments of time left over, and since the hand is not in two places at once we know each step must have its own time value, so this must be the case.
Therefore, when moving in steps like this, one will run out of infinite steps before they run out of moments in time to complete those steps. Hence, any finite distance can be traversed in this way over some bounded interval of time. QED.
Basically, you can traverse any distance in any time interval as long as physics allows you to move at a fast enough speed. Even if it doesn’t, there may be a limit to how fast you can traverse the distance, but it is still bounded. You can traverse any finite distance like this before existence runs out of time.
(I’m still learning. So if there’s an error in my proof please be gentle lol)
I think this idea works, and it’s consistent with the way the ‘paradox’ was presented. I just need to point out that movements aren’t performed in steps, the intermediate positions are also continuous, so there are undoubtedly infinite positions to map with undoubtedly infinite instants. To me, that’s the “true” solution.
As I see it, the only reason the paradox presents halving steps is because it’s the easiest way to demonstrate that there are infinitely many steps. It doesn’t bother to show that there’s uncountably infinite steps because countable is sufficient.
But as I said, your proof (or disproof?) works for the argument as it was presented, so it’s good!
I much prefer the physics approach: disproof by experiment! “Look at my gavel. According to your theory, it has infinitely many steps to go through before it reaches the pad, so it can never hit the pad.” *bang* “Motion denied.”
I haven’t read the actual proof you wrote, but I think the paradox was shown to, well, not be a paradox multiple times in history. It was thought of long ago, when our understanding of mathematics and physics was more limited, and we didn’t have the tools to prove the flaws exist.
It’s pretty cool that a problem created so long ago can still be relevant, and that random people find it interesting enough to solve in Lemmy comments though! Keep learning and rock on!
I thought it was first halfway proved. And then an extra 1/4th proved. And then an additional 1/8th proved. And then the proof went an extra 1/16th. So if you can wait an infinite time, it will be proved.
https://en.wikipedia.org/wiki/Supertask