33.49 Let $x, y$ be positive integers, where $y$ is not a perfect square, and $y < x$. Now, let $a, b$ represent the fractional parts of $x+\sqrt{y}$ and $x-\sqrt{y}$, respectively. Then the value of $a+b$ is
(A) an irrational number.
(B) a proper fraction.
(C) 1.
(D) indeterminate.
(Anhui Province Hefei City Junior High School Mathematics Competition, 1985)
[Solution] Suppose $y < x$, then $\alpha < \sqrt{y} < \alpha + 1 \leqslant x$, where $\alpha$ is a positive integer.
Thus, $x + \sqrt{y} = x + \alpha + (\sqrt{y} - \alpha)$, where $0 < \sqrt{y} - \alpha < 1$, so we know $a = \sqrt{y} - \alpha$.
Also, $x - \sqrt{y} = x - (\alpha + 1) + [(\alpha + 1) - \sqrt{y}]$, where $0 < \alpha + 1 - \sqrt{y} < 1$,
$$
\begin{array}{lc}
\text { hence } & b = \alpha + 1 - \sqrt{y} . \\
\therefore & a + b = (\sqrt{y} - \alpha) + (\alpha + 1 - \sqrt{y}) = 1 .
\end{array}
$$

Therefore, the answer is $(C)$.