To prove that every finite skew-field is a field, we use strong induction on the size of the skew-field. Thus, let "A" be a skew-field, and assume that all skew-fields that are proper subsets of "A" are fields. Since the center "Z"("A") of "A" is a field, "A" is a vector space over "Z"("A") with finite dimension "n". Our objective is then to show "n" = 1. If "q" is the order of "Z"("A"), then "A" has order "qn". Note that because "Z"("A") contains the distinct elements 0 and 1, q>1. For each "x" in "A" that is not in the center, the centralizer "Zx" of "x" is clearly a skew-field and thus a field, by the induction hypothesis, and because "Zx" can be viewed as a vector space over "Z"("A") and "A" can be viewed as a vector space over "Zx", we have that "Zx" has order "qd" where "d" divides "n" and is less than "n". Viewing "Z"("A")*, "A*", and the "Z*x" as groups under multiplication, we can write the class equation