We will let N stand for the set of all natural numbers in the following listing of Peano's axioms. A1) 1 is a natural number. A2) Every natural number x has a successor (which we will call x') that is a natural number, and this successor is unique. A3) No two natural numbers have the same successor. That is, if x, y are natural numbers, with x != y, then x' != y'. A4) 1 is not the successor of any natural number. A5) Let M be a subset of natural numbers with the property that (i) 1 is in M. (ii) Whenever a natural number x is in M, then x' is in M. Then, M must contain all natural numbers, that is, M = N. Must every natural number be the successor of some number? Clearly, this is not the case for 1 (why not?), but what about other numbers? Consider the statement "If x != 1, then there is a number u with u' = x." Let M be the set consisting of the number 1 plus all other x for which this statement is true. Use the axiom of induction to show that M = N and, thus, that every number other than 1 is the successor of some number.