Suppose u is an element in a ring. Recall that units can be canceled: If r and s are elements of R satisfying rs = 1, then r = s^-1. Observe that the logically equivalent contrapositive takes the following form: If r and s are elements of R satisfying r ≠s^-1, then rs ≠1. Suppose that R has only finitely many elements, and that there is a complete list of them with no repetitions. Deduce that the set {ur | r is an element of R} is also a complete list of the elements of R, with no repetitions. We might describe this result by saying that multiplication by u shuffles R. Give an example to show that in contrast, if a is an arbitrary nonzero element of R, the set {a^n | n is an element of Z} may not be a complete list of elements of R. (Again note that a = 0 is a trivial example. Also, note that some elements may be repeated in this list, and some may be omitted.) You need only produce a single ring R and a single nonzero element a in order to demonstrate this. Thus, multiplication by an arbitrary element may not shuffle R.