4. Arrange $\frac{131}{250}, \frac{21}{40}, 0.5 \dot{2} \dot{3}, 0.52 \dot{3}, 0.5 \dot{2}$ in ascending order, the third number is $\qquad$
4. Solution: $\frac{131}{250}=0.524, \frac{21}{40}=0.525$, so: $0.5 \dot{2}<0.5 \ddot{2} \dot{3}<0.52 \dot{3}<\frac{131}{250}<\frac{21}{40}$, the third smallest number is $0.52 \dot{3}$