Prove or disprove. The vertices of every regular tetrahedron in 3-space have at least two irrational coordinates.

Question

Prove or disprove. All four vertices of every regular tetrahedron in $ \mathbb{R}^3$ have at least two irrational coordinates.

This question arose from my inability to construct a tetrahedron in $\mathbb{R}^3$ with all the coordinates $$ M_x,M_y,M_z,N_x,N_y,N_z,P_x,P_y,P_z,Q_x,Q_y \;\mathrm {and}\; Q_z, $$ of its vertices $M$, $N$, $P$ and $Q$ being rational numbers.

Answer 1

You can place one of the vertices at the origin. A second one can be at $(1,0,0)$

Answer 2

I think you want to prove that there doesn't exist a regular tetrahedron such that all its vertices have only rational coordinates. Assume the contrary, let $ABCD$ be such a tetrahedron. Consider the plane $(ABC)$ as a complex plane with origin in $A$ and let the points $B$ and $C$ have associated complex numbers $a+bi$ and $c+di$. By our assumption, we would have that $a,b,c,d\in\mathbb{Q}$. We know that the points associated to complex numbers $z_1,z_2,z_3$ form an equilateral triangle if and only if $z_1+\varepsilon z_2+\varepsilon^2z_3=0$ where $\varepsilon=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ is the root of order 3 of the unity. By plugging our numbers, we obtain $$a+bi+\varepsilon(c+di)+\varepsilon^2\cdot 0=0\Rightarrow$$ $$\Rightarrow a+bi+\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)(c+di)=0\Rightarrow$$ $$\Rightarrow a+bi-\frac{c}{2}-\frac{d}{2}i+\frac{\sqrt{3}c}{2}i-\frac{\sqrt{3}d}{2}=0\Rightarrow$$ $$\Rightarrow a-\frac{c}{2}-\frac{\sqrt{3}d}{2}=0\Rightarrow d=0$$ also $$b-\frac{d}{2}+\frac{\sqrt{3}c}{2}=0\Rightarrow c=0$$ thus $A=C$, so we don't actually have a triangle $ABC$ hence the conclusion.

Answer 3

Choose four suitable vertices of the unit cube $[0,1]^3$, and you have a regular tetrahedron.

Answer 4

False.

Proof. Here is one possible solution:

• vertex 1: (1 0 0)
• vertex 2: (0 0 1)
• vertex 3: (0 1 0)
• vertex 4: (1 1 1)

You can check that:

• all coordinates of all vertices are rational;
• and that all distances from one vertex to another are equal (thus this is a regular tetrahedron).