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Zeno's Paradox (math)

Suppose that a man wants to cross to the far wall of a room that is 20-ft across. First, he crosses half of the distance to reach the 10-ft mark. Next, he crosses halfway across the remaining 10ft to arrive at the 5-ft mark. Dividing the distance in half again, he crosses to the 2.5- ft mark, and continues to cross the room in this way, dividing each distance in half and crossing to that point. Because each of the increasingly smaller distances can be divided in half, he must reach an infinite number of "midpoints" in a finite amount of time, and will never reach the wall.

Explain the error in Zeno's Paradox.

This is Dh's math question...anyone care to give it a go? i've no idea what the answer is, btw. (no idea what category to put it in either)

Answer Question

Asked by armywife43 at 11:30 PM on Mar. 22, 2011 in Just for Fun

Level 21 (11,516 Credits)
Answers (4)
  • Wow, I'm copying and sending it to my dad. Will come back later, probably tomorrow lol.

    Answer by Austinsmom35 at 11:35 PM on Mar. 22, 2011

  • If a number is continually divided in half, it will never reach zero. This would be a series of infinite midpoints; so no, the man will never reach the wall.

    Answer by mikesmom65270 at 11:48 PM on Mar. 22, 2011

  • Well... if reality applies, then with each midpoint, the distance becomes smaller and smaller, and the distance gets to a point where it is impossible for the man to cross only half the distance... in other words, he reaches a limit to the "half" distance and cannot go any smaller.

    But considering he got this as part of his math homework.... I'm sure reality is not supposed to apply! :)

    Answer by Busimommi at 11:55 PM on Mar. 22, 2011

  • It is the sum of an infinite series. An infinite series can have a finite value. One way to arrive at the solution is to define some distance traveled as say D(n)-distance at step n, where n is the number of "steps" taken, with each step being half the remaining distance to the wall. You can then define the distance traveled as D(n) = 10 + 10(1/2) + 10((1/2)^2) + 10 ((1/2)^4) + ... 10((1/2)^n). the ^ symbol means raised to the power of. Then take this equation and divide by 2 to get D(n)/2 = 10(1/2) + 10((1/2)^2) + 10((1/2)^4) + ... 10((1/2)^n-1). Subtract the 2 equations and almost everything on the right side cancels and you get D(n) - D(n)/2 = 10 + 10((1/2)^n). Multiply both sides of equation by 2 to get
    2D(n) - D(n) = 20 + 20((1/2)^n). Which leaves D(n) = 20 + 20((1/2)^n). Taking the limit as n goes to infinity D(n) = 20 + 0 = 20.

    Answer by confused969 at 12:18 AM on Mar. 23, 2011

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