Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote -4 down vote favorite
1

The area of a circle is calculated by the formula $\pi R^2$, in which the $\pi$ is an "approximate" number. This means we never get the exact area of a circle. I wonder why we can get the exact area of shapes like rectangles, squares, but not circles. Is there another mathematical system in which we can represent exact area of a circle? Why can't we get the exact area of a circle in our current mathematical system?

share|improve this question
13  
If $R$ is the exact radius of the circle, then $\pi R^2$ is the exact area of the circle. I don't understand the question. If you are concerned about the irrationality of $\pi$, then I suppose you would have the same problem finding the area of a rectangle with sidelengths (exactly) $e$ and $\sqrt 2$? (Its area would be exactly $e\sqrt 2$.) – Jonas Meyer Jun 30 '12 at 3:22
5  
Pi is not a "proximity number" (whatever this may mean...): it is just a good ol' number! – Mariano Suárez-Alvarez♦ Jun 30 '12 at 3:25
6  
$\pi$ is not approximate. $3.14159$ is approximate, but that's not $\pi$. If you want to start out with integer radius and get a rational area, you're out of luck, potentially unless you change some part of your definition of a circle. – Robert Mastragostino Jun 30 '12 at 3:31
1  
@didxga: What if $R = \frac{1}{\sqrt{\pi}}$, then the area is $1$. There is nothing "approximate" about this number. – fretty Jul 10 '12 at 17:27
1  
Just remembering that $\pi = \frac{C}{2R}$, where C is the length of the circumference. It follows that $A = \pi\times R^2 = \frac{C}{2R}\times R^2 = \frac{C\times R}{2}$. Conclusion: area $A$ is known if you know $C$ and $R$, where either $C$ or $R$ (or both) is irrational, since $\frac{C}{2R} \in \mathbb{I}$. Irrational numbers are incommensurable, and it often put people in a situation between logic and physical intuition. – Ian Mateus Jul 20 '12 at 21:56
show 7 more comments

closed as not constructive by Will Jagy, Norbert, Arkamis, Asaf Karagila, rschwieb Oct 28 '12 at 0:43

As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or specific expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, see the FAQ for guidance.

5 Answers

up vote 7 down vote

$\pi r^2$ is the exact area of the circle whose radius is $r$.

The point seems to be that $\pi$ is not known exactly. Just what that means is a question that can bear examination. However, the area of the circle is known as precisely as $\pi$ is known, since the area of the circle is exactly $\pi r^2$.

share|improve this answer
yes, Pi*R^2 is only a formula representation of exact area, but we never get exact result out of that formula, as you pointed exact value of Pi is unknown – didxga Jun 30 '12 at 3:38
1  
@didxga , I think you're confusing things here: that you (we) cannot write $\,\pi R^2\,$ as an integer number or as a rational number with a finite decimal expansion does not mean the formula is not "exact": it is as exact as you may hope to ever have it, you just won't be able to write it down as $\,7, 3.14\,\,or\,\,14/5\,$...*but it is exact!* – DonAntonio Jun 30 '12 at 3:45
1  
@didxga: Would your question be satisfied by a circle with radius $1/\sqrt{\pi}$? – Blue Jun 30 '12 at 3:49
@DonAntonio Your explanation make sense to me, but we can still get extremely and infinitely approximate result, as we only get extremely approximate value of Pi. – didxga Jun 30 '12 at 3:51
2  
@didxga: I do not see what limitation you are referring to. Hence I do not understand your question. I am done discussing this, but I wish you well in resolving your question. – Jonas Meyer Jun 30 '12 at 4:14
show 8 more comments
up vote 6 down vote

This is a philosophical question. $\pi$ is not known as 3 or 10 are known. When we say the area of a circle is $\pi r^2$, for given r, this is shorthand for a number whose distant digits are known only by dint of considerable labor, and whose remote digits will never be known. Yet we have the means to calculate with any required precision, and so we take it as a known quantity. We can bound it from above and below. $\pi$ is not unlike a trained bear. The trainer has certain expectations, but does not presume to understand the bear's thoughts. It's a pragmatic approach.

share|improve this answer
4  
The statement that $\pi$'s "remote digits will never be known" is misleading if not incorrect. There is no digit of $\pi$ that cannot be known given sufficient computation, and you say as much in the following sentence. (Of course, one cannot know all the digits of $\pi$, but every digit of $\pi$ can be known...) – Rahul Narain Jun 30 '12 at 6:50
Any digit can in principle be known. Many will never be known because there are an infinite number of digits and any attempt to run them down would be of finite duration. – daniel Jun 30 '12 at 12:45
1  
"Will never be known" is not the same as "cannot be known." The former is pragmatic. Also, +1 for "$\pi$ is not unlike a trained bear." – Neal Jun 30 '12 at 13:53
5  
I've never understood the fascination some people have with equating "number" with "infinite sequence of decimal digits". Doubly so when they only do it for certain numbers -- e.g. so that they can complain about the "infinite" part for numbers like $\pi$, but mysteriously forget about it for numbers like $3$ -- but don't seem to recognize that they are being inconsistent. (and $1/3$ could be in either category, depending on the person making the argument) – Hurkyl Jul 10 '12 at 9:06
it's wrong too say the area of the circle. the circle is a 1-dimensional object. the area of a disk – mohamez Jul 24 '12 at 17:17
show 1 more comment
up vote 1 down vote

The figure used to arrive at the solution to the area of a circle using pythagoras theorem for a right angled triangle in a special case for an isosceles triangle.

radius of the circle is 'r'.

OA is $perpendicular$ to YC

OB is $perpendicular$ to AC

$OB_{2}$ is $perpendicular$ to BC

$OB_{3}$ is $perpendicular$ to $B_{2}C$

$\ldots$

$OB_{n}$ is $perpendicular$ to $B_{n-1}C$

Let $n -> \infty$

TO BE CONTINUED $\ldots$

enter image description here

share|improve this answer
Can't you write an answer once and not in parts? – Asaf Karagila Oct 27 '12 at 23:47
up vote 0 down vote

Put another way, $\pi$ is defined to be one (in fact, both) of the following:

(1) the ratio of the circumference of a circle to its diameter (known to be independent of the choice of circle, or

(2) the ratio of the area of a circle to the square of its radius (again, independent of the choice of circle).

Now, it is known that there are uncountably many real numbers can never be completely described. Each such number is irrational, but not all irrationals are such numbers. $\pi$ and $\sqrt{2}$ are examples of describable irrationals.

share|improve this answer
This is a bit tangential, but… @didxga: I'm sorry if I am telling you something you already know, but I think it's worth mentioning. When Cameron uses the word "uncountable", it is not strictly the same thing as "infinite". This is because some infinities (including the amount of rational numbers) are called countable. So in a mathematically meaningful way, "almost every" real number is irrational and thus cannot be put into finite or repeating decimal notation; $\pi$ just happens to be one that people care a lot about. – Eric Stucky Jun 30 '12 at 9:23
up vote -1 down vote

An approximation for the area of a circle of radius $r$ is what follows:

$$f(n) = (2)(r^2)(2^n)\sqrt{2 -\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}}$$

Approximate area: $A = f(10)$

The bigger the $n$ the better the area.

$n$ is a natural number greater than or equal to 10.

The following is an alternative:

$$f(n)=(2+(1))(r^2)(2^n)\sqrt{2 -\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{1}}}}}$$

for $n = 3$, $$f(3) = (2)(r^2)(2^3)\sqrt{2 -\sqrt{2+\sqrt{2+\sqrt{2}}}}$$

for $n = 7$, $$f(7) = (2)(r^2)(2^7)\sqrt{2 -\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}}}}$$

share|improve this answer
Although we don't know all the digits of $\pi$, we can obtain any desired level of accuracy, and you formula $(1)$ is an example of this, which is derived by inscribing suitable $n$-agons in the circle. This is a very useful and interesting techinque used by ancient mathematicians to compute $\pi$. You could improve your answers by explaining a little more about the formulas you provide and addressing the question of the OP. – Peter Tamaroff Jul 24 '12 at 16:54
this is a method i'm going to publish soon in my new book. will give the name of the book and its publisher soon. – Rajesh K Singh Jul 28 '12 at 16:36
i will also put it here soon – Rajesh K Singh Jul 29 '12 at 12:16
let me know how do i depict a circle using latex or any other software – Rajesh K Singh Jul 29 '12 at 12:34
Using induction on the identity $\sin\left(\dfrac x2\right)=\sqrt{\frac12-\frac12\sqrt{1-\sin^2(x)}}\;$, it is not too difficult to derive $$ a_0=2\cos(x)\qquad a_{n+1}=\sqrt{2+a_n}\qquad2\,\sin\left(\frac{x}{2^{n+1}}\right)=\sqrt{2-a_n} $$ However, the question seems to be more about the ramifications of the irrationality of $\pi$ than whether we can approximate it. In $\LaTeX$, $\raise{3.5pt}{\circ} \raise{3pt}{\large\circ} \raise{2.5pt}{\Large\circ} \raise{1pt}{\huge\circ} \hspace{-2.5pt}\Huge\circ$ are limited. Others might be able to help you with TikZ or similar packages. – robjohn♦ Oct 4 '12 at 17:35

Not the answer you're looking for? Browse other questions tagged or ask your own question.