How many real numbers $a \in (1,9)$ such that the  corresponding number $a- \frac1a$ is an integer?

(A): $0$, (B): $1$, (C): $8$, (D): $9$, (E) None of the above.
1. Let \( k = a - \frac{1}{a} \). We need to find the values of \( a \) such that \( k \) is an integer and \( a \in (1, 9) \).

2. Rewrite the equation:
   \[
   k = a - \frac{1}{a} \implies k = \frac{a^2 - 1}{a}
   \]
   This implies:
   \[
   a^2 - ka - 1 = 0
   \]

3. Solve the quadratic equation for \( a \):
   \[
   a = \frac{k \pm \sqrt{k^2 + 4}}{2}
   \]
   Since \( a > 0 \), we take the positive root:
   \[
   a = \frac{k + \sqrt{k^2 + 4}}{2}
   \]

4. We need \( a \in (1, 9) \). Therefore:
   \[
   1 < \frac{k + \sqrt{k^2 + 4}}{2} < 9
   \]

5. Multiply the inequality by 2:
   \[
   2 < k + \sqrt{k^2 + 4} < 18
   \]

6. Subtract \( k \) from all parts of the inequality:
   \[
   2 - k < \sqrt{k^2 + 4} < 18 - k
   \]

7. Consider the left part of the inequality:
   \[
   2 - k < \sqrt{k^2 + 4}
   \]
   Square both sides:
   \[
   (2 - k)^2 < k^2 + 4
   \]
   Simplify:
   \[
   4 - 4k + k^2 < k^2 + 4
   \]
   \[
   4 - 4k < 4
   \]
   \[
   -4k < 0
   \]
   \[
   k > 0
   \]

8. Consider the right part of the inequality:
   \[
   \sqrt{k^2 + 4} < 18 - k
   \]
   Square both sides:
   \[
   k^2 + 4 < (18 - k)^2
   \]
   Simplify:
   \[
   k^2 + 4 < 324 - 36k + k^2
   \]
   \[
   4 < 324 - 36k
   \]
   \[
   36k < 320
   \]
   \[
   k < \frac{320}{36} \approx 8.89
   \]

9. Since \( k \) must be an integer, we have:
   \[
   1 \leq k \leq 8
   \]

10. For each integer value of \( k \) from 1 to 8, there is a corresponding \( a \) in the interval \( (1, 9) \). Therefore, there are 8 such values of \( a \).

The final answer is \( \boxed{8} \)