Draw a square with side length s. Label the corners clockwise A,B,C,D. To get from point A to point C around the perimeter you’ve traveled 2s. To get to point C along the diagonal AC you’ve traveled 1.414s. Now zig-zag along the diagonal to point C, traveling 0.1s horizontal then 0.1s vertical. You’ve traveled 2s. Travel 0.01s horizontal then 0.01s vertical; you’ve traveled 2s. Take a zig-zag pattern so small it looks just like a diagonal, you’ve traveled 2s. So, geniuses, when does it ever get to 1.414s? Is it just 2s or 1.414s and never anything in between? - 
As long as you’re dealing with “zig-zags” or straight lines you don’t have any choice. It’s still 180 degrees whether it’s two right angles or a straight line.
Now if you want to throw a curve in it then…that’s another story. 
This has come up before. It never gets to sqrt(2).
http://boards.straightdope.com/sdmb/showthread.php?s=&threadid=131714
I myself consider it a paradox. The limit of the path is the diagonal, but the limit of the path length is not the length of the diagonal.
There’s the signpost up ahead! You’ve just crossed over into Taxicab Space !
I don’t quite follow your logic, here: If I understand you correctly,
all you’ve done is made small, incremental strides to “C”. In the process, you’ve simply moved a total distance of 2c in integrated steps, so to say, regardless of how you slice it, right? Overall, you’ve moved a displacement of 1.414s, although the total distance travelled is 2s. It won’t matter of you zig-zag to “C” in increments of 0.00001s horixontal and 0.00001s vertical!
It’s like that old joke: Would you like your pizza cut in 8 slices or 6 slices? Better make it 6 slices…I don’t think I can eat 8!
As the “Snickers” bar once claimed, any way you slice it, you come up peanuts!
In other words, no matter how small the increment, you’re still ultimately moving a total distance of 1s horizontally and 1s vertically. You won’t get 1.414s, sorry. Think of it in terms of moving along the legs of a 3-4-5 triangle. No matter what the increments, you’ve moved a total distance of 7 units - along the legs. (You’re just making the legs into tinier “sub-legs” which will add up to the total length of each leg at the end.)
Does this help? - Jinx
a) Pardon the typos, Harpos, Grouchos, and spaghettios in my previous post above.
b) But seriously, for clarity, my example of the 3-4-5 triangle above is just another application of the OP…while the triangle in question would actually be a 1-1-1.4 triangle. - Jinx
P.S. I don’t see the paradox, doc!
The confusion is similar to that of the calculus class that wants to approximate the lateral surface area of a cone with a Riemann sum consisting of lateral surface areas of thin cylindrical disks (stacked up like the Tower of Hanoi). They then turn the Riemann sum into a definite integral, whose value is not at all the lateral surface area Pirl they learned in geometry.
Setting up a Riemann sum was not enough. They needed to be certain that it would converge to the quantity they were trying to measure.
That’s why when we use calculus to compute the arclength of a path or the surface area of a solid of revolution, we need the Pythagorean theorem to take into account the gradient of the function that describes our path or surface. It all boils down to the fact that we’re computing with the Euclidean metric: ds[sup]2[/sup] = dx[sup]2[/sup] + dy[sup]2[/sup]. On any scale, no matter how many partitions you care to divide the kathetes into, the Pythagorean theorem still determines the length of the differential hypotenuses, which are summed up to accurately reflect the total hypotenuse length, 1.414s.
Sorry for being thick, but I don’t get it. If you travel along the perimeter of the square from A to C you travel 2s. If you travel along the diagonal from A to C you travel sqrt(2)s. So if you zig-zag along the diagonal from A to C you travel somewhere between sqrt(2)s and 2s inclusive as long as your zigs are uniform and less than or equal to s. The smaller the zigs, the closer you get to sgrt(2)s.
No. Provided all your zigs are horizontal, and all your zags are vertical, you travel 2s, no matter which order you take them in, or how small they are. You could take 10,000,000,000 tiny little steps and it would still add up to 2s, even though from any distance it looks like you just traveled down the diagonal.
Make sense?
Any arguments now about quantization of space?
The argument you are trying to make is that the zigzag paths get closer and closer to the diagonal paths, therefore the (limit of the) lengths of them is equal to the length of the diagonal.
But you just can’t do this. You need to justify ‘x gets closer to y therefore lim f(x) = f(y)’ statements, and this one is wrong.
Take your unit square. It has an area of 1 and a perimeter of 4.
Put a step in the sides at the half way mark, and you end up with two squares joined at the corners. The area now enclosed is 0.5, but the perimeter is still 4.
Repeat the subdivision. Enclosed area becomes 0.25, perimeter remains 4.
As the number of subdivisions increases, enclosed area goes to zero, and the figure looks just like a diagonal line. But the perimeter remains 4.
I don’t get what the paradox is supposed to be.
If you added another square, with diagonal, to the bottom right to extend the diagonal line…then add another…and then a million more…then stand a mile back so it looks like a straight line instead of a load of boxes. This is essentially the same as making the zigzag smaller.
Now, what possible difference does it make how far back you stand? Just because it looks like a straight line? I don’t get it.
Google “Weyl tile”.
Heck. Here you go.
Hmm, I always thought that going the most direct route was the shortest, but apparently my thinking was crooked.
If you are going to walk to a location that is 10 block west and 10 blocks north it will take the same amount of time to zig zag 1 west and 1 north as it will to walk 10 west and 10 north? I had always approximated the distance saved as a rough diagonal. I now realize that the only distance saved will be any true diagonal paths taken through parks and green spaces.
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In this example I could save only the equivalent of 1.758 blocks of travel by taking the highlighted route (each park counts as 1.414 blocks instead of 2). I am walking on the south and west side of each block shown and diagonally through the parks.
Achenar
I don’t see it as a paradox. When you say that the limit of the path is the diagonal, what that means is that by one means of measuring the difference between them, they are getting closer and closer. Well, arclength is different means of measuring the difference between them. So there’s no need for the two means to concur. I think that part of the problem people have with this is that they are used to thinking of “the limit of f is l” as meaning that eventually (“after infinity”, as if that means anything"), f becomes equal to l. That’s not what it means. It means that according to a particular measure, it gets arbitrarily close. By another measure, it may never get close.
Suppose you have a two countries in which the number of apples produced is getting closer and close to each other, but the number of oranges is going all over the place. Paradox?
Let’s say you start at A and do one zig and one zag down the diagonal and arrive exactly at C, then you would have traveled along the perimeter or 2s.
Now lets say you do two zigs and two zags down the diagonal, since a zig is the same length as a zag, what you have done is formed two right triangles along the diagonal. The sum of the two new hypotenuses must equal the length of the original hypotenuse, namely sqrt(2)s. Using the Pythagorean Theorem we can say the sum of the two hypotenuses of the two new triangles must equal the original hypotenuse in this manner (let “a” equal one zig or one zag):
2a^2 + 2a^2 = (sqrt(2)s)^2
or
a = s/sqrt(2). So the total distance traveled is 4a or 4(s/sqrt(2)) or 2s * sqrt(2) which is actually more than 2s.
For 3 zig-zags it would be
2a^2 + 2a^2 + 2a^2 = (sqrt(2)s)^2
or
a = s/sqrt(3) so 6a is 3s * sqrt(3) which is more than your two zig-zag journey of 2s sqrt(2).
Then in general, the more zig-zags your take along the diagonal, the further you will travel. If you do N zigs-zags, you travel 2s * sqrt(N).
Taking zillions of zig-zags isn’t going to get you closer to sqrt(2)s, the more zig-zags, the further you travel.
Any errors with this argument? Comments and luke warm flames welcome.
[QUOTE]
*Originally posted by ccwaterback *
**Then in general, the more zig-zags your take along the diagonal, the further you will travel. If you do N zigs-zags, you travel 2s * sqrt(N).
[QUOTE]
2s * sqrt(N) should be Ns * sqrt(N)
Yeah, pretty much everything.
Just think about it. Say each side of the square is 2 units long. The diagonal is 2*sqrt(2) units long. We want to get from the top left to the bottom right corner, so we zig one unit to the right, zag one unit down, zig one unit to the right, zag one unit down (so length of zig=length of zag=1 unit).
Your equation:
2a^2 + 2a^2 = (sqrt(2)s)^2
(a=1, s=2)
now becomes:
2 + 2 = 8
I’d say that’s a problem. Your other equations have the same basic problem.
Think about it some more. It really is quite trivial if you bother to do that.