Equilibrium Posted January 5, 2011 Share Posted January 5, 2011 so i saw this paradox on this website, http://mathforum.org/isaac/problems/zeno1.html , but i don't know what the solution is, or is there a solution at all? A runner wants to run a certain distance - let us say 100 meters - in a finite time. But to reach the 100-meter mark, the runner must first reach the 50-meter mark, and to reach that, the runner must first run 25 meters. But to do that, he or she must first run 12.5 meters. Since space is infinitely divisible, we can repeat these 'requirements' forever. Thus the runner has to reach an infinite number of 'midpoints' in a finite time. This is impossible, so the runner can never reach his goal. In general, anyone who wants to move from one point to another must meet these requirements, and so motion is impossible, and what we perceive as motion is merely an illusion. Where does the argument break down? Why? Link to comment Share on other sites More sharing options...

ydoaPs Posted January 5, 2011 Share Posted January 5, 2011 Humans don't travel like that. It ignored gait. Link to comment Share on other sites More sharing options...

Equilibrium Posted January 5, 2011 Author Share Posted January 5, 2011 Yes, but you still move forward right? I think It's kind of trying to say, when you start to move forward, where is the start? Link to comment Share on other sites More sharing options...

D H Posted January 5, 2011 Share Posted January 5, 2011 (edited) Where does the argument break down? Why? It breaks down because an infinite series can have a finite sum, something of which Zeno was unaware. That lack of knowledge is not Zeno's fault; he was born 2000 years or so too early. In math, [math]\frac 1 2 + \frac 1 4 + \frac 1 8 \cdots = \sum_{n=1}^{\infty} \frac 1{2^n} = 1[/math] Edited January 5, 2011 by D H Link to comment Share on other sites More sharing options...

Sisyphus Posted January 5, 2011 Share Posted January 5, 2011 Just as the finite space traveled is infinitely divisible, so is the finite time you take to travel it. You have "infinite" time in the same way you have to travel an "infinite" distance. If you took the same amount of time for each step and that amount was finite, you would of course never reach your goal, but just as the distance in each step decreases by half, so does the time, and the sum of each infinite series is finite. Hence, you arrive at your destination in a finite time. Link to comment Share on other sites More sharing options...

md65536 Posted January 5, 2011 Share Posted January 5, 2011 (edited) Since space is infinitely divisible, we can repeat these 'requirements' forever. "Forever" is entirely misleading, because it suggests a span of time, when the sum of time it takes to meet the infinite "requirements" is finite. It would be like saying that it would take an infinite amount of time to describe this part of the problem, when really you described it in a single sentence. But that's just semantics. The real answer to your question is... Thus the runner has to reach an infinite number of 'midpoints' in a finite time. This is impossible, That's where the argument breaks down. Of course it's possible. There's not even a justification for why it might be considered impossible. With any movement, no matter how small, you would pass through an infinite number of "midpoints". This is related to the concept of there being no smallest positive non-zero number. Any real number can be divided into smaller numbers, just as (as you assumed) any distance can be divided into smaller distances with midpoints. The argument would be similar to saying "Any number is infinitely large," -- which is obviously not true -- "because it can be divided into an infinite number of smaller numbers" -- which is true. Edited January 5, 2011 by md65536 Link to comment Share on other sites More sharing options...

TonyMcC Posted January 6, 2011 Share Posted January 6, 2011 (edited) Make it your intention to run twice the distance (say 200 metres) and then change your mind when you are half way there! Edited January 6, 2011 by TonyMcC Link to comment Share on other sites More sharing options...

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