Proof that pi=4

In this context you might need specific descriptions of what is happening at the boundary points to answer this question.

The vectors are as simple as one would presume. You can zoom in to any extent you wish and see the same convergence in the middle.

Anymore than that and I would be answering my own question.

Quote from: JoshuaZ on March 17, 2011, 01:45:53 PM

In this context you might need specific descriptions of what is happening at the boundary points to answer this question.

The vectors are as simple as one would presume. You can zoom in to any extent you wish and see the same convergence in the middle.

Anymore than that and I would be answering my own question.

That doesn't work. Also I'm still confused about why you need vectors and not just sets. In any event, if I zoom in on the very center is that red or black? To make this more precise, let's assume that we have cartesian coordinates (so a standard x and y plane) with the lower left-hand corner as (0,0) and the upper right-hand corner as (1,1). Is (1/2,1/2) red or black?

if I zoom in on the very center is that red or black? To make this more precise, let's assume that we have cartesian coordinates (so a standard x and y plane) with the lower left-hand corner as (0,0) and the upper right-hand corner as (1,1). Is (1/2,1/2) red or black?

You're asking me to assign a color value to the infinitesimal boundary which is essentially like coloring the grid black or red to expand that colors area by a zero amount. This request is nonsensical.

Also, the two squares are the same size so it would have made more sense to make the total height/width an even number.

Also I'm still confused about why you need vectors and not just sets.

If pixels were mathematically perfect squares with no spaces between them, I'd be fine with it.

Quote from: JoshuaZ on March 17, 2011, 02:11:43 PM

if I zoom in on the very center is that red or black? To make this more precise, let's assume that we have cartesian coordinates (so a standard x and y plane) with the lower left-hand corner as (0,0) and the upper right-hand corner as (1,1). Is (1/2,1/2) red or black?

You're asking me to assign a color value to the infinitesimal boundary which is essentially like coloring the grid black or red to expand that colors area by a zero amount. This request is nonsensical.

It isn't nonsensical. Are you saying that (1/2,1/2) is not in either set?

Also, the two squares are the same size so it would have made more sense to make the total height/width an even number.

If you prefer scale everything by a factor of 2. It makes no difference. Or if you want make it 2 by 2 and shift things down and to the left so that the center is at (0,0). It makes no difference.

ITT: JoshuaZ asks you what colour this is:


Alternatively: ITT: JoshuaZ thinks two identical squares can have different heights.

Now, to answer your question:
y(0;0.5] x(0;0,5] is red
y(0.5;1] x(0;0,5] is black
y(0;0.5] x(0.5;1] is black
y(0.5;1] x(0.5;1] is red
For this scale (as you said, it changes nothing) x=0 and y=0 do not belong to the picture, because the squares are of identical dimensions.
It's very, very simple.

Ok. So the center is red? Then by most definitions of touching the red squares are touching and the black squares are not. The relevant technical term here is "path connected". I'm think (although I'm not very good with topology) that by the standard weaker notion of connectedness the black components are not connected. hence, I'd be inclined to say that they are not touching .But how one formally defines touching matters. 

If you prefer scale everything by a factor of 2. It makes no difference. Or if you want make it 2 by 2 and shift things down and to the left so that the center is at (0,0). It makes no difference.

Ok. So the center is red? Then by most definitions of touching the red squares are touching and the black squares are not. The relevant technical term here is "path connected".

Of course it can be shifted and scaled on the graph while maintaining it's properties. Thing is, PizzaPlanet's example used the origin in the lower left. While the open and closed intervals work to preserve the relative sizes and positions of the squares, you cannot conclude that the point at (0.5,0.5) is simply red. That result is a function of choosing the arbitrary origin and consequently which intervals are open or closed. The midpoint was chosen as (0.5,0.5) because the edges extended from the origin to 1. Not including the origin in the ranges allows the midpoint to assume a color identity.

Selecting only red should have been an intuitive mistake since the whole problem has mirror and rotational symmetry along the 'midpoint' and the actual colors used are non-consequential. Basically, if the red squares were touching, the black ones would have to be touching, and vice versa.

You could make the same argument by measuring the curvature of all the atoms currently on your arm. It doesn't have to be pixels.

I was using the word pixels as an analogy. I was positing that maybe particles can only exist in certain spaces. like on a grid. I never said this was true though

Presume this image represents fields defined by vectors, not just an arrangement of pixels. Are the black squares touching?

[/douchebaggery]

yes it is touching, at an infinitesimal spot, the vertex.

Quote from: JoshuaZ on March 17, 2011, 03:36:16 PM

If you prefer scale everything by a factor of 2. It makes no difference. Or if you want make it 2 by 2 and shift things down and to the left so that the center is at (0,0). It makes no difference.

Quote from: JoshuaZ on March 17, 2011, 06:55:01 PM

Ok. So the center is red? Then by most definitions of touching the red squares are touching and the black squares are not. The relevant technical term here is "path connected".

Of course it can be shifted and scaled on the graph while maintaining it's properties. Thing is, PizzaPlanet's example used the origin in the lower left. While the open and closed intervals work to preserve the relative sizes and positions of the squares, you cannot conclude that the point at (0.5,0.5) is simply red. That result is a function of choosing the arbitrary origin and consequently which intervals are open or closed. The midpoint was chosen as (0.5,0.5) because the edges extended from the origin to 1. Not including the origin in the ranges allows the midpoint to assume a color identity.

Selecting only red should have been an intuitive mistake since the whole problem has mirror and rotational symmetry along the 'midpoint' and the actual colors used are non-consequential. Basically, if the red squares were touching, the black ones would have to be touching, and vice versa.

What? No. Pizza's choice had nothing to do with where he choose the origins. He declared the point to be red. He could move the whole thing over and have had that point be black or kept it there and had it be black.  In that case it has an answer. If that point isn't red or black then the same analysis as that I gave for black applies to both colors. If it is both (I presume you don't want this given your setup) then both sets are path connected.

The rotational and reflective symmetry has nothing to do with anything. And the claim about the black touching being equivalent to the red touching is incorrect. What happens at the boundaries and how you define touching matters. But leaving the center point with an ill-defined color simply leaves things as just that: ill-defined. Don't make a problem with a poor set-up and then be act like it is a big deal that you can get nonsensical results out of that. There's a related xkcd.

Ok. So the center is red? Then by most definitions of touching the red squares are touching and the black squares are not. The relevant technical term here is "path connected".

Wrong. There is no center point. The amount of points is divisible by 4 for two dimensions. The amount of points for each side of the large square is divisible by 2. In any other case the squares would not be of equal dimensions. Only then would there be a "center point"

Wouldn't the center just be a smaller version of the original? Divided into two blacks and two reds?

Wouldn't the center just be a smaller version of the original? Divided into two blacks and two reds?

Yes, hence the:

ITT: JoshuaZ asks you what colour this is:

But that, obviously, is not one point.

Quote from: Trekky0623 on March 17, 2011, 09:14:17 PM

Wouldn't the center just be a smaller version of the original? Divided into two blacks and two reds?

Yes, hence the:

Quote from: PizzaPlanet on March 17, 2011, 04:06:13 PM

ITT: JoshuaZ asks you what colour this is:

But that, obviously, is not one point.

Does there have to be one?

Can a point have qualities of all four sectors? I would assume so. I think the single "center" 0,0 point would be red and black exactly as above, but I honestly don't know.

What? No. Pizza's choice had nothing to do with where he choose the origins.

If you are omitting the origin from the interval ya it does. He could have typed:

y(0;-0.5] x(0;0.5] is black
y(-0.5;-1] x(0;0.5] is red
y(0;-0.5] x(0.5;1] is red
y(-0.5;-1] x(0.5;1] is black

Can a point have qualities of all four sectors? I would assume so. I think the single "center" 0,0 point would be red and black exactly as above, but I honestly don't know.

No, that would mean the squares have common points. In that case they're not just touching, but overlapping. There is no single center point. There is a set of four points in the center.

Quote from: JoshuaZ on March 17, 2011, 08:39:11 PM

What? No. Pizza's choice had nothing to do with where he choose the origins.

If you are omitting the origin from the interval ya it does. He could have typed:

y(0;-0.5] x(0;0.5] is black
y(-0.5;-1] x(0;0.5] is red
y(0;-0.5] x(0.5;1] is red
y(-0.5;-1] x(0.5;1] is black

Yup, that is essentially the same and preserves the properties of the shape.

Oh btw, I will initially define "touching" to mean that if all the squares were one color, it would be completely continuous.

That still doesn't define anything. If you made the black squares red you'd get a continuous red square. That much is obvious.

That still doesn't define anything. If you made the black squares red you'd get a continuous red square. That much is obvious.

Yeah well, I thought the definition of 'touching' was obvious.  


...I'm going to bed, so here's a primer for the many conversations:

1. Does a horizontal line with a hole in it render the line noncontinuous?
2. Does that hole have width? (a point has zero dimensions).

Yes. Take, for example, y = x^2/x. The line is not continuos because it is undefined at 0. And yes, it does have a hole. You can't plug in 0.

Yes. Take, for example, y = x^2/x. The line is not continuos because it is undefined at 0.
And yes, it does have a hole. You can't plug in 0.

If there is a gap, what is its width?  

_{*For the purposes of constructing two level boundaries you can use a horizontal line with a hole too... like y=2(x-2)/(x-2).}

Quote from: JoshuaZ on March 17, 2011, 08:39:11 PM

Ok. So the center is red? Then by most definitions of touching the red squares are touching and the black squares are not. The relevant technical term here is "path connected".

Wrong. There is no center point. The amount of points is divisible by 4 for two dimensions. The amount of points for each side of the large square is divisible by 2. In any other case the squares would not be of equal dimensions. Only then would there be a "center point"

Huh? I don't follow this. First of all, the number of points (technically cardinality) is dividisible by four in the same that it is divisible by 3 or 2 or 5 or 1001. Any infinite set can be divided up into any finite number of sets all of the same cardinality. If the meaning of this isn't clear, I suggest reading the Wikipedia article on cardinality. The number of dimensions isn't relevant.  Any square has a center point (if this isn't obvious to you, draw the lines connecting the opposite corners. Consider the point that is the intersection of those two lines. That point is the center.)

Quote from: Trekky0623 on March 17, 2011, 09:59:29 PM

Yes. Take, for example, y = x^2/x. The line is not continuos because it is undefined at 0.
And yes, it does have a hole. You can't plug in 0.

If there is a gap, what is its width? 

_{*For the purposes of constructing two level boundaries you can use a horizontal line with a hole too... like y=2(x-2)/(x-2).}

A gap can have zero width and still be a gap.

1. Does a horizontal line with a hole in it render the line noncontinuous?
2. Does that hole have width? (a point has zero dimensions).

Sigh. Yes, it renders it non-continuous. This isn't obvious when one is using the rough intuitive notion of continuity. But when you try to make it more precise in terms of limits. (That is, a function f(x) is continuous at a point a, if f(a) exists, lim f(x) as x->a exists, and lim f(x)=f(a)) then it becomes immediately obvious that a line with a point missing is not continuous.

Quote from: ﮎingulaЯiτy on March 17, 2011, 10:06:09 PM

Quote from: Trekky0623 on March 17, 2011, 09:59:29 PM

Yes. Take, for example, y = x^2/x. The line is not continuos because it is undefined at 0.
And yes, it does have a hole. You can't plug in 0.

If there is a gap, what is its width?  

_{*For the purposes of constructing two level boundaries you can use a horizontal line with a hole too... like y=2(x-2)/(x-2).}

A gap can have zero width and still be a gap.

I agree completely, which is why the those squares can have a color boundary that doesn't have any width. If we conceive of that boundary as a linear extension of a hole (extending parallel to the y axis), it suggests that even the black and red squares that are side by side can be separate.

Sigh. Yes, it renders it non-continuous. This isn't obvious when one is using the rough intuitive notion of continuity. But when you try to make it more precise in terms of limits. (That is, a function f(x) is continuous at a point a, if f(a) exists, lim f(x) as x->a exists, and lim f(x)=f(a)) then it becomes immediately obvious that a line with a point missing is not continuous.

Good. If you recognize the non-continuity there, you can apply it between the squares even though there is zero distance between them.

Huh? I don't follow this.

Consider three equidistant points in one dimension: a, b, and c. What is the middle point? b
Consider four points, same principles: a, b, c, d. What is the middle point? There isn't one. What are the middle points? b and c
For one dimension, the amount of points has to be odd for there to be a center point.

Now, let's go 2d:
a1, b1, c1,
a2, b2, c2,
a3, b3, c3;
What is the middle point? b2
a1, b1, c1,
a2, b2, c2;
No single middle points. Arguably, b1 and b2 could be multiple middle points.
a1, b1,
a2, b2;
Again, no middle points. You will find that this will happen to any multiple of 4 in two dimensions (not just that, but this is the relevant example).

Whether or not the square is an infinite amount of points is irrelevant. It can still either be or not be divisible by four. In this case, each of the little squares is of exactly the same dimensions. Therefore, each of them has the area of A. The large square therefore has the area of 4A. We know that the large square consists of four times some amount of points (an infinite amount of points within each square's boundaries, of course). This is also why your original coordinate suggestions were wrong. Assuming that the squares do not have common points, plotting it from (0,0) to (1,1) would give us squares of areas (0.5+ε)^2 for two of them and 0.5^2 for the other two due to the inclusion of x=0 and y=0 (or, if you prefer x=1 and y=1)

If you disagree with this, well, you're wrong (I remind you of our boundaries! What's the middle item of (0;1]? No, it's not 0.5, that would be valid for [0;1], and I refer you to the above for why that can't be correctly split in four). However, even if you stubbornly cling to it, it doesn't help your case in the slightest. If we hold to your hypothesis, the "colour" of the middle "point" is this:

Now, ask yourself your own question: Is the middle "point" red or black?

We have a winrar.

I remind you of our boundaries! What's the middle item of (0;1]? No, it's not 0.5

ems that you are getting confused by the subtle things that it infinite sets can do. Let's say I have line segment, L. L might be (0,1), [0,1], [0,1), or (0,1]. Further, let us supose that a left part of it is red and a right part is black. Further suppose that if I'm at a black point and I move right I must hit a black point or the end of the segment. And similarly, assume the same for red and going left. Now, somewhere in the middle, there's a point X such that anything to the right of X is black, and anything to the left of X is red. It doesn't matter if X is "in the middle" or not, X is either red or black. We can choose either way but we must choose one. There's a similar choice in the box example. How you choose matters. (If this is confusing you might want to read up on open and closed sets.)