Error in a proof for the area of the circle?

[MegaMoh]

I saw multiple of those visual proofs, given that the circle's circumference is 2* pi * r, that the area of the circle is pi * r^2 and did this proof which gave me 2 * pi *r^2, twice the area?!

 

[lev888]

Your operation of transforming the circle into the rectangle involved some stretching - the center of the circle became a line (side of the rectangle). So, the area increased. It's interesting that it exactly doubled.

 

[Dr.Peterson]

What you are doing is replacing many thin approximate triangles with rectangles, whose area is twice that of the triangles. So the actual area is twice what you have calculated.

For a collection of similar visual "proofs", all of them correct, see this entry in my blog (which is based on answers to questions on my former site). Of course, all of these need a lot more (as in calculus) to turn them into actual proofs.

 

[JeffM]

My answer is fundamentally the same as Dr. P's.

You simply assumed that the area of the rectangle equals the area of the circle. Actually, proving statements about infinite sums is a bit complex, and your specific statement is false.

Morevever, you talked about the "next" point without even attempting to define what that might mean. If you were to deal with this as a Riemann sum, you would find, as Dr. P said, that you were summing the areas of triangles, not rectangles. By assuming that they were rectangles, you doubled the area. The center of the circle was, as lev said, "stretched" to form the fourth side of your rectangle.

What is interesting is that you came close to replicating Archimedes's method for estimating the value of pi, which involved using inscribed and circumscribed polygons.

 

[LCKurtz]

You didn't say whether you have had calculus or not. If not, you can ignore the following. If you were going to calculate the area of a circle of radius [MATH]R[/MATH], you might choose to integrate over the region [MATH]x^2+y^2 \le R^2[/MATH], which in rectangular coordinates would give

[MATH] \text{Area} =\int_{-R}^R\int_{-\sqrt{R^2 - x^2}}^{\sqrt{R^2-x^2 }}1~dydx[/MATH]
You could work it with a trig substitution to get the correct area [MATH]\pi R^2[/MATH]. More likely, you would choose to change it to polar coordinates using [MATH]x = r\cos\theta,~y = r\sin\theta[/MATH]. Remember when you change variables in a double integral you need to include the Jacobian:

[MATH]J = \frac{\partial(x,y)}{\partial(r,\theta)} = \left | \begin {array}{c,c} x_r & x_\theta \\ y_r & y_\theta \end{array} \right | = \left | \begin {array}{c,c} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{array} \right | = r [/MATH]
giving the resulting integral

[MATH]\text{Area}= \int_0^{2\pi} \int_0^R r ~drd\theta[/MATH]
which also comes out the correct area. The thing to notice is that this is NOT

[MATH]\text{Area}= \int_0^{2\pi R} \int_0^R 1 ~drd\theta[/MATH]
which is the area if the rectangle you have drawn in the [MATH]r,~\theta[/MATH] space. The stretching that others have mentioned is accommodated by the inclusion of the Jacobian.

 

[MegaMoh]

Your operation of transforming the circle into the rectangle involved some stretching - the center of the circle became a line (side of the rectangle). So, the area increased. It's interesting that it exactly doubled.

What you are doing is replacing many thin approximate triangles with rectangles, whose area is twice that of the triangles. So the actual area is twice what you have calculated.

For a collection of similar visual "proofs", all of them correct, see this entry in my blog (which is based on answers to questions on my former site). Of course, all of these need a lot more (as in calculus) to turn them into actual proofs.

I didn't actually mean to transform them to rectangles. I said I'm taking a line, the one that has no width, and I always hear about how a rectangle contains infinitely many lines(yes, the one with no width) so if I could extract infinitely many lines and line them up together, they would form a rectangle which has an area of 2*pi*r^2

My answer is fundamentally the same as Dr. P's.

You simply assumed that the area of the rectangle equals the area of the circle. Actually, proving statements about infinite sums is a bit complex, and your specific statement is false.

Morevever, you talked about the "next" point without even attempting to define what that might mean. If you were to deal with this as a Riemann sum, you would find, as Dr. P said, that you were summing the areas of triangles, not rectangles. By assuming that they were rectangles, you doubled the area. The center of the circle was, as lev said, "stretched" to form the fourth side of your rectangle.

What is interesting is that you came close to replicating Archimedes's method for estimating the value of pi, which involved using inscribed and circumscribed polygons.

On the circumference of a circle, there are infinitely many points. By the "next point" I meant the one adjacent to the point chosen at first that is closest to it. And actually, I saw Archimedes's proof of it and wondered what would happen if instead of triangles it were lines(the ones with no width). Also see my reply to Dr.Peterson and lev888 above about the rectangles point.

You didn't say whether you have had calculus or not. If not, you can ignore the following. If you were going to calculate the area of a circle of radius [MATH]R[/MATH], you might choose to integrate over the region [MATH]x^2+y^2 \le R^2[/MATH], which in rectangular coordinates would give

[MATH] \text{Area} =\int_{-R}^R\int_{-\sqrt{R^2 - x^2}}^{\sqrt{R^2-x^2 }}1~dydx[/MATH]
You could work it with a trig substitution to get the correct area [MATH]\pi R^2[/MATH]. More likely, you would choose to change it to polar coordinates using [MATH]x = r\cos\theta,~y = r\sin\theta[/MATH]. Remember when you change variables in a double integral you need to include the Jacobian:

[MATH]J = \frac{\partial(x,y)}{\partial(r,\theta)} = \left | \begin {array}{c,c} x_r & x_\theta \\ y_r & y_\theta \end{array} \right | = \left | \begin {array}{c,c} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{array} \right | = r [/MATH]
giving the resulting integral

[MATH]\text{Area}= \int_0^{2\pi} \int_0^R r ~drd\theta[/MATH]
which also comes out the correct area. The thing to notice is that this is NOT

[MATH]\text{Area}= \int_0^{2\pi R} \int_0^R 1 ~drd\theta[/MATH]
which is the area if the rectangle you have drawn in the [MATH]r,~\theta[/MATH] space. The stretching that others have mentioned is accommodated by the inclusion of the Jacobian.

I'm not actually asking for a proof for the area of the circle(although I am searching for one that proves the circumference pls) I just wanted to know what's wrong with that "visual" proof I made.

 

[JeffM]

I didn't actually mean to transform them to rectangles. I said I'm taking a line, the one that has no width, and I always hear about how a rectangle contains infinitely many lines(yes, the one with no width) so if I could extract infinitely many lines and line them up together, they would form a rectangle which has an area of 2*pi*r^2



On the circumference of a circle, there are infinitely many points. By the "next point" I meant the one adjacent to the point chosen at first that is closest to it. And actually, I saw Archimedes's proof of it and wondered what would happen if instead of triangles it were lines(the ones with no width). Also see my reply to Dr.Peterson and lev888 above about the rectangles point.




I'm not actually asking for a proof for the area of the circle(although I am searching for one that proves the circumference pls) I just wanted to know what's wrong with that "visual" proof I made.

Dealing with infinity is tricky. There is a whole field of mathematics called analysis that deals with what you can legitimately say about infinite sums and what you cannot.

Between any two distinct points on the circumference of a circle, there is a segment of the circumference, right? And there are an infinite number of points in that segment. So there is no "next point," which is one reason that your proof is invalid.

Moreover, whoever told you that an infinite number of lines of no width add up to a finite width gave you very misleading information.

It is important to note that Archimedes did not say that he had found the exact value of pi. Rather he said that pi was somewhere between two exact values but not equal to either.

 

[MegaMoh]

Dealing with infinity is tricky. There is a whole field of mathematics called analysis that deals with what you can legitimately say about infinite sums and what you cannot.

Between any two distinct points on the circumference of a circle, there is a segment of the circumference, right? And there are an infinite number of points in that segment. So there is no "next point," which is one reason that your proof is invalid.

Moreover, whoever told you that an infinite number of lines of no width add up to a finite width gave you very misleading information.

It is important to note that Archimedes did not say that he had found the exact value of pi. Rather he said that pi was somewhere between two exact values but not equal to either.

That wasn't really an answer, just saying "whoever told you that an infinite number of lines of no width add up to a finite width gave you very misleading information" doesn't specify what is true. A line, by definition, has no width, and a rectangle does have infinitely many lines so putting infinitely many lines together makes a rectangle. Why would it not be true?

 

[JeffM]

Consider a rectangle with a height of 2 meters and a width of 3 meters. It contains an infinite number of lines perpendicular to the width so, according to your logic, infinity times 3 = 6. The rectangle also contains an infinite number of lines perpendicular to the height so again according to your logic, infinity times 2 must = 6.

[MATH]\therefore \infty \times 3 = 6 = \infty \times 2 \implies \dfrac{\infty \times 3}{\infty \times 2} = \dfrac{\infty \times 2}{\infty \times 2} \implies[/MATH]
[MATH]\dfrac{\cancel {\infty} \times 3}{\cancel {\infty} \times 2} = 1 \implies \dfrac{3}{2} = 1 \implies 3 = 2.[/MATH]
Reasoning about infinity is hard because no human has any experience with infinity. Extrapolating from the finite to the infinite leads to many errors. As I said, there is a whole course devoted to this topic.

 

[MegaMoh]

Consider a rectangle with a height of 2 meters and a width of 3 meters. It contains an infinite number of lines perpendicular to the width so, according to your logic, infinity times 3 = 6. The rectangle also contains an infinite number of lines perpendicular to the height so again according to your logic, infinity times 2 must = 6.

[MATH]\therefore \infty \times 3 = 6 = \infty \times 2 \implies \dfrac{\infty \times 3}{\infty \times 2} = \dfrac{\infty \times 2}{\infty \times 2} \implies[/MATH]
[MATH]\dfrac{\cancel {\infty} \times 3}{\cancel {\infty} \times 2} = 1 \implies \dfrac{3}{2} = 1 \implies 3 = 2.[/MATH]
Reasoning about infinity is hard because no human has any experience with infinity. Extrapolating from the finite to the infinite leads to many errors. As I said, there is a whole course devoted to this topic.

Multiplying by infinity wasn't at all related to the area. There are infinitely many lines perpendicular to the width of 3 meters where each of the lines has a length of 2 meters. Multiplying 3 by 2 gives the area which is those infinitely many lines summed together. It kind of seems to me as it's more like a [math]0\times\infty[/math] situation because the area of a line is 0, infinitely many of them is [math]0\times\infty[/math]

 

[LCKurtz]

Multiplying by infinity wasn't at all related to the area. There are infinitely many lines perpendicular to the width of 3 meters where each of the lines has a length of 2 meters. Multiplying 3 by 2 gives the area which is those infinitely many lines summed together. It kind of seems to me as it's more like a [math]0\times\infty[/math] situation because the area of a line is 0, infinitely many of them is [math]0\times\infty[/math]

You haven't told us whether you have had any calculus. "Situations" like [MATH]0\cdot \infty[/MATH] and [MATH]\frac 0 0[/MATH]arise frequently in the study of limits in calculus. Making sense out of such "situations" is what calculus is all about. You might enjoy calculus, assuming you haven't already studied it.

 

[MegaMoh]

You haven't told us whether you have had any calculus. "Situations" like [MATH]0\cdot \infty[/MATH] and [MATH]\frac 0 0[/MATH]arise frequently in the study of limits in calculus. Making sense out of such "situations" is what calculus is all about. You might enjoy calculus, assuming you haven't already studied it.

Yeah, I didn't. I just have an idea about differentiation and integration and limits.

 

[MegaMoh]

I still don't get the error.

 

[JeffM]

Your error is that you cannot add up infinite sums as though they are finite sums.

 

[Dr.Peterson]

The error is that you can't do what you did without changing the area. You stretched the point at the center into a line; you changed regions that look like triangles into what look like rectangles; and you tried to mix the idea of area with the idea of infinity, which doesn't work.

Or, to put it another way, in a proof (even a visual "proof" that only suggests something), you can only do things for which you have adequate justification, and you did things without showing why they are valid. You know that the fact that you got a false result means that you did something wrong; we've told you several things that are wrong. The valid proofs you have seen along similar lines demonstrate what has to be done differently.

 

[MegaMoh]

Your error is that you cannot add up infinite sums as though they are finite sums.

Isn't a circle the "limit" of all polygons, that's related to infinity, so there isn't anything "finite" about this.

The error is that you can't do what you did without changing the area. You stretched the point at the center into a line; you changed regions that look like triangles into what look like rectangles; and you tried to mix the idea of area with the idea of infinity, which doesn't work.

Or, to put it another way, in a proof (even a visual "proof" that only suggests something), you can only do things for which you have adequate justification, and you did things without showing why they are valid. You know that the fact that you got a false result means that you did something wrong; we've told you several things that are wrong. The valid proofs you have seen along similar lines demonstrate what has to be done differently.

My justification was that a rectangle has infinitely many lines in its area and you can't deny that, so I thought if I got infinitely many lines, they'd make a rectangle. Others answers just denied that a rectangle has infinitely many sides which isn't actually correct.

 

[Dr.Peterson]

Area is not found by adding infinitely many lines or points. The number of points or lines in a region is unrelated to area.

These two lines, _____ and ___________________, both contain the same number of points (infinite). They do not have the same length. By your reasoning, they would.

 

[MegaMoh]

Area is not found by adding infinitely many lines or points. The number of points or lines in a region is unrelated to area.

These two lines, _____ and ___________________, both contain the same number of points (infinite). They do not have the same length. By your reasoning, they would.

But an infinity greater than another infinity. That's why [math]\frac{\infty}{\infty}[/math] is conventionally undefined.

 

[LCKurtz]

But an infinity greater than another infinity. That's why [math]\frac{\infty}{\infty}[/math] is conventionally undefined.

You are just wasting our time here. Come back after you have taken a calculus course or, better, a more advanced analysis course.

 

[lev888]

I still don't get the error.

Do you know why Greenland is larger (relatively) on maps than on globes?