Mathematics
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.
Whole Numbers
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Answer
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.
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