$1 \cdot 126$ Using $1-y$ as an approximation for $\frac{1}{1+y}$, $|y|<1$, the relative error is
(A) $y$.
(B) $y^{2}$.
(C) $\frac{1}{1+y}$.
(D) $\frac{y}{1+y}$.
(E) $\frac{y^{2}}{1+y}$.
(16th American High School Mathematics Examination, 1965)
［Solution］The relative error is
$$
\frac{\frac{1}{1+y}-(1-y)}{\frac{1}{1+y}}=-\frac{\frac{y^{2}}{1+y}}{\frac{1}{1+y}}=y^{2} \text {. }
$$

Therefore, the correct choice is $(B)$.