# How can an object move from point A to point B?

If I was at point $A$ and I wanted to walk directly to point $B$, I would have to walk half way to point $B$, but before that I would have to walk half way to halfway to halfway to point $B$ and half of that again and so on and so fourth. if I halved this distance an infinite amount of times then there would be an infinite amount of actions I would need to perform in order to cross from $A$ to $B$. Therefore, if each action took any quantity of time at all, then it would take me an infinite amount of time to cross from $A$ to $B$, even if $A$ and $B$ were only a few centimetres apart!

Because I am not infinitely old and I can move, there must be a flaw in this logic. Where is it?

• Do you know about the Zeno's Dichotomy Paradox? – 299792458 Apr 16 '15 at 6:26
• no i have not, i'll look into it. – ziggy Apr 16 '15 at 6:27
• The site has lots of questions related to Zeno's paradoxes – John Rennie Apr 16 '15 at 6:29
• ok i understand the paradox, but how can it be solved? – ziggy Apr 16 '15 at 6:31
• it is solved by knowing that doing an infinite amount of tasks does not have to take an infinite amount of time. – Jakob Apr 16 '15 at 9:21

You have stumbled onto one of Zeno's many paradoxes - the so-called Dichotomy paradox.

The resolution lies in the fact that sum of terms in an infinite series do not necessarily add up to produce an infinity, so the basic premise is flawed. In particular, the series

$$S = \sum_{n=1}^\infty \left(\frac{1}{2}\right)^n = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} \ldots$$

is a convergent infinite series. The sum is

$$S = 1 \, .$$

Since Zeno belonged to the BC era, this paradox stood as a paradox for many years. In fact, Zeno was satisfied with the reasoning than it had an infinite number of steps, but didn't find the sum. However, things changed with the development of Analysis and Calculus in the 19th century, and with these convergence arguments, it is settled now.

• But, for every action to complete, there must be last step, isn't it? Here you find no last step mathematically, but you find the action being completed in reality. I think this is the logic which has to be broken in Zeno's paradox and not the time argument. I think it is not settled yet. – Immortal Player Apr 16 '15 at 16:42
• @Feynman: No, when the steps are infinitely divisible, there is no last one, because the last step is always divisible in two, one of which is “laster”. – Jan Hudec Apr 17 '15 at 11:31
• @JanHudec: Yes, that is my point, there is no last step. So, you can't complete the series of steps because there is no last step. – Immortal Player Apr 17 '15 at 15:32
• @Feynman: “But, for every action to complete, there must be last step, isn't it?” No, it does not. The sequence only has to have a finite sum (size). (i.e. I am responding to your premise, not the point). – Jan Hudec Apr 17 '15 at 20:28
• – Immortal Player Apr 18 '15 at 9:46

I will try to explain this in simple english. Since the number of steps are infinite, then the distance you travel at each step is infinitly short. As such, the time teaken in each step is close to zero but not zero. Therefore the total time in taking this infinite number of steps does not necessary equal to infinite.

Mathematically the total time take is limit(n*t) where n-> infinite and t -> zero but not zero. When t is sufficiently small, this expression has a real value (as n increases, t will decrease) depends on the speed you move.

## protected by Qmechanic♦Apr 16 '15 at 19:36

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