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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

# Sums, differences, products and quotients of an irrational and a rational are irrational.

Prove that if with , and , then

are all irrational, i.e., are all in .

Proof. Since , we know there are integers and such that . Now we consider each of the elements we wish to show are irrational.

1. Suppose otherwise, that . Then, there exist such that

This contradicts our assumption that is irrational.

2. Since is irrational, so is (by part (a), the sum of a rational and an irrational cannot be rational and since is rational, cannot have rational). But then, and by part (a) this sum must be irrational.
3. Suppose otherwise, that . Then there exist such that . Further, since , we know exists.

Contradicting our assumption that is irrational.

4. First, we since is irrational, we have , and thus exists. Further, since is rational, and is irrational, by (c) we must have irrational as well. Then by (c), since , we must have irrational.
5. By (d) we know is irrational, and since is rational, by (c), we must have 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

# Prove equivalence of different forms for additive inverses of fractions

Prove that if then

Proof. We can use the first exercise of this section (Section I.3.3, Exercise #1) and the previous exercise (Section I.3.3, Exercise #8) to compute

Then for the other equality, similarly, we have