# Prove at least one of the length, width, height of the box must be rational.

Assume there is a big box be combined by finite small boxes and that the small boxes are not necessarily be the same.

The statement in my note is " If there is at least one of length, width, height of each small box is rational, then at least one of the length, width, height of the big box must be rational. "

If all small boxes are the same, then there is nothing be proved.

I wonder this question can be proved by contradiction. That is, assume all of the length, width, height of the box are irrational. But how to prove there is at least one of the small box having irrational length, width, and height？

• sum of two positive irrational numbers cannot be a negative rational number to offset one rational dimension of a small box – Vasya Dec 19 '18 at 14:29
• In two dimensions the problem is well known. I suggest trying to adapt the arguments in that case. – lulu Dec 19 '18 at 14:33
• here is an article containing $14$ proofs in the planar case. – lulu Dec 19 '18 at 14:34

The key is to notice that an arbitrary box $$B$$ aligned with the x-y-z axes in space has the property that one of its dimensions is rational if and only if there exists an integer $$q$$ such that $$\int\int\int_B \sin(2\pi qx)\sin(2\pi qy)\sin(2\pi qz)dxdydz=0\tag{1}$$

This equivalence is not difficult to prove. Indeed, if $$B$$ has one side of rational length $$\frac p q$$ (say, along the $$x$$ axis), then that triple integral is the product of three integrals. One of these integrals is $$\int_0^{\frac p q}\sin(2\pi qx)dx=0$$ Conversely, suppose the triple integral is $$0$$, then at least one of the single integrals that forms its product must be $$0$$. Assume it's true for the one along the $$x$$ axis. This implies that the domain is integration must be an interval whose length is a multiple of the period of $$\sin(2\pi qx)$$, so it's a rational number.

With that equivalence proven, Let $$B$$ be the large box, and $$\{B_i\}_{1\leq i \leq n}$$ be the small boxes. So all we have to do is to show that $$B$$ satisfies property $$(1)$$ for some integer $$q$$.

We know that each box $$B_i$$ has at least one dimension that's a rational number $$\frac {p_i}{q_i}$$.

Let $$q=\prod_{1\leq i \leq n}q_i$$ We can prove that property $$(1)$$ holds for that value of $$q$$.

The triple integral over $$B$$ can be decomposed as the sum of integrals over the small boxes $$\int\int\int_B \sin(2\pi qx)\sin(2\pi qy)\sin(2\pi qz)dxdydz=\sum_i \int\int\int_{B_i} \sin(2\pi qx)\sin(2\pi qy)\sin(2\pi qz)dxdydz$$

Since property $$(1)$$ is true with $$q$$ for each small box, each integral in the sum is $$0$$. This implies the integral over $$B$$ is also $$0$$, which implies the desired property.