Select from the given books two groups of three books each and put them on the opposite cups of the scale. If they weigh the same, the fake is among the remaining two books, and weighing these two books will identify the lighter fake. If the first weighing does not yield a balance, the lighter fake is among the three lighter books. Take any two of them and put them on the opposite cups of the scale. If they weigh the same, it is the third coin in the lighter group that is fake; if they do not weigh the same, the lighter one is the fake. Thus the problem is solvable in two weighings. It's not quite clear why the problem has been set up for 8 and not 9 books. The above solution shows that 1 weighing is sufficient to detect a lighter fake among 3 books. Splitting 9 books into three groups and thinking of each as a "big coin", reduces the problem to the previous one: it takes just one weighing to detect the lighter group. An additional weighing finds the fake coin among the three in the lighter group.