Porters and Sudents

Submitted by cxm on

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This is the thread for discussing the problem posed by Prof Leader. Any ideas on how infinitely many porters can be both enough and not enough can be discussed here.

Porters and Students

Submitted by The indefinite ... (not verified) on

I believe the paradox lies within there being infinite porters modelled as points. If 2 porters are next to each other at pheta1 and pheta2, they cannot be different points otherwise the student could leave at (pheta1+pheta2)/2 a new point where there is no porters. the only way there is no distance between pheta 1 and pheta2 is and only if pheta1=pheta2 , using this method and induction we can produce the statement pheta1=phetaN

even with infinite porters the only way they could have no distance between themselves for the students to escape is if an infinite amount of porters exist on the same point and in which case the student could pick a different point and would win?

can anyone see any flaws in my arguement i am desperate to be proved wrong as i love a student being told off.

Porters and students

Submitted by Jack K. (not verified) on

So far as I can tell, the problems lies in the fact that the two examples have different numbers of porters.
In the example where the student can escape, they do so by escaping them in order - P1, then P2, then P3, and so on. Thus, as the student can always reduce the amount of time it takes to escape the next porter by some fixed ratio, we can say that at time N they will have escape a porter for every positive integer.
Now how many porters were there?

You might simply say "infinitely many", but this is maths, so we need to be more precise. The cardinality - that is, number - of positive integers turns out to be called "countable infinity", or aleph-null. Thus, assuming the professor's proof was valid, we deduce only that:

Countably infinitely many porters are not enough to prevent the student from escaping.

Now what about the second case, in which a porter stands still at every point on the circumference of the circle? Clearly the student cannot escape the circle as this would require them passing through a point on the circumference, which must be occupied by a porter. How many porters were there in this case?

We can imagine that each porter is named for the angle (WLOG measured anti-clockwise from a horizontal line from the centre to the right) they make when a line is drawn from them to the centre. So for example, the porter at the top of the circle would be called "Porter pi/2", or "Porter 90 degrees".
For there to be nowhere for the student to escape, there must be a porter for every angle between 0 and 2pi (or 0 and 360). So we want to know the cardinality of the numbers x such that 0 < x < 2pi. Specifically the real numbers in this range. (This distinction is important.)
So how many real numbers are there between 0 and 2pi?

The cardinality of the real numbers in a range turns out, interestingly, to be greater than the countable infinity we met earlier, and is instead uncountable (See Cantor's diagonalisation proof for why this must be the case). Thus, assuming the professor's argument in this case was also valid, then we may deduce only that:

Uncountably infinitely many porters are enough to prevent the student from escaping.

As we now see, these two statements, although seeming contradictory, when looked at closer, are in fact not contradictory at all.

Countably infinitely many porters (a number of porters corresonding to the number of positive integers) are not enough to prevent the student from escaping.

Uncountably infinitely many porters (a number of porters corresonding to the number of real numbers)are enough to prevent the student from escaping.

QED

Jack K.

My thoughts exactly.

Submitted by ewokonfire (not verified) on

This is exactly the argument I came here to give. I think it's the only meaningful answer to the problem. Good job.

Porters and Student

Submitted by rjpds (not verified) on

I believe the problem lies in the fact that an infinite amount of porters filling every point on the circle would leave no room for movement from point to point by a porter without inevitably causing a 'gap' to appear, therefore ruling out any pursuit of the student by a porter. Assuming all the porters stay on one point, implying a lack of any space between porters, this would mean there would be no gap for any student to move through despite evading every porter down to a minute scale. As the student evades one porter in one direction he moves towards another. Another flaw I saw in the argument is that assuming it is 'game over' at time 2 is impossible as the student moves in time increments of 1, 1/2, 1/4, 1/8 and so on, although the student moves towards the edge each time the fractions in question will never in fact add up to a final time of 2, the student will never complete the journey.

if there is no gap between

Submitted by The indefinite ... (not verified) on

if there is no gap between the porters could they exist on two separate points? for any two porters next to each other you could take there angles , average them and find a new point unoccupied between the porters? the only way this cant happen is if anf only if the two porters exist on the same point???

The problem here is that it

Submitted by Jack K. (not verified) on

The problem here is that it makes no sense to talk about two "adjacent" porters when there are infinitely many, because between any two porter there will be infinitely many between them.
It's like asking the question "What is the first real number greater than 1.5?"

Porters and Students

Submitted by Shakira (not verified) on

I wish I (or another) had asked Professor Leader what his opinion was; whether he though infinite or more than infinite was enough. :/

Porters & Students

Submitted by Username (not verified) on

I think that the first theory about "infinite porters are not enough" is wrong due to the fact that if we had a cirle that had a circumference of 30m and each porter had 30cm space where they just stand still, shoulder-to-shoulder with the other porters, we would need 100 porters to cover the whole circle. As a result of them standing shoulder-to-shoulder, there is no place for them to move and if the kid tries to do the method described in the first theory, he would reach a dead end as there will be no way out (like a wall).

So basically, what I'm trying to say is that the second part of what the professor explained as in the second part, the porters all build up to become a wall. In theorem one, it only works if you have less than infinity porters. Like we only used 5 or 6 porters. But if there was as many porters covering the entire circle, then the porters will act like a wall and the kid has no chance. So I believe that the answer is that Infinity porters are enough.

PS: In my opinion, you just have to think of this logically instead of using deep maths however, I might be wrong so be free to make me change my mind :D

They don't have shoulders

Submitted by i... That's as ... (not verified) on

These porters aren't people, they're theoretical points on a diagram, they only have one dimension. So if the guys do decide to stand next to eachother the student can still slip passed them.

I disagree with the first comment in that the idea isnt for the porters to have no space between them (even though your logic stands up) the idea is that every point has a porter on it so between Θ1 and Θ2 is Θ1.5 and so on. So if the infinite porters exist on a point because they must to have no spasce between then the student will just move to another point and be blocked by the infinite porters there.

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Also this is just in theory, it's not supposted to work perfectly. Even in the infinity isn't enough theory (with the porters tracking)

The Professor said that move 1 took 1 minute

Then move 2 took 0.5 minutes

Then move 3 took 0.25 minutes

etc...
         ∞
So      Σ      [2^-(n-1)] >2
        n=1

So even though we say the student wins at 2 minutes he can never get to 2 minutes

Zeno's Paradox

Submitted by Apple314 (not verified) on

I think that the problem lies with the first argument, where 'infinite porters are enough' for the reason that the student will never reach the edge of the circle. Although we were told that "2 minutes will always come about," when you really think about it, the 2 minutes (and the student's escape) will never come to pass in this situation.

Much like Zeno's Paradox of Achilles and the Tortoise, if the distance the student crosses keeps getting smaller and smaller and smaller- tending towards zero- then the time it takes him to travel will follow the same pattern.

Therefore as distance --> 0 , time --> 2.
However the student will never travel a distance of 0, and so will never reach a time of 2 minutes.

--[Earlier some of us were talking about 'countable infinity', but the Professor used the concept of 'infinity' and so that is how many Porters I am assuming are trying to catch the student.]--

Zeno's Pardox continued

Submitted by Apple314 (not verified) on

Summary: the second argument, where infinite porters are enough is true, and the first argument is a paradox. The statement 'P

The statement 'P

Submitted by Apple314 (not verified) on

The statement 'P

P < infinity

Submitted by Apple314 (not verified) on

P < infinity

Dividing by Zero

Submitted by Leipzig (not verified) on

If we plot a graph of n, the number of porters, against t, time in minutes, we get an asymptote of t=2, along a curve described roughly by n = something/(t-2)

Put another way, as t --> 2, n --> infinity; but at t=2, n =/= infinity, because this would involve division by zero. 1/0 =/= infinity (or else 0 x infinity would be equal to 1).

I believe the Professor was using division by zero to reach a contradiction, much like the 'proof' for 2=1. He was talking about n at t=2 as though it were infinity, and not undefined.

Jack K has got it.

Submitted by cxm on

The basic answer is that the two arguments depended on different "sized" infinities. The first one relied on a countably infinate number of porters (I.e. we can put them in a row and start counting them - just like the natural numbers). In the second argument the number of porters was uncountable (because between any two porters there is always another porter so we cannot count them). Both arguments are correct, but the word infinity was used to mean two different things, the infinity in the second argument being "bigger" than the first.

Claire Metcalfe (STEP CC HQ)

Re: Jack K has got it.

Submitted by Jack K. (not verified) on

Yay!

I wondered whether there was something wrong with the arguments on their own but realised that they could both be true if they weren't talking about the same infinity. A bit of wikipedia-ing later and I suspected I'd found the error.

Thanks for such an interesting problem!