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Prove that the rationals satisfy the Archimedean property, but not the least-upper-bound axiom.

Recall, the Archimedean property states that if and is arbitrary, then there exists an integer such that .
Further, recall that the least upper bound axiom states that every nonempty set of real numbers which is bounded above has a supremum.
Now, prove that satisfies the Archimedean property, but not the least-upper-bound axiom.

First, we prove that satisfies the Archimedean property.
Proof. This is immediate since if are arbitrary elements in with , then they are also in since . Thus, by the Archimedean property in we know there is an such that . Hence, the Archimedean property is satisfied in

Next, we prove that does not satisfy the least-upper-bound axiom.
Proof. Let

Then is non-empty since . Further, is bounded above by 4 since . (Of course, there are better upper bounds available, but we just need any upper bound.)
Now, we must show that has no supremum in to show that the least-upper-bound property fails in (since this will mean is a nonempty set which is bounded above, but fails to have a least upper bound in ).
Suppose otherwise, say with . We know (I.3.12, Exercise #11). Thus, by the trichotomy law we must have either or .
Case 1: If , then there exists such that (since the rationals are dense in the reals, see I.3.12, Exercise #6). But then, (since ) and implies with , contradicting that is an upper bound for . Hence, we cannot have .
Case 2: If , then there exists such that . But then, , so if we have ; hence, is an upper bound for which is less than . This contradicts that is the least upper bound of . Hence, we also cannot have .
Thus, there can be no such (since by the trichotomy exactly one of must hold, but we have shown these all lead to contradictions).
Hence, does not have the least-upper-bound property

Prove there is no rational number which squares to 2.

Prove that there is no such that .

Proof. Suppose otherwise, that there is some rational number such that . Then, since , we know there exist integers not both even (I.3.12, Exercise #10 (e) such that . Then, since we have

But, by I.3.12, Exercise #10 (d), we know implies both and are even, contradicting our choice of and not both even. Hence, there can be no such rational number

Prove there is an irrational number between any two real numbers.

Let be given with . Prove that there exists an irrational number such that .

Note: To do this problem, I think we need to assume the existence of an irrational number. We will prove the existence of such a number (the ) in I.3.12, Exercise #12.

Proof. Since the rationals are dense in the reals I.3.12, Exercise #6, we know that for with there exist such that

Now, assume the existence of an irrational number, say (see note preceding the proof about this). Since we know and from the order axioms exactly one of or is positive ( is nonzero since ). Without loss of generality, let . Then, since , we know there exists an integer such that

Also, since , we have ; thus, .
Then, by I.3.12, Exercise #7 we have irrational and hence irrational.
Thus, letting , we have with irrational

The rationals are dense in the reals.

Prove that the rational numbers are dense in the reals. I.e., if with , then there exists an such that . It follows that there are then infinitely many such.

Proof. Since , we know . Therefore, there exists an such that

We also know (I.3.12, Exercise #4) that there exists such that . Putting these together we have,

Since we have . Hence, letting we have found such that

This then guarantees infinitely many such rationals since we can just replace by (and note that ) and apply the theorem again to find such that . Repeating this process we obtain infinitely many such rationals

There is exactly one integer between any real number x and x+1.

Prove that if , then there is exactly one such that .

Proof. By I.3.12, Exercise #4 we know that for there is exactly one such that

If , then we already done since .
If , then we have and so . But we already have ; hence,

So, is an integer with the desired properties. Uniqueness follows from the uniqueness of our choice of

Prove that any real number lies between exactly one pair of consecutive integers.

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

For a positive real x, there is always a positive integer whose reciprocal is smaller than x.

Prove that if , then there is an such that .

Proof. Since , we know that for any there exists an such that (Theorem I.30, p. 26 of Apostol). Let , then we have and so

Prove that every fixed real number is between two integers.

Prove that if is fixed, then there exist such that .

Proof. Since the set of positive integers is unbounded above (Example #1, p. 24 of Apostol) we know there exists an such that (otherwise would be an upper bound on ).
Then, by the same logic there is some such that ; hence, . Since , we know . Thus, we have found such that

There is a real number between any two given real numbers.

Prove that if with , then there exists such that .

Proof. (This is about the point in the blog at which I’m just going to use the basic properties of without much comment. As usual, if something is unclear, leave a comment, and I’ll clarify.)

Then, since and (since and using I.3.5, Exercise #4) and so their product is also greater than 0.

So, letting , we have with

as requested

Prove a squeeze-like property for reals.

If and for all , then .

Proof. Suppose otherwise, that for all , but . Then, we have and implies , so is a positive real number. Then let (since this is a valid choice of ). By assumption we then have (since for all ), but this is a contradiction since (equality is symmetric) implies . Therefore, we must have