Three rational numbers x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that all the rational numbers x, y, z must be simultaneously zero.

Given,

⇒ x + y + z = 0 …(i)

⇒ xy + yz + zx = 0 …(ii)

We use the algebraic identity :

⇒ (x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

Substituting the value of (i) and (ii) in above equation:

⇒ 0² = x² + y² + z² + 2(0)

⇒ 0 = x² + y² + z²

⇒ x² + y² + z² = 0.

Since the square of any rational number is non-negative (i.e., x² ≥ 0, y² ≥ 0, z² ≥ 0), the only way their sum can be zero is when each square is zero.

⇒ x² = 0, y² = 0, z² = 0.

⇒ x = 0, y = 0, z = 0.

Hence, all the rational numbers x, y, z must be simultaneously zero.