$13 \cdot 1$ Among the following numbers, the smallest positive number is
(A) $10-3 \sqrt{11}$.
(B) $3 \sqrt{11}-10$.
(C) $18-5 \sqrt{13}$.
(D) $51-10 \sqrt{26}$.
(E) $10 \sqrt{26}-51$.
(34th American High School Mathematics Examination, 1983)
[Solution] Note the following fact: if $m>n>0$, then $\sqrt{m}>\sqrt{n}>0$. From $10^{2}=100>99=(3 \sqrt{11})^{2}, \quad 18^{2}=324>2600=(10 \sqrt{26})^{2}$,
so among these five differences, only $(A)$ and $(D)$ are positive.
Also, by $a-b=\frac{a^{2}-b^{2}}{a+b}$, we have $10-3 \sqrt{11}=\frac{1}{10+3 \sqrt{11}} \approx \frac{1}{20}$.
And $51-10 \sqrt{26}=\frac{1}{51+10 \sqrt{26}} \approx \frac{1}{102}$.
Clearly, $10-3 \sqrt{11}>51-10 \sqrt{26}$, so the answer is $(D)$.