Homework Problems Some Even with Solutions

To say that n is the rank of Q is to assert that Q : An → A. So suppose that n and m are natural numbers and Q has rank m as well as rank n. We need to prove that n = m. It follows that An = Am since both An and Am turn out to be the domain of Q. Now in general AX is the set of all functions from X into A. Recalling that n = {0, 1, . . . , n−1} and m = {0, 1, . . . ,m−1} we can understand more clearly what An and Am are. Now, using that A is not empty, pick a ∈ A. The function with domain n having a as its constant value is just {(0, a), (1, a), . . . , (n− 1, a)} ∈ An. Since An = Am we see that for some b0, b1, . . . , bm−1 ∈ A we have {(0, a), (1, a), . . . , (n− 1, a)} = {(0, b0), (1, b1), . . . , (n− 1, bm−1)}. But the set on the left has exactly n distinct elements while the set on the right has exactly m distinct elements. Therefore n = m, as desired.