$1 \cdot 60$ Given that $a, b$ and $\sqrt{a}+\sqrt{b}$ are all rational numbers, then we can conclude
(A) $\sqrt{a}, \sqrt{b}$ are both rational numbers.
(B) $\sqrt{a}, \sqrt{b}$ are both irrational numbers.
(C) $\sqrt{a}, \sqrt{b}$ can be both rational numbers, or both irrational numbers.
(D) One of $\sqrt{a}, \sqrt{b}$ is a rational number, and the other is an irrational number.
（1st "Five Sheep Cup" Junior High School Mathematics Competition, 1989)
[Solution 1] Given that $\sqrt{a}+\sqrt{b}$ is a rational number, then by
$$
\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}},
$$

we know that $\sqrt{a}-\sqrt{b}$ is a rational number.
And $\sqrt{a}=\frac{1}{2}[(\sqrt{a}+\sqrt{b})+(\sqrt{a}-\sqrt{b})]$ is also a rational number.
Similarly, $\sqrt{b}$ is a rational number.
Therefore, the answer is (A).
[Solution 2] If $a, b, \sqrt{a}+\sqrt{b}$ are all rational numbers, given that $\sqrt{a} \geqslant 0, \sqrt{b} \geqslant 0$, we know that $\sqrt{a}, \sqrt{b}$ are both rational numbers.

Otherwise, if $\sqrt{a}, \sqrt{b}$ are both irrational numbers, noting that $a, b$ are rational numbers, then $\sqrt{a}+\sqrt{b}$ must be an irrational number;
if not, suppose $\sqrt{a}+\sqrt{b}=r$ is a rational number, then
$\sqrt{a}=r-\sqrt{b}$, i.e., $a=r^{2}+b-2 r \sqrt{b}$,
since $a, b, r$ are all rational numbers, the above equation cannot hold, noting that $\sqrt{b}$ is irrational. If one of $\sqrt{a}, \sqrt{b}$ is rational and the other is irrational, then $\sqrt{a}+\sqrt{b}$ is also irrational. Therefore, the answer is (A).