Lothar i==0 ???????????

[ccb056]

maybe some of you math gurus can help

a few days ago I can accross the equation: e^(2pi*i)-1=0
therefore
e^(2pi*i)=1
ln(e^(2pi*i))=ln(1)
2pi*i=0

since 2 and pi are both constants, you divide both out and are left with: i=0

but 0 is a real number
but i is an imaginary number
how can an imiganary number be the same as a real number?

[Krom]

And your posting this in the Tech Forum because?

[ccb056]

because it would look dumb in the ethics and commentary

[DCrazy]

Well, i does equal zero in the real plane. i = 0 + 1i, and 0 = 0 + 0i. Technically as far as reals go, the real portion of i equals the real portion of zero.

[Jeff250]

For e^(k*i), there are multiple solutions so that it equals 1, including k = 0, 4pi, 6pi, 8pi, etc. That's about all that I can say, but if you look at it as a trigonometric function, with k restricted to the interval [0,2pi), it should work according to your method, replacing 2pi with the equivalent, 0.

[STRESSTEST]

because it would look dumb in the ethics and commentary

It looks dumb here too. Off to the cafe' with ya

[Tricord]

a few days ago I can accross the equation: e^(2pi*i)-1=0

Who said that equation was actually accurate?

[Tetrad]

Who says imaginary numbers have to follow traditional algebraic thought?

[Gooberman]

Like others have said, you need to use complex algebra. Lothar may know an easier way, this is how I learned it.

your equation,
e^(2pi*i)=1+0i

Standard form z=x+iy, so let z = 1+0i
so your equation reads,

e^(2pi*i)= z

convert to polar (tends to be easier that way),
z=r*e^(i*theta), where r =sqrt(x^2+y^2) and theta = tan^-1(y/x)

after doing that, take ln(z)

ln(z)= ln[r*e^(i*theta)]

note: ln(x*y)=ln(x)+ln(y)

So for complex numbers,

ln(z)=ln(r)+ ln[e^(i*theta)]]
ln(z)= ln(r)+ i(theta+2*n*pi),

where n equals any integer. This extra part 2*n*pi is becasue, as jeff said, the phase angle theta can go around any integral multiple of 2*pi without changing z, its a circle so it comes back to its origional value.

Use this, and what you did above, and you should be able to show that it works with n=1.

[Dedman]

I used to know how to use imaginary numbers. But my imagination isn't that good anymore.

[Lothar]

e^(2pi*i)-1=0
therefore
e^(2pi*i)=1
ln(e^(2pi*i))=ln(1)
2pi*i=0

Here's a simpler example that contains the same mistake:

-2^2 = 4
sqrt(-2^2) = sqrt(4)
-2 = 2

The mistake is in the line I made bold -- namely, that square root is a multi-valued function, and so is ln(x) in the complex plane.

That is, sqrt(x) has 2 values. The positive one is called the "principal value", and the other is the negative of the principal value. (Actually, in complex numbers, this gets even messier -- for example, what's sqrt(i)? The principal value is (1+i)/sqrt(2), and the other is the negative of that.) Now, just like sqrt(x) has 2 values... ln(x) has an infinite number of values -- the "principal value" has imaginary part between -pi and pi, and then you can add any multiple of 2*pi*i to get another value.

Another way to say this is to say that e^x and x^2 are not invertible functions -- if I take 2^2 and -2^2, I get a 4, so if I say 4=x^2 you can't uniquely identify x. By convention, we take sqrt(4) = 2 -- the positive root. And by convention, we take ln(1) = 0, rather than 2*pi*i or 86*pi*i or -36*pi*i. But in reality, sqrt(4) gives us multiple values: +2 and -2. And in reality, ln(1) gives us multiple values: 0 + 2*k*pi*i for k=0,+-1, +-2, ...

Does that help? I know it's still a little fuzzy -- and it'll remain that way until you take a complex analysis class. But hopefully that shows you where your mistake happened.

[ccb056]

I understand the theory, but the math is a little beyond me.

[AceCombat]

ehhhh............yah same here. i hate algebra and anything above that

[Sting_Ray]

And I thought pre calc was "complex" algebra.

You guys make me feel stupid.

[whuppinboy]

who thinks ccb056 is getting y'all to do his homework with his last few topics?

[ccb056]

actually, that wasn't homework, I was just curius

I can do my homework by myself thank you very much

[Tricord]

Nice, Lothar.

Here's some more math trivia. Maybe Drakona will want a piece of this one

Take a circle. Everyone knows you can draw regular polygons in a circle, where each vertex is exactly on the circle itself. If you increase the amount of edges by one in order to get the next regular polygon, there is one more vertex on the circle, and the polygon becomes a better approximation of the circle. My question is: if you let the amount of edges (and thus vertices) go to infinity, do you cover the entire circle? Regardless of your answer, I want a sound mathematical proof to support it

[woodchip]

No, because there will still be a side (chord) to the polygon even though the polygons side is infinitly small. In short a polygon is straight sided and a circle's side is curved. I'll let Lothar or Drakona give the nuts and bolts answer.

[Drakona]

It depends greatly on what you mean by "letting things go to infinity."

I am very sure that the idea of a regular infinite-sided polygon inscribed in a circle doesn't make sense. For suppose you examine the first side of the polygon: it has a point at each end of the line segment, and since the points cannot be the same, the line segment must have positive, nonzero length. But then, a positive length will only go so many times around the circle--certainly not an infinite number of times!

If you can't get an infinite-sided non-self-intersecting regular polygon, what exactly do we mean by "letting things go to infinity?"

One thing that is sometimes meant by letting things go to infinity is to find a limiting curve. We would say that the sequence of polygons converges uniformly to the circle if, as the number of sides increased, the polygons got closer and closer to the circle and never farther away. This, of course, turns out to be true, but you knew that already.

Your question suggests that you're looking for an answer in the case that you already have the infinite-sided polygon in hand. Such a construction doesn't really make sense, though. Suppose we took the polygon to be inscribed in the unit circle in the complex plane. Your points are then at e^(2*pi*i/k) where k is the number of sides of the polygon. Notice what happens to this as you let k go to infinity--your points all end up at e^0, or 1. In other words, the polygon collapses to a point. This is most likely not what you intend to mean by letting things go to infinity.

Even if you did have an infinite-sided polygon in hand that you considered the end of your sequence of polygons, it still wouldn't cover the circle. Note first that none of the sides will be on the circle, since the ends are a finite, non-zero distance apart--however small, it is enough to make the chord be not on the circle. The vertices by themselves, even though there are infinitely many, cannot be enough to exhaust the circle: since you added them one at a time, there are only a countably infinite number of them, and the circle is uncountably infinite. (For the non-mathematicians whose eyes haven't glazed over yet, the circle is a bigger infinity than the number of points. Yes I know, it boggles the mind. )

I suspect what you really mean is a polygon with infinitesimal side-lengths. This, however, is more mathematically difficult than I care to explain since I'm already late for work . However, you can take my word that such a thing would not work even in the spaces I know of that incorperate infinitesimals. The basic reason is that you cannot use a sequence (even an infinite sequence!) of infinitsimals to get from "here" to "there". At least, not in the hyperreals.

All of that is to say, it depends. If you're willing to take a limiting curve, the answer is "yes, of course, you knew that already." If you're looking for an infinite-sided polygon, the answer is, "there ain't no such beast, and if there were, the answer would still be no."

I suspect that was more confusing than enlightening, but I have sketched proofs as you asked... hope my math was right.

-Drak

( w00t. You know my specialization on my master's degree is geometry, right? Now ask Lothar an amath question... )

[Krom]

God I must be a nerd, I was not confused by that and it makes sense to me!

[Dedman]

You math folk are funny I bet Lothar and Drakona stay up late on friday nights and talk math to each other.

[Lothar]

not just Friday nights.

[Dedman]

I suspected as much but didn't want to push it. Although I do understand. My wife (also an engineer) and I talk shop a lot too.

[Tricord]

Lol Drakona! I knew I would get you nicely confused on this one, no matter how good you are

Anyway, here's the reasoning. We know that between every two rational numbers, there is at least one irrational number to be found, and vice versa. We also know that the set of rational numbers is enumerable, but the set of irrational numbers is not. By it's very nature, the set of points on the circle is not enumerable. The amount of vertices on an infinite regular polygon are enumerable, thus we must be missing a lot of points.

You see, the definition of the infinite polygon does not allow coverage for all points of the circle, but your reasoning that the distance between two vertices is always a real number is biased. If you were to establish a formula that gives the distance between two vertices for a n-edged polygon, and you take the limit of n -> infinity, you will get zero as outcome. The reason this is possible is that the rational numbers, while an infinite set, are a distinct subset of the irrational numbers.

[Lothar]

Tricord, I think you're the one who's "nicely confused" by thinking you "got her" ;)

the definition of the infinite polygon does not allow coverage for all points of the circle.... The reason this is possible is that the rational numbers, while an infinite set, are a distinct subset of the irrational numbers. -Tricord

The vertices by themselves, even though there are infinitely many, cannot be enough to exhaust the circle: since you added them one at a time, there are only a countably infinite number of them, and the circle is uncountably infinite. -Drakona

So she already covered that point -- if you only count the vertices of the polygon, you cannot cover the circle.

However, what you're missing is an explanation of why you only count the vertices and not the edges. You're just assuming you don't count the edges -- that they don't cover any part of the circle -- while she gave two distinct possibilities based on whether or not you count them: first, that you can recognize that the edges are converging to the circle, and therefore treat the place they converge to as "covered" (meaning the whole circle is covered by the limiting curve); second, that you can treat the edges as not on the circle, and therefore treat the irrational-argument points on the circle as uncovered, and therefore, the circle not completely covered.

Whether it's "covered" or not depends very much on what you're looking at as "going to infinity" -- what you're treating as the limiting case. If you look at the limiting case as the locations of the vertices, and ignore the edges, you get an uncovered circle. If you look at the limiting case as the limiting curve (the limit of the vertex *and* edge locations) you get a covered circle.

The mistake you made is that you said to "let things go to infinity" but didn't specify what "things" are going to infinity or what you're considering as "covered" in the limit. That is, you specified that a limit must be taken, but didn't specify whether you intended for the limit to include only where the vertices ended up, or for the limit to include where the edges converge to. Your question was ambiguous. In response, she dealt with both cases, while you dealt with only one.

So, in summary, you're nicely confused :)

[Tricord]

Maybe it's just me not being used to interpret mathematical terms in English, because if what you wrote offers nothing more than what she wrote, I must have misunderstood her. I have not caught that she found my question ambigious, that the distinction between edges and vertices needed to be made. In my opinion she was being vague about what the infinite-order polygon would be or not be.
I never said "let go things to infinite" I distinctly mentioned the amount of edges and vertices, the order of the polygon, or whatever you use to denote this . I realise now there is a distiction to be made, and that in view of this question only the vertices are to be considered.

At any rate, I think we all agree it was an interesting problem. Blame the misunderstanding on the fact that I have had mathematics in dutch my entire life.

[DCrazy]

ΜΑΤΗ ΙΖ ΡΗμη

^^ No English

[Lothar]

Ahh, yes, if only the vertices are to be considered, then it's a trivial problem: Q is strictly smaller than R and we're done. But then, to even say that the number of vertices is the size of the rationals requires you to assume some things about convergence...

But why only consider the vertices when you take the limit as n->infinity (which, unless otherwise specified, is what I'll mean by "limit" from here on out)? You're treating it as if you could somehow create an infinite-sided polygon that, nonetheless, only had its vertices on the circle. That's calc-1-student limit-taking, not analyst limit-taking.

Look at what we have for a regular polygon with n sides, inscribed in a circle of radius R:
1) The inner radius (to an edge midpoint) is r = R*cos(pi/n)
2) The distance between vertices (that is, the edge length) is 2*R*sin(pi/n)
3) The area is n/2 * R^2 * sin(2*pi/n)
4) The peremiter is 2*n*R*sin(pi/n)

Now, in the limit as n->infinity (recalling that sin(x)/x -> 1 as x->0):
1) r -> R
2) edge length -> 0 (but linearly in 1/n)
3) area -> pi*R^2
4) peremiter -> 2*pi*R

So in the limit, the n-gon converges to the circle (and, in fact, it does so uniformly.) For any point on the circle, I can find a sequence of points on edges that (in the limit) approach that point along a radial line. So, in the limit, I can say that point is covered -- there is a sequence of points on n-gons that converge to every point on the circle (and since I can force my sequences to be on radial lines, I can furthermore guarantee that each sequence is unique.) The points in the sequences just happen to be on the edges of the n-gons, rather than at the vertices, for any points on the circle with arguments that are irrational multiples of pi.

You're working from a different fact, namely, that any n-gon only has n points on the circle. But, you're using that to conclude that in the limit, the countably-infinite-gon has a countably-infinite number of points incident with the circle. I don't think you can say that, though -- there's not necessarily such thing as a countably-infinite-gon (if you can prove such a thing exists, though, I'll relent.) The limit of n-gons, based on the construction you've given, is not a polygon with an infinite number of sides; the limit of n-gons is a circle. So whether or not any n-gon has a finite number of points incident with the circle, the limiting curve is identically equal to the circle.

(In other words, when you take limits, you can't just assume things converge, and you can't assume everything converges in the same way -- n is countable, but the limiting curve need not have a countable number of vertices. It need not have *any* vertices, in fact.)

[Drakona]

You didn't confuse me, Tricord, but here's something you should consider: the original question doesn't actually make sense. You can't always just say "let the number of edges go to infinity" and assume you're talking sense. The reason I gave several different answers to the question was that I was trying to work out what you might have meant by that.

In your post, you seem to assume that there is such a thing as an infinite-sided polygon that makes sense as the limit of a sequence of regular polygons. But exactly what polygon are you thinking of? And how exactly are you taking the limit? As I showed above, if you take the limit based on the location of the curve in space, you get a circle. On the other hand, if you take the limit of the locations of each of the points, your polygon collapses to a point. It would not surprise me to learn that if you did other things, the limit simply did not exist. Such is the case when working with infinities and limits--strange things can happen, and you cannot simply assume that everything always works out.

In this case, I'm not sure how you would get something that you can rightly call a polygon. To make you aware of the problem, try to picture in your head the polygon you are thinking of as infinite-sided, the limit of a sequence of regular n-gons. Now, the definition of a polygon is a sequence of points, each one connected to the one before by a line segment, and the last one connected to the first. But if you think about your infinite polygon, take one point on it. Where is the next point? You'd like it to be infinitely close to the one you just took--but there's no such thing as two points that are infinitely close. The construction simply doesn't make sense.

It takes some mathematical maturity to realize that you can't always just take a limit, or let things go to infinity, and get something sensible. Just because you have a sequence of regular polygons doesn't mean that if you take a limit and let things go to infinity you'll get a regular polygon back. Or a polygon at all. It all depends on how exactly you let things go to infinity--what process you mean those words to imply.

I hope that helps a little...

To return to the original question, I have been thinking about it little more throughout the day, and I have nearly (but not quite) convinced myself that you cannot cover the circle with a regular polygon with an uncountably infinite number of sides. There is one way you might be able to do it, though, which I still have to investigate: take a chord of the circle which covers an angle with a rational number of radians. Repeat that chord around the circle, around and around and around... can you make the thing go an uncountably infinite number of times and close up? Or is that nonsense? I haven't decided.

Somewhat surprisingly, if you don't require the polygon to be regular, and don't require it to close, I can cover the circle. Here's how to do it:

Take the interval [0, 1] and a well-ordering on it. (Here we are using the well-ordering principle, so the axiom of choice as well.) Let the first point of the polygon be the number which is smallest in the set [0, 1] under the well-ordering. Let the next point be the number smallest in the set [0, 1] minus that point above. And so forth, for all the points in the set. Thus we have the point X, for any X in [0, 1] connected to the point Y, where Y is the smallest point under the well-ordering in the set [0, 1] minus { A : A <= X under the well-ordering}. Such a point is guaranteed to exist by the well-ordering principle.

So we have edges defined for every point in [0, 1]. Simply take a function mapping [0, 1] continuously to the unit circle (letting 0 and 1 overlap). The thing will look like an unbelievable spider web inside the circle, but it should hit every point on the edge.

The problem is that that 'polygon' won't close. Under the well-ordering principle there's no guarantee that the set has a "greatest" member, only a "least" member, and I'm pretty sure you can't do anything simple to resolve that. So what I just stated does not show that there is a polygon which can be inscribed in the circle and also covers it. I have hopes that the ordinals might be of use here, but I'll have to get home and look in my set theory book. Anyway, it's been a while since I've done any analytic stuff like this, so it's kind of fun for me.


==============================

On a completely different topic, is anybody here the sort of doodlebug that always tries to disprove the four-color theorem any time someone talks about it? I've got another theorem like that that I was supposed to prove for homework last week. It's called the Motzkin-Rabin Theorem, and it goes like this:

Suppose there are some red and green points in the plane, and they aren't all on one line. Then there is a line through at least two points of one color which doesn't pass through any points of the other color.

If you like doodling on math problems, try to prove that. (Or try to prove it false). I spent nearly a week on it, and while I didn't succeed in proving it, I did make some pretty spastic pictures in the process

[El Ka Bong]

... phew.... Survived that read,..... what are you all on, ...?! .. Fractint... ? or something ...?

[Tricord]

I would beg to differ. I have been taught that a limit always makes sense, but that the outcome may vary. If you have a definition with an enumerable attribute, you can always examine the limit for this attribute to reach infinity. If not, then most proofs by induction would have to be reviewed, for one thing. If we take the definition of polygon and let increase the n value (I will refer to this as the "order" of the polygon due to lack of a better term), we can enumerate the polygons. A triangle, a square, a pentagon, hexagon, etc... I am by all means justified to take the limit to infinity and treat the result as a mathematical entity, AND still consider it a polygon (with some nuances, though). The best analogy I can make, is the expansion of carthesian geometry to the complex numbers. If we take the definition of circle, i.e. x² + y² = R², the object which is represented by this analytical notation is well-known: all points which have a distance R to the origin. Nothing prevents us from considering the complex field instead of the irrational field, in which case we find more points which answer to the analytical definition of circle. We can even write the equation of circles with an imaginary radius, or we can always find two intersection points for a circle and a line, even if they are not visually intersecting.
This is the interesting thing about mathematics, and I am surprised you seem to be hesitant to recognise it. You have a definition, then you expand the area on which you can apply the definition, and you see what the results are. Complex geometry is no more difficult than real geometry in it's analytical form, yet it is not to be drawn visually.

Also, I am not convinced about covering the circle using the well-ordening principle.

Let the first point of the polygon be the number which is smallest in the set [0, 1] under the well-ordering. Let the next point be the number smallest in the set [0, 1] minus that point above.

Take a number, any number, or in particular the smallest number from the set you describe, and despite the well-ordening principle, I will always be able to give you a smaller one. I can even give an arbitrary method to construct a smaller number for any number you might come up with, for instance using f(x) = x/2.
I can also counter your reasoning with my initial answer. You are still enumerating the points, "let the next point be this one". You are constructing an enumerable set of points, yet we agreed that the amount of points on a circle are not enumerable and cannot be covered all by an enumerable method.

[Tricord]

Suppose there are some red and green points in the plane, and they aren't all on one line. Then there is a line through at least two points of one color which doesn't pass through any points of the other color.

I think this will need more elaboration as well, because I can give a counter example just from the top of my head. Either I'm missing something from the quote, or the wording of the theorem is too vague.

Say, the green points are given by the following set:
{(x,y) | (x,y) elem R² and y = x}
The red points are given by the following set:
{(-1,0),(1,0)}

They are not all on the same line. I can not draw a line over the red points without having a green one.

[Lothar]

If you have a definition with an enumerable attribute, you can always examine the limit for this attribute to reach infinity.

Sometimes your examination leads to the conclusion that the attribute doesn't make sense as actually reaching infinity. It makes sense for any number N, but it doesn't make sense for infinity itself.

If not, then most proofs by induction would have to be reviewed

You're treating infinity like a number you can reach by successive addition -- if this was the case, you'd be right. But proofs by induction don't prove anything about what happens "at infinity"; they simply prove what happens for every natural number (even those approaching infinity.) You never actually reach infinity in induction.

Recall a basic inductive proof looks like this:
- statement f(k) is true for k=0
- if f(k) is true, f(k+1) is true
Tell me, what's f(infinity)? Inductive proofs don't say -- because you can't ever reach infinity by adding 1 to a finite number. All you can say is that f(k) approaches some limit as k approaches infinity; that doesn't tell you what happens if k is actually equal to infinity.

I am by all means justified to take the limit to infinity and treat the result as a mathematical entity, AND still consider it a polygon

No, you're not. As I demonstrated, the limit of polygons is not a polygon, it's a circle. And, as Catherine demonstrated, an infinite-sided polygon can't exist by definition (start at the first point. Give me coordinates for the next point. It can't be done.) Therefore, whatever mathematical entity you're considering an infinite-sided polygon... really isn't.

The best analogy I can make, is the expansion of carthesian geometry to the complex numbers.

Yes, we *can* expand certain geometries. That doesn't mean, though, that it's valid to always expand every construction. Based on the definition of a polygon, an infinite-sided polygon can't exist. Now, you may be able to create a modified definition for what a polygon is, and create an object with an infinite number of sides that qualifies under your modified definition, but under the currently accepted definition of "polygon" your construction doesn't lead to a valid mathematical object.

This is the interesting thing about mathematics, and I am surprised you seem to be hesitant to recognise it. You have a definition, then you expand the area on which you can apply the definition, and you see what the results are.

Right, and in this case, you recognize that the results are "this object doesn't make sense." But you're quite welcome to try to redefine "polygon" in such a way that the object does make sense.

Also, I am not convinced about covering the circle using the well-ordening principle.

Let the first point of the polygon be the number which is smallest in the set [0, 1] under the well-ordering. Let the next point be the number smallest in the set [0, 1] minus that point above.

Take a number, any number, and despite the well-ordening principle, I will give you a smaller one.

Well, first of all, you can't take "any number" in this case -- just numbers in [0,1]. Second, while your number may be smaller in the standard metric, the well-ordering principle defines an ordering under which you *can't* find a smaller number than the "smallest". You have to remain within that ordering -- and within that ordering, something like x/2 won't necessarily help you, because you don't know where x/2 occurs within the well-ordering. (The well-ordering principle doesn't say "take [0,1] in the normal order -- it defines an order that might be arbitrarily ugly.)

You are still enumerating the points, "let the next point be this one". You are constructing an enumerable set of points, yet we agreed that the amount of points on a circle are not enumerable and are not to be covered all by an enumerable method.

Actually, she's not constructing an enumerable set of points, because she's using the well-ordering principle on [0,1]. Well-ordering is based on the axiom of choice. You run into some weird things when you allow for the axiom of choice, and being able to construct something uncountable by simply choosing points is one of the weird things it leads to (though it can't hold a candle to Banach-Tarski, which literally allows you to create 2 complete balls from 1 ball of the same size. Like I said, axiom of choice leads to some odd results.)


edit:
with respect to the red and green points, the original problem stated that the set of points was finite (otherwise, you could fill the plane with red, then draw a green circle and 3 or more green rays coming out of it.) Also, it doesn't require you to be able to draw a line through only red points -- just that you can draw a line through only one color (either red, or green.) In your case, you'd be able to draw a line through only green points.

[Lothar]

I just thought of something:

Pierre, you said the limit of your n-gons is still a polygon. I'm going to assume you're fixing one vertex at (r,0).

So, tell me, is there a vertex at (-r, 0)? Is that point covered? When n=3, there isn't. When n=4, there is. When n=5, there isn't. When n=6, there is. In general, for n odd, there is no vertex there, and for n even, there is. So, when you take the limit as n goes to infinity, is there a vertex there or not? (Or, is there half a vertex, and if so, what does that mean?)

The same question holds for any rational fraction of the circle -- say, p/q of the way around (assuming no common factors in p and q). Whenever n/q is an integer, there will be a vertex there; whenever n/q is not an integer, there won't. So there will be a vertex at any rational point p/q exactly 1/q of the time. What does that mean in the limit? Is there a vertex there, or no vertex, or 1/q of a vertex?

In traditional calculus, you'd say the positions of vertices don't converge -- there is no "limit" of the positions of the vertices, and therefore, no "polygon" that you get out of the limiting process. There's a vertex at (r,0) in the limit, but all the the other points on the circle either have no vertices (if they're at irrational fractions of the circle) or have an undefined number of vertices. (If you don't mind working with non-convergent series, you can also say certain points have 1/q of a vertex, though that's an ambiguous statement as well!)

[Tricord]

That makes sense, the limit of the sinus function is undefined in the same way.

Ok, so I give up on the infinite order polygon

There is a lot more to this issue than I first had thought, so it's nice to learn stuff

[snoopy]

Ok, bear with me, I'm an engineer and don't really know all the much high level theoretical stuff, but:

if N is the number of vertices on a circle (all joined by lines), then what you all are saying (and I think is right) is the the limit of the geomerty formed by the points (and lines, assuming that lines are never part of the circle) as N approaches infinity is not equal to the circle itself. But, if you start with the lines tanget to the circle (circle inscribed in the polygon) and the number of sides goes to infinity (M), the polygon approaches the cirlce itself?

[Lothar]

not quite, no

If you start with the polygon EITHER inside OR outside the circle, its limit is the circle itself, as I demonstrated by showing the inner and outer radii of the polygon approach the same number.

However, Tricord was under the false impression that since a polygon with N sides would have N points in common with the circle (either vertices or edges, depending on which is inscribed how) then the limit would only cover a countable number of points (the points at rational divisions of the circumference), rather than the entire circle (including irrational points). This was based on a misunderstanding: essentially, he believed that the mathematical object of an "infinite-sided polygon" actually exists and behaves like a normal polygon. Since each polygon would only have vertices at rational points around the circle, he concluded the limit would only have vertices at rational points around the circle. I demonstrated this to be false by showing that taking the limit in such a way leads to points that are ambiguous -- for example, (-r,0) is a vertex of a polygon with an even number of sides but not of one with an odd number of sides, so in the limit as N->infinity, you can't say it definitely does or doesn't have a vertex there. Therefore, the limit of the polygons doesn't behave like a polygon, so the property of finite polygons (sharing only rational points with the circle) doesn't carry over when you take the limit.