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Suppose that I am trying to travel from location $(0, 0)$ to my destination $(1, 1)$. Unfortunately, I cannot walk there in a straight line — instead, I can either move horizontally (by taking a step of size $\delta$) or vertically (again, by taking a step of $\delta$). We may assume that 1 is a multiple of $\delta$ (so that it is possible for me to reach my desired destination). How far must I travel before I arrive?

On the one hand, the answer is clearly $2$. The reason is clear: I must travel a distance of at least $1$ to the right and a distance of at least one upwards. So my total distance travelled must be at least $2$. Note that this is true regardless of the value $\delta$.

On the other hand, it is curious that the distance does not approach $\sqrt(2)$ as $\delta$ becomes small. If $\delta \rightarrow 0$, then the grid to which I am restricted converges to the entire plane. Thus, it might seem (although this is incorrect) that my minimum distance converges to the minimum distance in the case where my movements are unrestricted, i.e. $\sqrt(2)$ (by the Pythagorean theorem).

Another way to think about this is the following. Suppose (for example) that I decide to travel as follows: take a step to the right, then a step up, then right, then up, etc. As $\delta \rightarrow 0$, this route converges (uniformly) to the 45 degree line (i.e. the shortest route if your movements are unrestricted). However, the total length of your route does not converge to the length of the 45 degree line — instead it is fixed at 2. Why?

afreelunch
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  • If you search hard enough, you will find this question already here in math.se at least once. Basically, the length of the limit is not equal to the limit of the length. And why should it be? – GEdgar Oct 23 '19 at 10:56
  • Here is one about the circumference of a circle https://math.stackexchange.com/questions/12906/the-staircase-paradox-or-why-pi-ne4 – GEdgar Oct 23 '19 at 11:08

2 Answers2

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One of the "intuitive explanations" I can think of is that, even if you approach something that appears to look like a straight line, at each step you reduce $\delta$ it has more and more turns, so it's defnitely something very different from a straight line.

Think that you can always make your path by going first to $(1,0)$ and from there to $(1,1)$. Any alternaive path is just a rearrangement of the same number of segments, so its length must always be $2$

David
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This is the known as the staircase paradox, or the diagonal paradox; the same reasoning is also used in the 'proof' of the "$\pi=4$" fallacy.

Probably the simplest way to see the flaw is to notice that at for any step length $\,\delta,\,$ the number of steps to the right and the number of steps up will both be will be $\,\frac{1}{\delta},\,$ for a total of $\,\frac{2}{\delta}\,$ steps. Then as $\,\delta \rightarrow 0,\,$ we'll have the total path length

$$\lim_{\delta\to 0} \; \delta \cdot \left(\frac{2}{\delta}\right) = \;\lim_{\delta\to 0}\; 2 = 2$$

A.J.
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  • I agree with this. However, perhaps it would be worth pointing out the flaw in the argument for $\sqrt(2)$? Of course, it is quite informal as stated but I suspect relies on confusing different notions of convergence – afreelunch Oct 23 '19 at 14:14