Squaring the circle with materials and tools
In 10th grade, my Algebra II teacher went on a short tangent, introducing the class to a famous math problem: squaring the circle. It’s been about five years since then and the problem appeared again in my abstract algebra textbook. Here I will chat about the problem and sketch an accessible proof of its solution.
The problem: given a circle, and using only a compass and straightedge, construct a square with precisely the same area as the circle. (Note, our straightedge is not graduated like a ruler!) This seems easy enough. After all, we have simple equations for the area of a square and circle. If \(s\) is a square’s side length, and \(r\) is a circle’s radius:
The natural step is to set these equal to each other and find a relation between the circle’s radius and the square’s side length:
So far so good. Now all we need to do is draw a square with side lengths \(\sqrt{\pi}r\). To simplify, we will assume the radius is 1, which reduces our problem into needing to draw a square with side lengths \(\sqrt{\pi}\). This ends up being an issue because \(\sqrt{\pi}\) is not “constructible” using a straightedge and compass.
To be constructible means that an object can be built using only provided “materials” and “tools”. If your materials are the numbers 2 and 3, and your only tool is multiplication, what are the numbers you can construct? There are lots: 2, 3, 6, 9, 12, and more. But there are also numbers you can’t construct. For example, try to make 1, 5, or 14.
In our case, the only material is the circle’s radius, 1. Our tools are straightedge and compass. With given materials and tools, it might be possible to create new or different tools out of what you already have. We shall do this to translate our geometric tools (like drawing lines or circles) into algebraic tools (like adding or multiplying). A fun puzzle that we won’t go into here is seeing that with straightedge and compass, we can construct the algebraic tools of addition, subtraction, multiplication, division, and square roots. Quite the arsenal, be can we construct \(\sqrt{\pi}\)?
Recall what our materials were: just the circle’s radius, 1. That doesn’t seem like much, but let’s see what we can construct with our tools:
 We can add or subtract 1 over and over to get any integer \((…3, 2, 1, 0, 1, 2, 3…)\).
 Then we can divide or multiply to get any fraction/rational number, like \(\frac{1}{2}, \frac{42}{71}, \frac{13}{5}…\)
 Square roots give us square roots of rationals, but if we take the square root of a square root we get 4th roots, 8th roots… In fact we have all “power of two” roots of rational numbers.
It’s still not clear if \(\sqrt{\pi}\) is constructible or not, so I present a strategy: if we add more tools and find that \(\sqrt{\pi}\) isn’t constructible, then it is still not constructible when we take the extra tools away.
Our extra tools shall be all the positive integer roots(cube, 4th, 5th roots…). We are no longer just restricted to “power of two” roots of rationals! You can convince yourself that with these extra tools, we can construct exactly the roots of polynomials. A number is called algebraic if it is the root of a polynomial. For example, \(2\) is algebraic since for \(f(x) = x + 2\), we have \(f(2) = 0\). \(\sqrt[3]{2}\) is algebraic since for \(g(x) = x^3  2\), we have \(g(\sqrt[3]{2}) = 0\).
With some effort it can be seen that \(\pi\) isn’t algebraic. Such numbers are called transcendental.
This means that \(\sqrt{\pi}\) also isn’t algebraic. In other words, it’s not constructible with our extra tools, hence it’s not constructible with our basic tools of straightedge and compass.
Extra thoughts

In the wonderful show “Adventure Time”, Finn the human is known for his strange exclamations. There are several instances where he yells the term “algebraic!”. I wonder if anyone has told him the definition.

Here is a Numberphile video on the same topic.

My algebra textbook didn’t present a proof that \(\pi\) was transcendental either. Their proof that the number isn’t constructible used that fact to show that \(\pi\) resides in an infinite degree field extension of \(\mathbb{Q}\), while we have at best a finite degree extension.

You can certainly construct an approximate square.