Problem 1218

9.5. Proof 1: If all the numbers are rational, the proposition naturally holds.

Now suppose that among the 10 numbers, there is an irrational number $a$, then the other numbers are of the form $p-a$ or $\frac{p}{a}$, where $p$ is a rational number.
We will prove that there are no more than 2 numbers of the form $p-a$.
Indeed, if there are 3 different numbers of this form, let's assume $b_{1}=p_{1}-a, b_{2}=p_{2}-a, b_{3}=p_{3}-a$. Then, it is easy to see that $b_{1}+b_{2}=p_{1}+\dot{p}_{2}-2 a$ is not a rational number. Therefore, $b_{1} b_{2}=p_{1} p_{2}-a\left(p_{1}+p_{2}\right)+a^{2}$ should be a rational number.
Similarly, $b_{2} b_{3}$ and $b_{1} b_{3}$ are also rational numbers.
This means,
$$\begin{array}{l} A_{3}=a^{2}-a\left(p_{1}+p_{2}\right), \\ A_{2}=a^{2}-a\left(p_{1}+p_{3}\right), \\ A_{1}=a^{2}-a\left(p_{2}+p_{3}\right) \end{array}$$

are all rational numbers, thus, $A_{3}-A_{2}=a\left(p_{3}-p_{2}\right)$ is a rational number. This is only possible when $p_{3}-p_{2}=0$. Therefore, $b_{2}=b_{3}$, which is a contradiction.
From the above, we know that there are more than 2 numbers of the form $\frac{p}{a}$.
Let $c_{1}=\frac{p_{1}}{a}, c_{2}=\frac{p_{2}}{a}, c_{3}=\frac{p_{3}}{a}$ be 3 such numbers. Clearly, only when $p_{1}+p_{2}=0$, $c_{1}+c_{2}=\frac{p_{1}+p_{2}}{a}$ can be a rational number, and $p_{1} \neq p_{3}$, so $c_{3}+c_{2}=\frac{p_{3}+p_{2}}{a}$ is an irrational number. Therefore, $c_{3} c_{2}=\frac{p_{3} p_{2}}{a^{2}}$ is a rational number, which implies that $a^{2}$ is a rational number.
Proof 2: Consider any 6 of the numbers.
Construct a graph, placing these 6 numbers on its 6 vertices. If the sum of any 2 numbers is a rational number, connect the corresponding 2 vertices with a blue edge; if the product of any 2 numbers is a rational number, connect the corresponding 2 vertices with a red edge. It is well known that in such a graph, there exists a triangle with all three sides of the same color. We will discuss the various cases.
(1) If there is a blue triangle, then there exist 3 numbers $x, y, z$ such that $x+y, y+z, z+x$ are all rational numbers. Therefore,
$$(x+y)+(z+x)-(y+z)=2 x$$

is a rational number, i.e., $x$ is a rational number.
Similarly, $y, z$ are also rational numbers.
Now consider any other number $t$. Clearly, whether by the rationality of $x t$ (the problem states that all numbers are non-zero), or by the rationality of $x+t$, it can be deduced that $t$ is a rational number.
Therefore, in this case, all 10 numbers are rational.
(2) If there is a red triangle, then there exist 3 numbers $x, y, z$ such that $x y, y z, z x$ are all rational numbers. Therefore, $\frac{(x y)(z x)}{y z}=x^{2}$ is a rational number.
Similarly, $y^{2}, z^{2}$ are also rational numbers.
If at least one of $x, y, z$ is a rational number, then by discussing the previous case, it can be concluded that all 10 numbers are rational.
Now assume $x=m \sqrt{a}$, where $a$ is a rational number, and $m= \pm 1$.
Since $x y=m \sqrt{a} y=b$ is a rational number, we have,
$$y=\frac{b}{m \sqrt{a}}=\frac{b \sqrt{a}}{m a}=c \sqrt{a},$$

where $c \neq m$ is a rational number.
Now consider any other number $t$. If $x t$ or $y t$ is a rational number, then after a similar discussion, it can be concluded that $t=d \sqrt{a}$, where $d$ is a rational number. Therefore, $t^{2}$ is a rational number.
If $x+t, y+t$ are both rational numbers, then
$$(x+t)-(y+t)$$

is a rational number. But in fact,
$$(x+t)-(y+t)=(m-c) \sqrt{a}$$

is an irrational number, which is a contradiction.
In summary, either each number is a rational number, or the square of each number is a rational number, which is exactly what we need to prove.