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Topic: 5 Minute Mile (Read 720 times) |
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william wu
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5 Minute Mile
« on: Dec 11th, 2007, 10:07am » |
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A runner runs a six mile track in 30 minutes. Prove that somewhere along the track, the runner ran a mile in exactly 5 minutes.
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(The hints below suggest one way of approaching the problem.)
Hint 1: Define a useful continuous function.
Hint 2: Use the pigeonhole principle.
Hint 3: Use the intermediate value theorem.
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| « Last Edit: Dec 11th, 2007, 10:22am by william wu » |
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Hippo
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Re: 5 Minute Mile
« Reply #1 on: Dec 11th, 2007, 12:40pm » |
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Standard technique for continous functions on real numbers ...
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Let we measure distance d(t) run at 5 minutes starting at an arbitrary time 0<=t<=25.
As the speed of light is bounded, d(t) is continous.
As 6=d(0)+d(5)+d(10)+d(15)+d(20)+d(25) it cannot be d(t)>1 for all t or even d(t)<1 for all t.
So there is d(a)<=1 and d(b)>=1. If a<b take supremum s of t<b such that d(t)<=1. Otherwise take
supremum s of t<a such that d(t)>=1.
In both cases d(s)<1 leads to contradiction as well as d(s)>1. |
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Random Lack of Squiggily Lines
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Re: 5 Minute Mile
« Reply #2 on: Dec 12th, 2007, 5:13am » |
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he ran 4
4
4
4
and 14
there you go, he didn't
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You can only believe i what you can prove, and since you have nothing proven to cmpare to, you can believe in nothing.
I have ~50 posts to hack a "R" into a "D". Which one?
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towr
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Re: 5 Minute Mile
« Reply #3 on: Dec 12th, 2007, 5:28am » |
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on Dec 12th, 2007, 5:13am, tiber13 wrote:he ran 4
4
4
4
and 14
there you go, he didn't |
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The claim isn't that he ran one of the 6 consecutive miles in 5 minutes; but that he ran 1 mile in 5 minutes somewhere in the race, not necessarily between two milestones. It might be from 1.2746 miles in to 2.2746 miles, rather than from 0-1 or 1-2 or 2-3 etc.
Also, you seem to have only accounted for 5 miles.
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| « Last Edit: Dec 12th, 2007, 5:30am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Aryabhatta
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Re: 5 Minute Mile
« Reply #4 on: Dec 12th, 2007, 4:12pm » |
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A proof which _might_ be more easily understandable by non-math people (though in essence it is not much different from Hippo's):
Consider a clone of the runner who starts exactly 5 minutes after our runner and follows the exact same running pattern as the first one.
Considering their positions in 5 minute intervals of the 30 mins (0,5,10,...,25,30), at some point among those times between the two runners is <= 1 mile and at some point it is >= 1 mile.
Thus at some point in time (not just 0,5,...), distance between the runners must be exactly 1 mile.
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| « Last Edit: Dec 12th, 2007, 4:17pm by Aryabhatta » |
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Hippo
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Re: 5 Minute Mile
« Reply #5 on: Dec 13th, 2007, 3:39am » |
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on Dec 12th, 2007, 4:12pm, Aryabhatta wrote:A proof which _might_ be more easily understandable by non-math people (though in essence it is not much different from Hippo's):
Consider a clone of the runner who starts exactly 5 minutes after our runner and follows the exact same running pattern as the first one.
Considering their positions in 5 minute intervals of the 30 mins (0,5,10,...,25,30), at some point among those times between the two runners is <= 1 mile and at some point it is >= 1 mile.
Thus at some point in time (not just 0,5,...), distance between the runners must be exactly 1 mile.
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Yes, it's exactly same proof, expect you didn't mention the speed of light (no "teleporting") and didn't prove the intermediate value theorem. ... The things which may confuse nonmathematicans.
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Aryabhatta
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Re: 5 Minute Mile
« Reply #6 on: Dec 13th, 2007, 10:56am » |
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on Dec 13th, 2007, 3:39am, Hippo wrote:
Yes, it's exactly same proof, expect you didn't mention the speed of light (no "teleporting") and didn't prove the intermediate value theorem. ... The things which may confuse nonmathematicans. |
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Umm.. yes. To "prove" it rigorously we do require mathematics...
That is why I said _might_ and perhaps I should have called it argument instead of proof.
The statement should probably have read:
"An argument which might convince non-math people... "
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cool_joh
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on Dec 13th, 2007, 3:39am, Hippo wrote:
Yes, it's exactly same proof, expect you didn't mention the speed of light (no "teleporting") and didn't prove the intermediate value theorem. ... The things which may confuse nonmathematicans. |
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Is it really same? I don't understand Hippo's solution. But Aryabhatta's solution is much more understandable.
Thanks Aryabhatta! 
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Hippo
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Re: 5 Minute Mile
« Reply #8 on: Dec 16th, 2007, 12:52pm » |
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on Dec 14th, 2007, 2:18am, cool_joh wrote:
Is it really same? I don't understand Hippo's solution. But Aryabhatta's solution is much more understandable.
Thanks Aryabhatta! |
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Probably you cannot imagine what you should be avare. ... If the function would not be continous, the distance function can jump from say 0.9 to 1.1 and back without entering 1. This is nothing you can observe in real word as the teleporting and time travelling appears in sci fi (and quantum physic?) only. Just a small teleporting would suffice for the statement not to hold.
The other required thing was the continuity of time ... if say the time would gain only rational values, the function t^2-1 would never get value 1. Using such tricks you can construct function never using value 1 but "summing" to 6 miles.
The main property of real numbers is that each bounded set has its supremum in reals (minimal upper bound) if we suppose time uses real numbers, everything is OK.
The most complicated part was the proof of "intermediate value theorem" (holding for reals and continous functions):
Detail which was omited in the proof was why the value at supremum must be 1. ... If the value differ from 1, there is value neibourhood not containig 1.
As the function is continous, there must be also domain neibourhood of s which is mapped into the value neighbourhood. But in that case s either is not upper bound or is not minimal upper bound.
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| « Last Edit: Dec 16th, 2007, 12:53pm by Hippo » |
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