Mathematical Paradox

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If the frog takes the same finite amount of time for each jump, then of course it will never reach the finish-line.

Xeno's paradox, as I remember, was about a continuously steadily moving person or animal.

In that case, each increment, while shorter, is also taking proportionately less time, and so a turtle walking 1/2 foot per second will reach a 10-foot-distant finish-line in 20 seconds, no matter how you choose to count its progress.

Michael829

We all responded to the original post with the same answer that you all are giving now on this last page (now that the thread has been reivived). The answer being that the frog would go on forever, and never reach the opposite wall.

But theoriginal poster insisted that there WAS an alternative answer. Some finite integer number that could be deduced from an equation. But he never gave the answer, nor even attempted to point us in the direction of any kind of math that would give his alternative to infinity answer. In fact when I tried to pin him down about it at one point in the thread, because he was already angry at me for some reason, he just gave me a snooty response.

If the frog jumps an infinite number of times then it will reach its destination but that's theoretical

I like this a lot. It takes the pure math, and uses it to calculate the maximum in reality.

you are certainly correct. and also correct in saying it is theoretical (unless i disseminate that and see a flaw)


it is useful to consider infinity as a quanta simply to see it is not quantifiable (rather than just swallow the stuff one reads)

even if the toad jumped 1 quadrillion times per second for 1000 trillion years, it would still never get there.
(although the physical frog has dimensions that are unhelpful in assessing the matter if factored in, so i consider the "frog" to be a mathematical point with no length or width or height).


so one can see it is absurd to even try to consider infinity.
the divisor applied to the length of each jump must be 2 in order for this perfect chase to arise.
if it is less than 2, the frog will reach it's destination, and if it is greater than 2, then the frog's trajectory lands it at a number greater than zero after an infinite amount of jumps which is actually an absurdity when considered carefully.

if the divisor is exactly 2, then the result will eventually be zero after an infinite amount of recursions, and this is called "binary decomposition", and it is very useful to use in the design of fractals.

whatever.

good thinking 13 year old.

except for that it is not correct.
there are no precepts to the meaning of his variable assignations, and in truth, the frog, given his initial example example, would never reach one.

so he says that the smallest finite distance possible is the planck length which is the smallest measurable quantity of distance according to subatomic dimensions.

but take that number and halve it.
then halve it again........etc....

you never get zero.
zero is indefinite, but is also definite in that it is the starting point fro all consideration.

I also undertook to include the perception of the frog. However, unless I want to include the frog jumping and moving backwards while thinking it has jumped forwards, the frog has to reach the goal by jumping one planck length at a time some arbitrarily large finite number of times.

It seemed hopeless, but... once we stipulate anything being able to jump precisely one planck length, I think we've put that thing into the quantum realm in some respects. I put it to you that we could either say that the frog had jumped one planck length, or say in which direction it had jumped, but not both. In part, because any apparatus we have that could measure a jump of that length would affect the jump.

The frog is not making smaller and smaller jumps. You're just splitting time into smaller chunks. The speed of the frog in the problem is constant and is greater than zero.

For those who think that the frog will never get there, consider this.

Pick three distinct points, A, B, and C, in a line such that B is midway between A and C.

If the frog is jumping from point A to point C and can never get there but does get to B, then are you claiming that if it were jumping from A to B it could never get to B? If it can never get from A to C in finite time, then it can never get to any point between A and C and so it must not be moving at all.

You are overlooking time. Shorter distances are covered faster given a constant rate.

For example, say your walking rate is 100 feet per minute. So, if it takes 1 minute to walk 100 feet then it takes 30 seconds to walk 50 feet and 15 seconds to walk half of that. If you add up the times, no matter how many divisions, you will end up with 1 minute. No matter how finely you divide 100 feet, it still adds up to 100 feet.

There is no paradox here. Only a complete misunderstanding of math by some that makes them think there is a paradox.

There is no reason to get hung up on adding shorter and shorter intervals, each 1/2 of the previous interval. It is simple to show that the sum of 1/2 + 1/4 + 1/8 + 1/16 + ... = 1.

Let x = 1/2 + 1/4 + 1/8 + 1/16 + ...
Then 2x = 1 + 1/2 + 1/4 + 1/8 + 1/16 + ...
Subtract the first from the second and we will get
2x - x = 1 + 1/2 + 1/4 + 1/8 + 1/16 +... - 1/2 - 1/4 - 1/8 - 1/16 - ...
x = 1 + 1/2 - 1/2 + 1/4 - 1/4 + 1/8 - 1/8 +1/16 - 1/16 + ...
= 1.

Thus 1/2 + 1/4 + 1/8 + 1/16 + ... = 1.

So you get the frog jumps x/2 + x/4 + x/8 + x/16 + ... = x units of distance in t/2 + t/4 + t/8 + t/16 + ... units of time.
That is, it jumps x units of distance in t units of time.