47.

Consider an apportionment problem where we will compare two methods, the Hamilton method and the method. (We do not specify what the method is, and we leave the apportionment problem also unspecified.) Our objective is to show that the maximum absolute deviation (MAD) for the apportionment is at least as large as the MAD for the Hamilton apportionment.

Since the Hamilton method satisfies the quota condition, we will assume that the method has given each state either its upper or lower quota; otherwise, its MAD would be greater than 1, and thus worse than Hamilton’s. If the methods give different apportionments, they must differ for more than one state; otherwise, they would not have the same house size.

Let be the state that has the largest absolute deviation with the Hamilton method, and let be its quota. The apportionment for is either the lower quota, , or the upper quota, , so the absolute deviation for (and the MAD for the Hamilton apportionment) is either .

If the apportionment for the state is the same as the Hamilton apportionment for , then the MAD for the apportionment is at least equal to the absolute deviation for , so it cannot be less than the MAD for the Hamilton apportionment.

Suppose that the apportionment for differs from the Hamilton apportionment: Say Hamilton assigns its lower quota and assigns its upper quota. There must be another state to which Hamilton assigns the upper quota and assigns the lower quota. Since the Hamilton method rounds down and up, the fractional part of is greater than or equal to the fractional part of . The absolute deviation in the apportionment for state is the fractional part of , so the MAD for the apportionment, which must be at least the absolute deviation for state , is greater than or equal to the MAD for the Hamilton method in this apportionment problem.

If the Hamilton method awards state its upper quota, we can reason as before and reach the same conclusion—that the apportionment has a MAD no less than that of the Hamilton method.

A-36