# Are the area of a circle inscribed in a square and the area of the “spandrels” (the four corners that remain) commensurable?

And how would you demonstrate that most simply?

See the beginning of my blog post for a little more: http://seekecho.blogspot.fr/2013/02/different-ilks.html

• You should include the definition of the spandrel in your question. – coffeemath May 16 '13 at 11:08

So you want to show that the area of the circle, i.e. $\pi$, is not commensurable with the area of “square minus circle”, i.e. $4-\pi$, right? Two things are commensurable if their fraction is a rational number. Now look at it this way:

$$\frac{4-\pi}\pi = \frac4\pi-1 = r\not\in\mathbb Q$$

Suppose $r$ were a rational number, then all the following would be rational numbers as well:

\begin{align*} \frac4\pi &= r + 1 \\ \frac1\pi &= \frac{r+1}4 \\ \pi &= \frac4{r+1} \end{align*}

But as you know that $\pi$ is irrational, you know that $r$ cannot be rational.

Given two incommensurable numbers $a$ and $b$, the set $\{pa+qb\;\vert\;p,q\in\mathbb Q\}$ can be interpreted as a vectorspace over $\mathbb Q$ with dimension two. You can check all the vector space axioms to see that this is true. The switch from “circle and square” to “circle and spandrel” is simply a change in basis vectors, but does not make the vectors linearily dependent. So as a general conclusion, two numbers $p_1a+q_1b$ and $p_2a+q_2b$ will still be incommensurable unless they are linearily dependent in the vector space sense, i.e. unless

$$\begin{vmatrix} p_1 & p_2 \\ q_1 & q_2 \end{vmatrix} = p_1q_2 - p_2q_1 = 0$$

• Thanks for that, MvG. The first bit explains it beautifully - embarrassingly - simply. The second part I'll have to research to understand... – Simon G Jun 25 '13 at 6:16
• @SimonG: If you look at the Wikipedia lemma for vector space, you will find “A field extension over the rationals ℚ can be thought of as a vector space over ℚ” (link added by me) which expresses this concept. More info in the surrounding text. In your case you have the field extension $\mathbb Q(\pi)\subsetneq\mathbb R$ obtained via adjunction. Also look up “Linear independence”, “Basis (linear algebra)” and “Dimension”. – MvG Jun 25 '13 at 8:38