Geometry is a powerful source of beauty and inspiration, perhaps in part because of its instant accessibility. Using only a straight-edge and a compass, it's possible for anyone to harness geometry and create tremendous works of art and form. These aesthetics, symmetries, and constructions are so moving that people have built entire personal cosmologies around the concept of geometry as a sacred science. Throughout the past, as scientists and philosophers used the power of math to play with shape, certain figures grew in recognition as being especially relevant and foundational -- notably, the square and the circle.

It became naturally relevant to explore the relationship between these two important forms. Armed with a straight-edge we can tackle lines, and a compass gives grasp over angles and curves. These were the principal geometric tools of the ancients, and past curious minds wanted to know if, equipped with these, we could transform a circle into a square. Specifically, they wondered if given a circle with a particular area, is it possible to construct a square which has the exact same area? Circles and squares are such simple, commonplace figures -- it should be easy enough, right? But the question remained unresolved throughout thousands of years of humanity's mathematical development. Surprisingly, the answer came not from geometry but from an entirely different area of mathematics: number theory. Our eventual resolution to the millennia-old problem of squaring the circle turned out to be equivalent to whether or not one of nature's most prevalent and mysterious numbers, \(\pi\), is transcendental.

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What is a transcendental number?

There are many different classes of numbers, and when doing math we typically choose to work in the particular number realm which is relevant to the problem at hand. For instance, if the problem is counting the number of people who work at a company, we would only consider non-negative integers, or whole numbers. In this context it wouldn't make sense to talk about \(-2\) people or \(\frac{3}{4}\) of a person.

Mathematicians often choose to work within special kinds of number sets called fields\(^{(1)}\), and sometimes equations arise that cannot be solved by numbers in that particular field. For instance, within the realm of the rational numbers, \(x^{2} - 2 = 0\) is not solvable because \(\sqrt{2}\) is not a rational number -- in fact, no rational number exists which can be substituted for \(x\) to make the equation true. If we move instead to the field of the real numbers, \(x^{2} - 2 = 0\) does have a solution because \(\sqrt{2}\) is a real number, and so \(x = \sqrt{2}\) solves the equation (as does \(x = -\sqrt{2}\)). But the real numbers have their own limitations, as \(x^{2} + 1 = 0\) is unsolvable within the reals because \(\sqrt{-1}\) is not a real number. This equation becomes solvable, though, when you look at it from within the complex numbers, where \(x = \pm i\) is the solution. So the solvability of an equation depends both upon the equation and upon the number field within which you're working.

However, given a particular starting field, there are ways to extend that field in order to expand its capabilities. A field extension is a special type of construct that lets you take an existing field and add additional elements to it, expanding your pool of available potential solutions and allowing for a wider range of equations to be solvable. These new, added-in numbers fall into two categories within the context of the original field, depending on whether or not the new number is the solution to a special type of equation (in particular, whether or not the new number is the root of a polynomial whose coefficients are elements from the original field). If the new number is the solution to such an equation, then the number is said to be "algebraic" over the original field. If the number is not a solution to any such equation, then the number is "transcendental" over the field. Typically, the field we are considering is the field of rational numbers. So when we talk about a transcendental number, we're conventionally referring to a number that doesn't solve any polynomial with rational coefficients.

The proof that \(\pi\) is transcendental arose in the late 1800's as a direct consequence to another mathematical result: The Lindemann theorem. This theorem says that if \(a\) is an algebraic number, then \(e^{a}\) is a transcendental number. The proof that \(\pi\) is transcendental (and \(e\), as well) follows immediately from this.

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The transcendental nature of \(\pi\) over the field of rational numbers:

Proof by contradiction: Assume that \(\pi\) is algebraic.

Let's explore \(i\pi\).

\(i\) is algebraic over the rational numbers because it is a root of the polynomial \(x^{2} + 1\).

The set of algebraic numbers is itself a field, and so it is closed under multiplication (i.e., the product of any two algebraic numbers is also algebraic). So, \(i\pi\) is algebraic.

Therefore, by the Lindemann theorem, \(e^{i\pi}\) is transcendental.

But, Euler's identity, considered to be one of the most beautiful and moving equations within mathematics, tells us that \(e^{i\pi} + 1 = 0\), or, equivalently: \(e^{i\pi} = -1\).

\(-1\) is clearly algebraic over the rationals, as \(-1\) is itself a rational number and a root of the polynomial \(x + 1\).

So \(e^{i\pi}\) is algebraic.

A number cannot be both algebraic and transcendental, so we've reached a contradiction.

Therefore, our assumption that \(\pi\) is algebraic must be false. Hence, \(\pi\) is a transcendental number.

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So what does this have to do with squaring a circle?

Let's consider a simple example: a circle whose radius measures \(1\) (also known as the unit circle). The area of a circle is \(\pi r^{2}\), where \(r\) is the length of the circle's radius, so the area of the unit circle is \(\pi 1^{2} = \pi 1 = \pi\). Now, we want to make a square whose area is also equal to \(\pi\). The area of a square is computed by squaring the length of any of its sides. So if each side of the square has length, say, \(s\), then the area of the square is \(s^{2}\). Therefore, a square whose area is \(s^{2} = \pi\) will have sides whose lengths are each \(s = \sqrt{\pi}\).

For a line with length \(l\) to be drawable using a straight-edge and a compass, \(l\) must be a constructible number\(^{(2)}\), and constructible numbers are always algebraic. The transcendental nature of \(\pi\) also makes \(\sqrt{\pi}\) transcendental, and so the lines of length \(\sqrt{\pi}\) that we would need in order to create a square whose area is equal to \(\pi\) simply can't be drawn. Thus, there is no way to draw a square whose area is \(\pi\) (or any multiple of \(\pi\), for that matter). Since \(\pi\) is an inherent component in the area of all circles, no circle can be drawn which may be manipulated to generate a square\(^{(3)}\). We cannot, therefore, square the circle.

- - - - - - - - - -

There are endless connections between the array of seemingly separate disciplines within mathematics. Geometry and algebra each have their own set of rules and formulations, but many problems can be solved from within either arena. The proofs look different and reference distinct concepts and theorems particular to each discipline, but somewhere deep down the truths are the same. What is fundamental? We see that the same problem can take on any number of forms, depending upon the realm of mathematics from within which the problem is approached. Similarly, the solvability of an equation depends not only upon the equation itself but also upon the field within which the equation is being evaluated. Is there a limit to how far we can push our field extensions? Is there a highest level field which encompasses them all and which contains every solution? Is there something which is ultimate and all-encompassing? Or do we find an infinite nesting of individually limited fields?

In overcoming life's challenges, it's important to consider not only the nature of the issue but also the resources that are available to you. Your resources are the field within which you're solving your problem. Examine what you're working with. Do you need to find ways to extend your resource field to include new solutions? From where can you source these additional resources?

Squares and circles are such simple forms. It seems counterintuitive that we should be so limited in our ability to manipulate their constructions. What is complexity? Does it arise within the nature of a thing or is it woven by the interactions of that thing with the rest of the universe? How are complexity and connection related?

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(1) A field is a special status given to a set of numbers which exhibits particular behaviors. Familiar examples of fields are the rational numbers, the real numbers, and the complex numbers, but there are many other fields such as \(Z_{5}\) (the set of integers modulo \(5\)) or the quotient ring \(Q[x]/(x^{2}+1)\).

(2) Algebraically, these are numbers which can be written as a finite combination of sums, differences, multiplications, divisions, and square roots of integers.

(3) Note that we could try to circumvent the roadblock by setting the radius of our circle to something like \(\frac{1}{\sqrt{\pi}}\), so that our area works out to be \(1\). We could then draw a square whose sides have length \(1\) and whose area is therefore equal to \(1\). However, now the circle becomes impossible to draw because we would need to generate it by constructing a radius whose length is \(\frac{1}{\sqrt{\pi}}\), which, as noted, we simply cannot draw.

It became naturally relevant to explore the relationship between these two important forms. Armed with a straight-edge we can tackle lines, and a compass gives grasp over angles and curves. These were the principal geometric tools of the ancients, and past curious minds wanted to know if, equipped with these, we could transform a circle into a square. Specifically, they wondered if given a circle with a particular area, is it possible to construct a square which has the exact same area? Circles and squares are such simple, commonplace figures -- it should be easy enough, right? But the question remained unresolved throughout thousands of years of humanity's mathematical development. Surprisingly, the answer came not from geometry but from an entirely different area of mathematics: number theory. Our eventual resolution to the millennia-old problem of squaring the circle turned out to be equivalent to whether or not one of nature's most prevalent and mysterious numbers, \(\pi\), is transcendental.

- - - - - - - - - -

What is a transcendental number?

There are many different classes of numbers, and when doing math we typically choose to work in the particular number realm which is relevant to the problem at hand. For instance, if the problem is counting the number of people who work at a company, we would only consider non-negative integers, or whole numbers. In this context it wouldn't make sense to talk about \(-2\) people or \(\frac{3}{4}\) of a person.

Mathematicians often choose to work within special kinds of number sets called fields\(^{(1)}\), and sometimes equations arise that cannot be solved by numbers in that particular field. For instance, within the realm of the rational numbers, \(x^{2} - 2 = 0\) is not solvable because \(\sqrt{2}\) is not a rational number -- in fact, no rational number exists which can be substituted for \(x\) to make the equation true. If we move instead to the field of the real numbers, \(x^{2} - 2 = 0\) does have a solution because \(\sqrt{2}\) is a real number, and so \(x = \sqrt{2}\) solves the equation (as does \(x = -\sqrt{2}\)). But the real numbers have their own limitations, as \(x^{2} + 1 = 0\) is unsolvable within the reals because \(\sqrt{-1}\) is not a real number. This equation becomes solvable, though, when you look at it from within the complex numbers, where \(x = \pm i\) is the solution. So the solvability of an equation depends both upon the equation and upon the number field within which you're working.

However, given a particular starting field, there are ways to extend that field in order to expand its capabilities. A field extension is a special type of construct that lets you take an existing field and add additional elements to it, expanding your pool of available potential solutions and allowing for a wider range of equations to be solvable. These new, added-in numbers fall into two categories within the context of the original field, depending on whether or not the new number is the solution to a special type of equation (in particular, whether or not the new number is the root of a polynomial whose coefficients are elements from the original field). If the new number is the solution to such an equation, then the number is said to be "algebraic" over the original field. If the number is not a solution to any such equation, then the number is "transcendental" over the field. Typically, the field we are considering is the field of rational numbers. So when we talk about a transcendental number, we're conventionally referring to a number that doesn't solve any polynomial with rational coefficients.

The proof that \(\pi\) is transcendental arose in the late 1800's as a direct consequence to another mathematical result: The Lindemann theorem. This theorem says that if \(a\) is an algebraic number, then \(e^{a}\) is a transcendental number. The proof that \(\pi\) is transcendental (and \(e\), as well) follows immediately from this.

- - - - - - - - - -

The transcendental nature of \(\pi\) over the field of rational numbers:

Proof by contradiction: Assume that \(\pi\) is algebraic.

Let's explore \(i\pi\).

\(i\) is algebraic over the rational numbers because it is a root of the polynomial \(x^{2} + 1\).

The set of algebraic numbers is itself a field, and so it is closed under multiplication (i.e., the product of any two algebraic numbers is also algebraic). So, \(i\pi\) is algebraic.

Therefore, by the Lindemann theorem, \(e^{i\pi}\) is transcendental.

But, Euler's identity, considered to be one of the most beautiful and moving equations within mathematics, tells us that \(e^{i\pi} + 1 = 0\), or, equivalently: \(e^{i\pi} = -1\).

\(-1\) is clearly algebraic over the rationals, as \(-1\) is itself a rational number and a root of the polynomial \(x + 1\).

So \(e^{i\pi}\) is algebraic.

A number cannot be both algebraic and transcendental, so we've reached a contradiction.

Therefore, our assumption that \(\pi\) is algebraic must be false. Hence, \(\pi\) is a transcendental number.

- - - - - - - - - -

So what does this have to do with squaring a circle?

Let's consider a simple example: a circle whose radius measures \(1\) (also known as the unit circle). The area of a circle is \(\pi r^{2}\), where \(r\) is the length of the circle's radius, so the area of the unit circle is \(\pi 1^{2} = \pi 1 = \pi\). Now, we want to make a square whose area is also equal to \(\pi\). The area of a square is computed by squaring the length of any of its sides. So if each side of the square has length, say, \(s\), then the area of the square is \(s^{2}\). Therefore, a square whose area is \(s^{2} = \pi\) will have sides whose lengths are each \(s = \sqrt{\pi}\).

For a line with length \(l\) to be drawable using a straight-edge and a compass, \(l\) must be a constructible number\(^{(2)}\), and constructible numbers are always algebraic. The transcendental nature of \(\pi\) also makes \(\sqrt{\pi}\) transcendental, and so the lines of length \(\sqrt{\pi}\) that we would need in order to create a square whose area is equal to \(\pi\) simply can't be drawn. Thus, there is no way to draw a square whose area is \(\pi\) (or any multiple of \(\pi\), for that matter). Since \(\pi\) is an inherent component in the area of all circles, no circle can be drawn which may be manipulated to generate a square\(^{(3)}\). We cannot, therefore, square the circle.

- - - - - - - - - -

There are endless connections between the array of seemingly separate disciplines within mathematics. Geometry and algebra each have their own set of rules and formulations, but many problems can be solved from within either arena. The proofs look different and reference distinct concepts and theorems particular to each discipline, but somewhere deep down the truths are the same. What is fundamental? We see that the same problem can take on any number of forms, depending upon the realm of mathematics from within which the problem is approached. Similarly, the solvability of an equation depends not only upon the equation itself but also upon the field within which the equation is being evaluated. Is there a limit to how far we can push our field extensions? Is there a highest level field which encompasses them all and which contains every solution? Is there something which is ultimate and all-encompassing? Or do we find an infinite nesting of individually limited fields?

In overcoming life's challenges, it's important to consider not only the nature of the issue but also the resources that are available to you. Your resources are the field within which you're solving your problem. Examine what you're working with. Do you need to find ways to extend your resource field to include new solutions? From where can you source these additional resources?

Squares and circles are such simple forms. It seems counterintuitive that we should be so limited in our ability to manipulate their constructions. What is complexity? Does it arise within the nature of a thing or is it woven by the interactions of that thing with the rest of the universe? How are complexity and connection related?

- - - - - - - - - -

(1) A field is a special status given to a set of numbers which exhibits particular behaviors. Familiar examples of fields are the rational numbers, the real numbers, and the complex numbers, but there are many other fields such as \(Z_{5}\) (the set of integers modulo \(5\)) or the quotient ring \(Q[x]/(x^{2}+1)\).

(2) Algebraically, these are numbers which can be written as a finite combination of sums, differences, multiplications, divisions, and square roots of integers.

(3) Note that we could try to circumvent the roadblock by setting the radius of our circle to something like \(\frac{1}{\sqrt{\pi}}\), so that our area works out to be \(1\). We could then draw a square whose sides have length \(1\) and whose area is therefore equal to \(1\). However, now the circle becomes impossible to draw because we would need to generate it by constructing a radius whose length is \(\frac{1}{\sqrt{\pi}}\), which, as noted, we simply cannot draw.