Prove that for fixed , there is exactly one such that .
First, we prove a lemma.
Lemma: 1 is the smallest element of .
Proof. Consider the set . Then, is an inductive set since and for all , . Thus, (since, by definition, is the set of elements common to every inductive set). But, for any , if , then and hence . Thus, 1 is the least element of .
Now, for the theorem of the exercise.
Proof. First, we prove existence.
Let . We know is non-empty since by I.3.12, Exercise #2 we know there exist integers and such that . This must be in our set . We also know is bounded above since is an upper bound by our definition of . Therefore, by the least upper bound axiom (Axiom 10, p. 25), we know exists.
Then, by Theorem I.32, we know that for any positive real number , there is some such that
Letting , we have
Therefore, since the supremum of is an upper bound, we know ; therefore, by definition of . But we already know , and so by definition of we must have . Therefore, we have found an integer such that
Now, for uniqueness. Suppose there are integers with this property. Then we have
Then, adding the terms of these inequalities we have,
Without loss of generality, assume . Then this inequality implies, . But since 1 is the smallest element of (from the lemma above) and is closed under subtraction, we must have , i.e., . Thus, any such is unique