[b]5/1/32.[/b] Find all pairs of rational numbers $(a, b)$ such that $0 < a < b$ and $a^a = b^b$. 
1. Given the equation \(a^a = b^b\), we start by taking the natural logarithm of both sides:
   \[
   \ln(a^a) = \ln(b^b)
   \]
   Using the property of logarithms \(\ln(x^y) = y \ln(x)\), we get:
   \[
   a \ln(a) = b \ln(b)
   \]

2. Since \(0 < a < b\), we need to find rational numbers \(a\) and \(b\) that satisfy this equation. We can express \(b\) as a rational power of \(a\). Let \(b = a^{\frac{p}{q}}\) where \(p\) and \(q\) are positive integers with \(q > p\).

3. Substitute \(b = a^{\frac{p}{q}}\) into the equation \(a \ln(a) = b \ln(b)\):
   \[
   a \ln(a) = a^{\frac{p}{q}} \ln(a^{\frac{p}{q}})
   \]
   Simplifying the right-hand side using the logarithm property:
   \[
   a \ln(a) = a^{\frac{p}{q}} \cdot \frac{p}{q} \ln(a)
   \]
   Dividing both sides by \(\ln(a)\) (assuming \(a \neq 1\)):
   \[
   a = a^{\frac{p}{q}} \cdot \frac{p}{q}
   \]

4. Divide both sides by \(a^{\frac{p}{q}}\):
   \[
   a^{1 - \frac{p}{q}} = \frac{p}{q}
   \]
   Let \(k = \frac{q - p}{q}\), then:
   \[
   a^k = \frac{p}{q}
   \]

5. Raise both sides to the power of \(\frac{1}{k}\):
   \[
   a = \left(\frac{p}{q}\right)^{\frac{1}{k}} = \left(\frac{p}{q}\right)^{\frac{q}{q-p}}
   \]

6. To find rational solutions, we need \(a\) to be rational. This implies that \(\frac{p}{q}\) must be a rational number raised to a rational power. We now seek pairs \((p, q)\) such that \(q - p = 1\).

7. If \(q - p = 1\), then \(p = q - 1\). Substitute \(p = q - 1\) into the equation:
   \[
   a = \left(\frac{q-1}{q}\right)^q
   \]
   and
   \[
   b = a^{\frac{q-1}{q}} = \left(\left(\frac{q-1}{q}\right)^q\right)^{\frac{q-1}{q}} = \left(\frac{q-1}{q}\right)^{q-1}
   \]

8. Therefore, the pairs \((a, b)\) are given by:
   \[
   (a, b) = \left(\left(\frac{q-1}{q}\right)^q, \left(\frac{q-1}{q}\right)^{q-1}\right)
   \]
   for integer \(q \geq 2\).

The final answer is \( \boxed{ \left(\left(\frac{q-1}{q}\right)^q, \left(\frac{q-1}{q}\right)^{q-1}\right) } \) for integer \(q \geq 2\).