[Solution] If $a, b$ are two irrational numbers that are opposites of each other, then (1) does not hold.
If $k<0$, then (2) does not hold.
If $a \neq 1, a \neq 0 ; \quad b$ is a rational number not equal to 0, and greater than -1 and less than 1, such as $a=2, b=\frac{1}{2}$, then (3) does not hold.
If $x<0, y<0$, then (4) does not hold.
According to the uniqueness of the choice, (5) is correct.
Therefore, the answer is $(A)$.
$6 \cdot 26$ Given five propositions:
(1) If $a, b, a-b$ are all irrational numbers, then $a+b$ is also an irrational number.
(2) If $k$ is an integer, then $2^{k}$ is also an integer.
(3) If $a, b$ are both rational numbers, then $a^{b}$ is also a rational number.
(4) For real numbers $x, y$, if $\lg (x y)$ is defined, then $\lg (x y)=\lg x+\lg y$.
(5) If $a>b>0$, then $a^{2}>b^{2}$.

The number of correct propositions among the above is
(A) 1.
(B) 2.
(C) 3.
(D) 4.
("Zu Chongzhi Cup" Junior High School Mathematics Invitational Competition, 1988)