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PeteH

Math problem: Zeno's paradox

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will be neck and neck at 11.1111111111 meters. One moment after that, Achilles will be ahead



not if a "moment" is defined as 0.00000000001 or smaller :P

O


Yeah, yeah, you know what I mean. At (10/9) m, they will have travelled and equal distances.


;)

FunBobby

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Actually, I believe you need calculus to "prove" the fallacy of Zeno's Paradox.



Or common sense... ;)

Zeno is going to where the tortoise was and stopping, while the tortoise is now up ahead.

Kind of like diving for the spot where the formation was when you left the plane... oops, they moved. Diving for the new spot... oops, they moved.

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I'm talking about a mathematical proof, which is something beyond common sense.

Using common sense, -obviously- it's a fallacy. To think otherwise would be just silly.



I don't think you need calculus to sum a convergent infinite series.
...

The only sure way to survive a canopy collision is not to have one.

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I'm talking about a mathematical proof, which is something beyond common sense.



I agree, no one believes that math makes any sense, but I couldn't convince my parents of that. :)



Like the sum of a conditionally convergent infinite series ... you can get any answer you like by reordering the terms.
...

The only sure way to survive a canopy collision is not to have one.

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8 pirates and a monkey are marooned on an island. The pirates collect coconuts and agree to divide them up in the morning.

During the night a pirate wakes up and thinks his colleagues will steal his share, so he takes and hides 1/8 of the coconutrs, throws a coconut to the monkey, and returns to sleep. Shortly after another pirate wakes up, takes 1/8 of the remaining coconuts and hides them, throws one of the remainder to the monkey, and goes back to sleep.

Then another pirate awakes and does the exact same thing. And then the next, and then the next, etc. until each pirate has done it.

In the morning they share the remaining coconuts equally, and finding a remainder of one, they give it to the monkey.

What is the smallest number of coconuts they could have started with?



I am still not sure how many coconuts there were B|[:/]:(
Dave

Fallschirmsport Marl

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maybe I'm misreading, but:

Let V1 = speed of tortoise;
Let V2 = Achillees speed
Let t = elaspsed time

Distance travelled by tortoise = 10m + V1t
Distance travelled by Achilles = V2 t

time to catchup ==> both distances are equal

10m + V1t = V2t

Since V2 = 10V1

10m + V1t = 10V1t

10m/9V1 = t

As V1 approaches zero, t approaches infinity.


The faulty logic? Zeno needed to quit yapping, sober up, and apply simple Algebra.

Walt

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re. the animal problem

I was not able to reduce it to a mathematical formula, so if that is possible, I wish someone would explain it to me.

Neverthelss by trial & error I got 5 cows, 1 pig, 94 hens. Total = 100 animals.

5 cows = $50, 1 pig = $3, 94 hens = $47. Total = $100
Speed Racer
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A variation on this problem:

A chemist and a chemical engineer were imprisoned in Saudi Arabia for offending the local caliph. After ten years in solitary confinement, the Caliph order the two men brought to him in his harem. They are placed on a chalk line drawn on the floor of the harem chamber, and two stunningly beautiful naked women are placed at the other end of the chamber, 100 feet away.
" I have a treat for the two of you!" the Caliph says. "Every time I clap my hands, you can advance half the distance between you and the women."
He claps and the Chemical engineer runs 50 feet across the floor, while the chemist stand where he started. The caliph claps again and the Chem. E. runs forward twenty five feet, while the chemist just stands there. Exasperated the Chem. E. asks the chemist why he isn't running.
"It's rigged" the chemist replies "the approach is asymptotic... you'll never get all th eway there."

"True" replied the Chem. E. "but I can get close enough for practical purposes."
"

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So .....
You have $100, you must spend it all.

You need to buy exactly 100 animals.

You can't buy half an animal, but you buy at least one of each.
On offer are Cows for $10 each
Pigs for $3 each
and Hens for 50 cents each.

How many of each animal must you buy?

Orders please....



Looks to me like there is no solution.

Let Nc = number of cows
Np = number of pigs
Nh = number of hens


Nc + Np + Nh = 100
10Nc + 3Np + .5Nh = 100

Therefore,
Nc + Np + Nh = 10Nc + 3Np + .5Nh

0 = 9Nc + 2Np + .5Nh

The only solution compatible with the conditons given is the trivial solution 0, which is NO solution.

Walt

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see my previous post. I figured it out, but I had the same problems as you trying to do it algebraicly. I just did it by sort of trial & error.



Yeah, I saw that after I posted and obviously there is a major flaw in my reasoning (which happens pretty often!).

Walt

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obviously there is a major flaw in my reasoning (which happens pretty often!).

Walt



I'll have to remember that quote for Speakers' Corner.



Feel free. I always reserve the right to learn and change the error of my ways.

Walt

p.s. How many people in Speakers' Corner will admit to that?!!!;)

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Therefore,
Nc + Np + Nh = 10Nc + 3Np + .5Nh

0 = 9Nc + 2Np + .5Nh



Here's where you went wrong. You're subtracting Nc +Np + Nh from the right side of the equation. The answer should read:

0 = 9Nc + 2Np - 0.5 Nh
Speed Racer
--------------------------------------------------

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