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