Three rational numbers satisfy and . Show that all the rational numbers must be simultaneously zero.

Question

Homi · Accepted Answer

Understand the Question

We are given two conditions for rational numbers $x, y, z$ $x, y, z$ :
- $x + y + z = 0$
- $xy + yz + zx = 0$
We can relate these quantities using the algebraic expansion identity: $(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$
Since the square of any real/rational number is non-negative ( $a^2 \ge 0$ ), the only way a sum of squares can equal zero is if each individual number is zero.

Step 1 · Use the Algebraic Expansion Identity

Given $x + y + z = 0$ $xy + yz + zx = 0$

Using the algebraic identity $(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$

Substitute the given values into the identity

\begin{aligned} (0)^2 &= x^2 + y^2 + z^2 + 2(0) \\ 0 &= x^2 + y^2 + z^2 \\ x^2 + y^2 + z^2 &= 0 \end{aligned}

Step 2 · Analyze the Sum of Squares

For any rational numbers $x, y, z$ , the square of a number is always non-negative $x^2 \ge 0, \quad y^2 \ge 0, \quad z^2 \ge 0$

The sum of non-negative quantities is zero if and only if each individual term is equal to zero

\begin{aligned} x^2 = 0 &\implies x = 0 \\ y^2 = 0 &\implies y = 0 \\ z^2 = 0 &\implies z = 0 \end{aligned}

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

Answer

Hence proved, $x = y = z = 0$ .

Common Mistakes

Assuming Cancellation in Sum of Squares: For real/rational numbers, $x^2 + y^2 + z^2 = 0$ cannot be satisfied by positive and negative numbers canceling out, because $a^2 \ge 0$ for all real $a$ .
Overcomplicating the System: Trying to express $z = -(x+y)$ and substituting into the quadratic equation rather than directly applying the identity $(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx)$ .

Question 14