Incommensurability of the side and the altitude of an equilateral triangle

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Prove that in an equilateral triangle the side and the altitude are incommensurable magnitudes, that is, their ratio cannot be expressed by a rational number.

Solution

Let $\ell$ be the side of the equilateral triangle and $h$ its altitude. As seen, the altitude falls on the midpoint of the base and, by the Pythagorean theorem, $h^2=\ell^2-\left(\frac{\ell}{2}\right)^2=\frac{3\ell^2}{4}\quad\Longrightarrow\quad h=\frac{\sqrt3}{2}\,\ell.$ The ratio of altitude to side is therefore $\frac{h}{\ell}=\frac{\sqrt3}{2}.$ Two magnitudes are commensurable if and only if their ratio is a rational number. Suppose, for contradiction, that $h$ and $\ell$ are commensurable: then $\dfrac{\sqrt3}{2}$ would be rational and, consequently, so would $\sqrt3=2\cdot\dfrac{\sqrt3}{2}$.

Let us show that $\sqrt3$ is irrational. If $\sqrt3=\dfrac{p}{q}$ with $p,q$ coprime integers and $q\neq 0$, then squaring gives $p^2=3q^2$. Hence $p^2$ is a multiple of $3$, and since $3$ is prime, $p$ too is a multiple of $3$: write $p=3k$. Substituting, $9k^2=3q^2$, that is $q^2=3k^2$, so $q$ is also a multiple of $3$. But then $p$ and $q$ would share the factor $3$, contradicting that they are coprime. The contradiction proves that $\sqrt3$ is not rational.

Consequently $\dfrac{h}{\ell}=\dfrac{\sqrt3}{2}$ is not rational: the side and the altitude of the equilateral triangle are incommensurable. $\;\blacksquare$