We are given that $x, y, z$ are rational numbers satisfying $x + y + z = 0$ and $xy + yz + zx = 0$. We must show that $x = y = z = 0$.

Because rational numbers are closed under addition and multiplication, and follow the laws of commutativity and distributivity, we can expand the square of their sum as follows:

$$(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$$

Now, we substitute the given conditions, $x + y + z = 0$ and $xy + yz + zx = 0$, into the expanded equation:

$$(0)^2 = x^2 + y^2 + z^2 + 2(0)$$ $$0 = x^2 + y^2 + z^2$$

For any rational number, its square is always non-negative (i.e., $x^2 \geq 0$, $y^2 \geq 0$, $z^2 \geq 0$). The sum of non-negative numbers equals zero only if each individual term is zero.

Therefore, we must have $x^2 = 0$, $y^2 = 0$, and $z^2 = 0$. This directly implies that $x = 0$, $y = 0$, and $z = 0$. Hence, all the rational numbers $x, y, z$ must be simultaneously zero.