It is to be observed that, if it were impossible to select one out of each pair of socks, it would follow that the socks could not be arranged in a progression, and therefore that there were not of them. This case illustrates that, if is an infinite number, one set of pairs may not contain the same number of terms as another set of pairs; for, given pairs of boots, there are certainly boots, but we cannot be sure of this in the case of the socks unless we assume the multiplicative axiom or fall back upon some fortuitous geometrical method of selection such as the above. Another important problem involving the multiplicative axiom is the relation of reflexiveness to non-inductiveness. It will be remembered that in Chapter VIII. we pointed out that a reflexive number must be non-inductive, but that the converse (so far as is known at present) can only be proved if we assume the multiplicative axiom. The way in which this comes about is as follows:— It is easy to prove that a reflexive class is one which contains sub-classes having terms. (The class may, of course, itself have terms.) Thus we have to prove, if we can, that, given any non-inductive class, it is possible to choose a progression out of its terms. Now there is no difficulty in showing that a non-inductive class must contain more terms than any inductive class, or, what comes to the same thing, that if is a non-inductive class and is any inductive number, there are sub-classes of that have terms. Thus we can form sets of finite sub-classes of : First one class having no terms, then classes having 1 term (as many as there are members of ), then classes having [Pg 127] 2 terms, and so on. We thus get a progression of sets of sub-classes, each set consisting of all those that have a certain given finite number of terms. So far we have not used the multiplicative axiom, but we have only proved that the number of collections of sub-classes of is a reflexive number, i.e. that, if is the number of members of , so that is the number of sub-classes of and is the number of collections of sub-classes, then, provided is not inductive, must be reflexive. But this is a long way from what we set out to prove. In order to advance beyond this point, we must employ the multiplicative axiom. From each set of sub-classes let us choose out one, omitting the sub-class consisting of the null-class alone. That is to say, we select one sub-class containing one term, , say; one containing two terms, , say; one containing three, , say; and so on. (We can do this if the multiplicative axiom is assumed; otherwise, we do not know whether we can always do it or not.) We have now a progression , , , ... sub-classes of , instead of a progression of collections of sub-classes; thus we are one step nearer to our goal. We now know that, assuming the multiplicative axiom, if is a non-inductive number, must be a reflexive number.