2.25 To satisfy the equation $\frac{a+b}{a}=\frac{b}{a+b}$, then $a$ and $b$ must
(A) both be rational numbers.
(B) both be real numbers but not rational numbers.
(C) not both be real numbers.
(D) one be real, the other not real.
(E) one real, one not real, or both not real.
(11th American High School Mathematics Examination, 1960)
[Solution] From $\frac{a+b}{a}=\frac{b}{a+b}$, we have $a^{2}+a b+b^{2}=0$. Or $\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1=0$, thus $\frac{a}{b}=\omega$ or $\omega^{2}$, where $\omega=\frac{-1+\sqrt{3} i}{2}$. Therefore, the answer is $(E)$.