Vtgyjfy

Consider the red path from A that zigzags to B, which takes $n$ even steps of length $w$. The path length of the route $P_n$ will be equal to:

$ P_n = P_x + P_y = frac{n}{2}times w + frac{n}{2}times w = n times w $

But $frac{n}{2}times w = 1$ beacuse it is the length of one of the sides of the triangle so:

$P_n = 2$

Which will be true no matter how many steps you take. However in the limit $n to infty, w to 0$ the parth length $P_infty$ suddenly becomes:

$P_infty = sqrt{1^2 + 1^2} = sqrt{2}$

Due to Pythagoras. Why is this the case? It seems the path length suddenly decreases by 0.59!

The length of the $n$ path is define by : $ int_{(a,b)} gamma_n'(t)dt $ where $gamma_n$ is your path. But you can't pass at the infinity because "you have too much point of discontinuity". For instance you have no simple convergence for $gamma'_n$...

I was just about to ask the exact same question (although phrased differently), but in the process of asking it, I figured it out:

Basically the point is no matter how small each of the smaller zigzag steps' edges get, you can draw a line across it that represents its hypotenuse. If you sum all of those together, you'll always get the original length ($sqrt{2}$).

It is very counterintuitive, though. The question came to me while driving a grid—am I better off wiggling left and right repeatedly to stay closest to the diagonal (apparently the shortest distance between two points), or driving down the outside edges of the grid? Some mental arithmetic showed that the distance travelled (presuming all corners are perpendicular) should be identical no matter how many times I turn (provided I don't double-back on myself):

 A                  B                  C

┌──────────────────┬──────────────────┐

│                  │                  │

│                  │                  │

│                  │                  │

│                  │                  │

│                  │                  │

│D                 │E                 │F

├──────────────────┼──────────────────┤

│                  │                  │

│                  │                  │

│                  │                  │

│                  │                  │

│                  │                  │

│G                 │H                 │I

└──────────────────┴──────────────────┘

To travel from point $A$ to $I$, the distance of driving $AG + GI$ is the same as $AD + DE + EH + HI$ (since $AD + EH = AG$ and $DE + HI = GI$).

All that's not so hard to grasp, but when you do this same task recursively on each grid square (as you describe), you quickly end up producing something that closely approximates half the box, but somehow has the same perimeter as the starting square:

This went beyond counterintuitive to me and became downright unacceptable to my brain. If you treat the square as a unit square (as you have), both of these shapes have a perimeter of 4, but the real triangle formed by joining the two diagonal corners has a perimeter of $2 + sqrt{2}$ (less than 3.5!). At some amount of resolution, those zig zags are going to become visually indistinguishable from a straight diagonal line, but somehow there's more than half an edge length extra hiding somewhere.

The solution though, as described above is simple: no matter how large $n$ gets, you can always imagine zooming right in to that "triangle", and you do indeed end up with a series of zigzags, never a diagonal line. And if you calculate the sum of all those little zigzags' hypotenuses (that is, $frac{n}{2} times wsqrt{2}$), you'll end up with the hypotenuse of the larger triangle ($sqrt{2}$, since $frac{n}{2} times w = 1$ as you stated).

There are some misconceptions here regarding magnitude of a vector versus its direction+magnitude.

First, label the bottom left corner of the triangle as $O$ and declare the length of $OB=OA=1$ (you seem to imply this given your statement). Each one of these $n$ mini-paths (assuming we go up and over by equal amounts of $1/n$) tracing up hypotenuse $AB$ of the triangle has position vector given by $(-1/n,1/n)$. The magnitude of this position vector is given by the Pythagorean theorem, namely: $||(-1/n,1/n)||=sqrt{2}/n$. Recall that there are $n$ such paths and so the total path $AB=underbrace{sqrt{2}/n+cdots+sqrt{2}/n}_n=sqrt{2}$.

Length of a path is defined in a very complicated way using Calculus. To define that, we first have to define distance in $mathbb{R}^2$. We seek a definition of distance from any point in $mathbb{R}^2$ to $mathbb{R}^2$, a function from $(mathbb{R}^2)^2$ to $mathbb{R}$ that satisfies the following properties.

• For any points $(x, y)$ and $(z, w)$, $d((x, y), (x + z, y + w)) = d((0, 0), (z, w))$
  


• For any point $(x, y)$, $d((0, 0), (x, y))$ is nonnegative


• For any nonnegative real number $x$, $d((0, 0), (x, 0)) = x$
  


• For any point $(x, y)$, $d((0, 0), (x, -y)) = d((0, 0), (x, y))$
  


• For any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$
  

Suppose a function $d$ from $(mathbb{R}^2)^2$ to $mathbb{R}$ satisfies those conditions, then for any point $(x, y)$, $d((0, 0), (x, y))^2 = d((0, 0), (x, y))d((0, 0), (x, y)) = d((0, 0), (x, y))d((0, 0), (x, -y)) = d((0, 0), (x^2 + y^2, 0)) = x^2 + y^2$ so $d((0, 0), (x, y)) = sqrt{x^2 + y^2}$ so for any points $(x, y)$ and $(z, w)$, $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ Now I will show that $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ actually satisfies those properties. It's trivial to show that it satisfies the first 4 conditions. It also satisfies the fifth condition because for any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = sqrt{(xz - yw)^2 + (xw + yz)^2} = sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = sqrt{(x^2 + y^2)(z^2 + w^2)} = sqrt{x^2 + y^2}sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$

As a result of this, from now on, I will define the distance from any point $(x, y)$ to any point $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$ and denote it as $d((x, y), (z, w))$. I will also use $d(x, y)$ as shorthand for $d((0, 0), (x, y))$.

Calculus defines the derivative of any function from a subset of $mathbb{R}$ to $mathbb{R}^2$. For some such functions, the derivative is undefined even where the original function is defined. The derivative of the function at any real number where it's defined can be called the velocity of that function at that real number. Speed is defined to be the square root of the sum of the squares of the components of the velocity. For any path that's topologically equivalent to a line segment, when there exists a function from a closed interval on $mathbb{R}$ to $mathbb{R}^2$ that's continuous and at some point travels along the path with a speed of 1 at all but finitely many points in that interval on $mathbb{R}$ and assigns to each end point of that interval, opposite ends of that path, the length of that path is defined to be the difference between the end points of that domain of $mathbb{R}$. Just because one path can be continuously transformed into another path doesn't mean its length continuously varies with time during the transformation.

That might seem so counterintuitive to you. That can be explained by the fact that statements about Calculus can be formalized as statements in the formal system of Zermelo-Fraenkel set theory and the formal system of ZF disproves the formalization of the intuitive statement that when ever a path topologically equivalent to a line segment gets continuously transformed, its length varies continuously with time.

Source: The validity of the proofs of the Pythagorean Theorem and the concept of area