Different Dice

If the same numbers repeats on the second turn then the product is a rational number. There are six of these cases. The remaining cases are ( $\sqrt{1}, \sqrt{4}$ ) and ( $\sqrt{4}, \sqrt{1}$ ). Totally 8 cases. Then the probability equals $\frac{8}{36}$ .

There are 36 possible combinations.

1. If you get the same number on the second turn, the product will be a rational number: $\sqrt{1} \times \sqrt{1} = 1 ; \sqrt{2} \times \sqrt{2} = 2 ; ...$ Altogether 6 combinations. $$

2. The only rational numbers on the dice are: $\sqrt{1}$ and $\sqrt{4}$ . If you multiply those with the irrational numbers, the product will be irrational. Therefore the only two combinations left are: $\sqrt{1} \times \sqrt{4} = 2 \times 1 = 2$ $\sqrt{4} \times \sqrt{1} = 1 \times 2 = 2$

A total of 8 combinations: $\boxed{ \frac{8}{36} = \frac{2}{9} }$

Sorry for mediocre English. (English is not my native tongue :D )