For a positive number such as 3.27, 3 is referred to as the integral part of the number and .27 as the decimal part. Find a positive number such that its decimal part, its integral part, and the number itself form a geometric progression.
1. Let \( x \) be the positive number such that its decimal part, its integral part, and the number itself form a geometric progression.
2. Let \( d \) be the decimal part of \( x \), and let \( n \) be the integral part of \( x \). Therefore, we can write \( x \) as:
   \[
   x = n + d
   \]
3. Since \( d \), \( n \), and \( x \) form a geometric progression, we can denote the common ratio by \( r \). Thus, we have:
   \[
   n = dr \quad \text{and} \quad x = dr^2
   \]
4. Substituting \( x = n + d \) into \( x = dr^2 \), we get:
   \[
   dr^2 = n + d
   \]
5. Substituting \( n = dr \) into the equation above, we get:
   \[
   dr^2 = dr + d
   \]
6. Dividing both sides by \( d \) (since \( d \neq 0 \)), we obtain:
   \[
   r^2 = r + 1
   \]
7. This is a quadratic equation in \( r \). Solving for \( r \), we rearrange the equation:
   \[
   r^2 - r - 1 = 0
   \]
8. Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -1 \), and \( c = -1 \), we get:
   \[
   r = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}
   \]
9. Since \( r \) must be positive, we take the positive root:
   \[
   r = \frac{1 + \sqrt{5}}{2}
   \]
10. Now, we need to find \( x \). Recall that \( x = dr^2 \) and \( n = dr \). Substituting \( r = \frac{1 + \sqrt{5}}{2} \) into \( r^2 \), we get:
    \[
    r^2 = \left( \frac{1 + \sqrt{5}}{2} \right)^2 = \frac{1 + 2\sqrt{5} + 5}{4} = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2}
    \]
11. Therefore, the number \( x \) is:
    \[
    x = dr^2 = d \cdot \frac{3 + \sqrt{5}}{2}
    \]
12. Since \( d \) is the decimal part, we can choose \( d = 1 \) for simplicity, giving us:
    \[
    x = \frac{3 + \sqrt{5}}{2}
    \]

The final answer is \(\boxed{\frac{3 + \sqrt{5}}{2}}\).