How to prove the infinite number of sides in a circle?
I was in geometry class today when I came across the following formula for the external angle of a regular polygon with n sides: $$Ea = \frac{360º}{n}$$
So I thought if $$ n\rightarrow\infty $$ then $$ Ea\rightarrow0$$ Thus if a circle is a polygon with an infinite number of sides it's external angles would approach 0. I then tried to do the reverse way, trying to figure out a way to prove the premise that a circle has n = infinity; however I could not prove it.
In this sense, how do you prove the infinite number of sides in a perfect circle?
geometry circles infinity
Bernardo MeurerBernardo Meurer
$\begingroup$ What are you talking about? Circles have one curved side. $\endgroup$ – John Feb 25 '15 at 17:47
$\begingroup$ @John, indeed, circles ave one curved side; however put some thought into it. A triangle (n=3) doesn't resemble a circle at all, a pentagon (n=5) is slightly closer, and an octagon (n=8) closer and so on. In this sense: $$n\rightarrow \infty thus \triangle\rightarrow\bigcirc$$ $\endgroup$ – Bernardo Meurer Feb 25 '15 at 18:34
$\begingroup$ Take a look at en.wikipedia.org/wiki/Megagon for example. $\endgroup$ – Bernardo Meurer Feb 25 '15 at 19:13
$\begingroup$ Arquimedes was an extraordinary believer on this. $\endgroup$ – Piquito Jun 14 '16 at 18:00
$\begingroup$ math.stackexchange.com/questions/720935/… $\endgroup$ – user301988 Jun 16 '16 at 5:48
In a way, there is no reverse way for you to prove. You must always be careful about what your definitions. That way, you will always have a clear mathematical way of writing what you are trying to prove and how to prove it, as well as how "reverse ways" of some theorems would look.
For example, in your case, you have defined the following:
regular polygons with $n$ sides.
The angles of these polygons.
What you have shown is this:
As $n$ approaches infinity, the external angle of a $n$-sided regular polygon approaches $0$.
What you Have not shown is this:
As $n$ approaches infinity, the $n$-sided regular polygon approaches a circle.
Why haven't you shown this? Well, let's see:
First of all, you haven't defined a circle. Sure, you can define a circle, but a circle will not be a regular polygon, at least not by the definition of a regular polygon. OK, that may be a problem you can overcome. We can say that a circle is some sort of curve, just like a polygon. However, there is another problem:
You did not define what it means for two curves to approach one another. Without defining exactly what it means for a polygon to approach a circle, you cannot say a circle approaches it, neither can you then say "How to prove the inverse of this statement?"
Even if you define what it means for two curves to approach each other, you are still miles away from showing the statement which you, ultimately, want to show:
A circle is a regular polygon with an infinite number of sides
Again, why could you not proven this statement? Well, it isn't a mathematical statement, so it cannot be proven in a mathematical way. For example, a polygon is defined as a collection of a finite amount of straight lines, so the concept of "a regular polygon with an infinite number of sides" does not exist yet. You can define it, sure, but if you just define it as a circle, then the statement becomes empty. You could define "generalized polygons" as such:
A curve is a generalized polygon if it is a limit of a sequence of (finite-sided) polygons.
In this case, you must, of course, define what a limit of a curve is, but that is possible (albeit not trivial).
If you decide to define it that way, you can now prove the statement:
If $P_n$ is a regular $n$-sided polygon, then the limit of the sequence $P_1, P_2, \dots$ is a circle.
This statement is, basically, the statement "$n$-sided polygons approach a circle as $n$ tends to infinity", in mathematical terms. However, notice what happened:
You can no longer speak of a "reverse" of this statement. The statement, by its nature, works only in one direction: the limit of this sequence is this curve.
In defining generalized polygons, you lost the ability to speak about the number of sides of a polygon. For example, a square is a $4$ sided polygon and has $4$ sides. You can say "the number of sides of this polygon is such and such". You cannot say the same thing about generalized polygons. You can define the number of sides as such:
The number of sides of a generalized polygon $P$ is $n$, if $P$ is a $n$-sided polygon, and is $\infty$ if, for all values of $n$, $P$ is not a $n$ sided polygon.
If you decide the number of sides that way, then the statement
A circle has an infinite number of sides
Becomes equivalent to the statement
A circle is a generalized polygon and for all values of $n$, a circle is not a $n$ sided polygon.
5xum5xum
$\begingroup$ May I use this (quoting you, for sure) when I have to explain and illustrate mathematical logics ? $\endgroup$ – Claude Leibovici Feb 25 '15 at 15:30
$\begingroup$ @ClaudeLeibovici I would be honored if you would. Thank you for the compliment. $\endgroup$ – 5xum Feb 25 '15 at 15:31
For all intents and purposes, infinity is not a number. It doesn't really make sense to say $n = \infty$. The integers have order, and if it were true that $n = \infty$ then we would get the following logic: $$\infty = n < n+1 = \infty \implies n = n+1$$ This is why a limit is useful (as you have denoted with the notation $n \to \infty$) because we can explore what happens to an expression if we allow $n$ to be as large of an integer as we want. A large enough number will cause our expression to behave similarly to infinity, which is why we know things like $\lim_{n \to \infty} \frac{360^\circ}{n}=0$. This limit is true in an algebraic sense, but geometrically we don't really make sense of an infinite sided polygon with an exterior angle of zero. After all, that would imply that $\infty \cdot 0 = 360^\circ$, as the sum of all the exterior angles must add to $360$ degrees. But is it possible that $\infty \cdot 0 = 360$? How do we know if instead it should be $\infty \cdot 0 = 0$ or $\infty \cdot 0 = \infty$? These are questions you will need to explore with calculus, where you can develop a much stronger notion of the limit. With the aid of calculus, you may be able to reverse-engineer the problem you are trying to work on.
Anyway, here is what I think is the main problem you are running into. The formula you have to calculate the external angle of a regular polygon is true for a polygon with $n$ sides. So, if you define a circle as "the limit of a regular polygon with $n$ sides", you cannot apply that formula because it is only true for those regular polygons with finitely many sides. As we've established, $n \neq \infty$, so you simply cannot use this formula for a polygon where $n = \infty$
graydadgraydad
$\begingroup$ In the cardinal numbers and in the ordinal numbers, there is an infinity. (There's actually more than one infinity in each of those number systems…) You might say those aren't numbers, but then I'll counter that you never defined "number." Is $i$ a number? $\endgroup$ – Akiva Weinberger Feb 25 '15 at 15:35
$\begingroup$ @columbus8myhw While your point is true, it is generally safe to assume that, in a large part of mathematics, a "number" is an element of real (sometimes complex) numbers. $\endgroup$ – 5xum Feb 25 '15 at 15:38
$\begingroup$ @columbus8myhw Yes I could get into that I suppose. But it seems irrelevant to this question. I wanted to establish that (at least at the level of geometry and algebra) you don't treat infinity like a number; you use it as a concept. I edited my first sentence to make my statement less encompassing. $\endgroup$ – graydad Feb 25 '15 at 15:39
$\begingroup$ About the n=infinity. I'm aware infinity can't be treated as a number, but I couldn't figure out how to use LaTeX without 'breaking' the line, thus I used '=' purely for aesthetics; but you're absolutely right to correct me. $\endgroup$ – Bernardo Meurer Feb 25 '15 at 16:10
$\begingroup$ Also, I asked here because my algebra teacher told me to wait until I learned calculus in university to understand it; but I was extremely curious and didn't want to wait. High-school level maths is very frustrating, I can't solve most of the problems I think of. $\endgroup$ – Bernardo Meurer Feb 25 '15 at 16:13
Here is my approach to a proof that a circle has infinitely many sides:
We consider an equilateral n-gon. The n vertices are equidistant from the center.
If n is infinity, there is an infinite number of vertices that are equidistant from the center. That is the definition of a circle.
For each adjacent pair of vertices on the n-gon, there is a side (edge) connecting them. Therefore, if there is an infinite number of vertices, there would have to be an infinite number of sides.
Phil ScopesPhil Scopes
While you can approximate a circle from within or from the outside with n-polygons, thus all the geometrical objects of the approximating series are n-polygons, the limit object circle is no n-polygon: There is clearly not a single line segment on the circle.
mvwmvw
Infinity was a number to both Kepler and Leibniz who spoke of a circle as an infinite-sided polygon. This point of view is useful in analyzing the properties of the circle, as well as more general curves, in infinitesimal calculus. Today we have rigorous mathematical frameworks incorporating both infinite and infinitesimal numbers, vindicating Leibniz's ideas.
Mikhail KatzMikhail Katz
Well if n = ∞ then the angle would not be 0, but an infinitesimal. This means that 360 divided by that infinitesimal would equal ∞.
If the outer angle was 0, then the shape would not be a shape at all; it would be a straight line in a sense, but that would be impossible since a straight line has an angle of 180 degrees along its edge. An outer edge of 0 is impossible. Since a circle is a shape, there has to be at least some angle to it's outer edge, hence it being an infinitesimal not 0.
The formula for a n sided regular polygon that inscribes a circle of radius 1 is 2n*sin(pi/2n). We can see this by drawing lines that connect the center of the polygon to the corners. This creates n isosceles triangles with side lengths 1 and the central angle is pi/n. Breaking each of those triangles in half and we see that half of each side length is sin(pi/2n). Therefore each side length is 2sin(pi/2n) and the whole perimeter is 2nsin(pi/2n). If you take the limit of that as n goes to infinity (applying l'hopital's rule) you get pi.
At the very least, an infinite sided polygon with distance from the center to the corners 1 has a perimeter/circumference of pi.
AGINSAGINS
$\begingroup$ In general it is helpful, when answering an older Question with an Accepted and/or upvoted Answer, to clarify what additional information your post is adding to the existing Answers. It seems you do little here except restate the information given in the Question using radians rather than degrees as a measure of angles. $\endgroup$ – hardmath Jun 14 '16 at 18:17
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