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Squaring the Circle. A proof for whatever its worth

I just proved that the circle is a perfect square. This probably belongs in the math philosophy forum as I'm sure it will be controversial.

I decided to post the proof here just as food for thought for anyone who might be interested.

This is hot off the presses as I only just did this proof tonight. In fact, if you find any typos in the proof I'd be glad to have them pointed out.

The proof is in the attached pdf document. I'm not about to type it in here in latex. Plus I use a lot of graphic illustrations too.

Let me know what you think.

And yes, I know it's not a perfect square of rational numbers, but that's the philosophical controversy part. ;)

What has been demonstrated is that it is indeed a perfect square. In other words, there exists a quantity that when squared will give the area of a circle. And this is what I have demonstrated.

Re: Squaring the Circle. A proof for whatever its worth

If you have a unit circle, then a square with side lengths $\displaystyle \sqrt{\pi}$ has the same area as the circle. I don't think this fact was ever in doubt. I believe the problem was to square a circle with a straight edge and compass. Since $\displaystyle \sqrt{\pi}$ is irrational, it means the best you can do is approximate a side of the square to an arbitrary degrees of precision.

Re: Squaring the Circle. A proof for whatever its worth

Hi Jakncoke,

Yes, I realize I haven't done anything profound. I'm just playing with geometry and studying Euclid's Elements. I'm actually doing this specifically in mind of using irrational and transcendental numbers for some of these lengths to see what I might learn from this exploration. I just went through this proof to see how it all works out. I was really kind of fascinated by how everything worked out so perfectly. This was of course, precisely because I started with a rectangle that had one leg as a multiple of $\displaystyle \pi$. And that surely played a large roll in making things work out nice.

Even so, I found this proof pretty interesting. I thought I'd share it since I have it all typed up anyway. I'm hoping to continue to do more "proofs" along these lines. Even if they do turn out to be trivial truths.

As far as the compass and straight go though, it seems to me the Archimedes and Euclid have indeed provided us a way of "Squaring the Circle" using compass and straight edge. Our "error" concerning the precise length of $\displaystyle \pi$ isn't going to be any more than our "error" trying to construct a line that is exactly 1 unit long in terms of drawing it with a compass and straight edge. We're not going to be right on the money with that either if you really stop and think about it.

Re: Squaring the Circle. A proof for whatever its worth

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Originally Posted by

**Zeno** As far as the compass and straight go though, it seems to me the Archimedes and Euclid have indeed provided us a way of "Squaring the Circle" using compass and straight edge. Our "error" concerning the precise length of $\displaystyle \pi$ isn't going to be any more than our "error" trying to construct a line that is exactly 1 unit long in terms of drawing it with a compass and straight edge. We're not going to be right on the money with that either if you really stop and think about it.

You may be thinking of a construction that, by an infinite number of steps, converges to a square of the same area as a given circle. That would be more Archimedes than Euclid. But that, in any case would not be "squaring the circle" which must be done in a finite number of steps.

And your remark about "our "error" trying to construct a line that is exactly 1 unit long in terms of drawing it with a compass and straight edge" is completely off the point. None of this has anything to do with **measuring** a length. That's why it is required that we use a "straight edge" rather than a "rule". We **can** in fact draw a segment of **exactly** "1 unit". We do that by using the straight edge to draw a straight line, marking any two points on it and **declaring** that to be our unit length. We can then construct, using straight edge and compasses, construct segments of any integer length, any rational number length, and **some** irrational lengths such as $\displaystyle \sqrt{2}$.

It was Pierre Wanzel, in the late 19th century that showed that only some numbers are "constructible" (x is "constructible" if **given** a unit length segment we can, using compasses and straight edge, construct a segment of length x- and, again, this has nothing to do with actually **measuring** the length). Specifically, he showed that a number is "constructible" if and only if it is "algebraic of order a power of 2".

A number is said to be "algebraic of order n" if it satisfies a polynomial equation of degree n, with integer coefficients, but no such equation of lower degree. Any rational number, x, can be written in the form x= m/n and so satisfies nx- m= 0, a polynomial equation, of degree 1, with integer coefficients and so is "algebraic of order 1". Conversely if x satisfies an equation of the form ax+ b= 0 then x= -b/a and so is a rational number. That is, the rational number are exactly the numbers that are "algebraic of order 1".

$\displaystyle \sqrt{2}$ is not rational and so cannot be algebraic of order 1 but does satisfy $\displaystyle x^2- 2= 0$ and so is algebraic of order 2 (and it is easy to construct a segment of length $\displaystyle \sqrt{2}$). In general, numbers of the form $\displaystyle \sqrt[n]{p}$, where p is a prime number, is "algebraic of order n". (If p is not prime then $\displaystyle \sqrt[n]{p}$ may be algebraic of order less than n.)

$\displaystyle \sqrt[4]{2}$ is algebraic of order $\displaystyle 4= 2^2$ and so is also "constructible" but $\displaystyle \sqrt[3]{2}$ is algebraic of order 3, not a power of 2, and so is not "constructible". Given a circle, if we take its radius as our "unit" it has area $\displaystyle \pi$. A square with that same area would have each side of length $\displaystyle \sqrt{\pi}$. But $\displaystyle \pi$ itself is "transcendental" (not algebraic of any order) so $\displaystyle \sqrt{\pi}$ is certainly NOT algebraic of order a power of 2. There cannot be any construction, using only compasses and straight edge for "squaring the circle".

Re: Squaring the Circle. A proof for whatever its worth

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Originally Posted by

**HallsofIvy** We can then construct, using straight edge and compasses, construct segments of any integer length, any rational number length, and **some** irrational lengths such as $\displaystyle \sqrt{2}$.

I wasn't aware that we can construct the irrational length of $\displaystyle \sqrt{2}$. This is quite interesting. I'm going to have to learn how to make that construction for sure.

Quote:

Originally Posted by

**HallsofIvy** It was Pierre Wanzel, in the late 19th century that showed that only some numbers are "constructible" (x is "constructible" if **given** a unit length segment we can, using compasses and straight edge, construct a segment of length x- and, again, this has nothing to do with actually **measuring** the length). Specifically, he showed that a number is "constructible" if and only if it is "algebraic of order a power of 2".

I was totally unaware of this. I'll definitely be looking into the work of Pierre Wanzel, it sounds quite interesting.

Would this include $\displaystyle \sqrt{5}$, is that considered to be a "power of 2", because it's also a square root? Or does the number inside the radical need to be a power of 2?

The reason I ask about $\displaystyle \sqrt{5}$ is because this is used in the Golden Ratio which I hope to be working with in geometric constructions too.

Quote:

Originally Posted by

**HallsofIvy** A number is said to be "algebraic of order n" if it satisfies a polynomial equation of degree n, with integer coefficients, but no such equation of lower degree. Any rational number, x, can be written in the form x= m/n and so satisfies nx- m= 0, a polynomial equation with integer coefficients and so is "algebraic of order 1". Conversely of x satisfies an equation of the form ax+ b= 0 then x= -b/a and so is a rational number. That is, the rational number are exactly the numbers that are "algebraic of order 1".

Yes, I'm familiar with this definition of "algebraic", but I wasn't aware that an algebraic irrational number could be geometrically defined using Euclid's methods of geometry. So I'm excited about that result to be sure.

Quote:

Originally Posted by

**HallsofIvy** Given a circle, if we take its radius as our "unit" it has area $\displaystyle \pi$. A square with that same area would have each side of length $\displaystyle \sqrt{\pi}$. But $\displaystyle \pi$ itself is "transcendental" (not algebraic of any order) so $\displaystyle \sqrt{\pi}$ is certainly NOT algebraic of order a power of 2. There cannot be any construction, using only compasses and straight edge for "squaring the circle".

Ok, I'm aware that $\displaystyle \pi$ and $\displaystyle e$ are both transcendental numbers in terms of having been proven not to satisfy any polynomial equations where the coefficients are rational numbers. But I didn't realize how that spills over to geometric constructions.

So I've learned something new and valuable here. Good thing I posted my "proof".

I should call my proof, "Squaring the Circle Transcendentally" ;)

I've downloaded some proofs that $\displaystyle \pi$ and $\displaystyle e$ are both transcendental numbers but I haven't been able to understand them yet. It appears that they have to do with derivatives and even limits or infinite series or sums. I haven't fully understood them yet. I'd like to get a handle on those particular proofs.

Re: Squaring the Circle. A proof for whatever its worth

Quote:

Originally Posted by

**Zeno** I wasn't aware that we can construct the irrational length of $\displaystyle \sqrt{2}$. This is quite interesting. I'm going to have to learn how to make that construction for sure.

I was totally unaware of this. I'll definitely be looking into the work of Pierre Wanzel, it sounds quite interesting.

Would this include $\displaystyle \sqrt{5}$, is that considered to be a "power of 2", because it's also a square root? Or does the number inside the radical need to be a power of 2?

Yes, $\displaystyle \sqrt{5}$ is "algebraic of order 2"

Quote:

The reason I ask about $\displaystyle \sqrt{5}$ is because this is used in the Golden Ratio which I hope to be working with in geometric constructions too.

Yes, the golden ratio, $\displaystyle \phi= \frac{1+\sqrt{5}}{2}$, satisfies the equation $\displaystyle \phi^2- \phi- 1= 0$, a second degree equation with constant coefficients (and clearly is not rational) and so is algebraic of order 2.

Quote:

Yes, I'm familiar with this definition of "algebraic", but I wasn't aware that an algebraic irrational number could be geometrically defined using Euclid's methods of geometry. So I'm excited about that result to be sure.

Ok, I'm aware that $\displaystyle \pi$ and $\displaystyle e$ are both transcendental numbers in terms of having been proven not to satisfy any polynomial equations where the coefficients are rational numbers. But I didn't realize how that spills over to geometric constructions.

So I've learned something new and valuable here. Good thing I posted my "proof".

I should call my proof, "Squaring the Circle Transcendentally" ;)

I've downloaded some proofs that $\displaystyle \pi$ and $\displaystyle e$ are both transcendental numbers but I haven't been able to understand them yet. It appears that they have to do with derivatives and even limits or infinite series or sums. I haven't fully understood them yet. I'd like to get a handle on those particular proofs.