Prove the following consequences of the order axioms.

- If and , then .
- If , then .
- If , then and are both both positive or and are both negative.
- If and , then .

* Proof. * First, by definition of we have

Then, by Axiom 8 (and since implies and implies ), we have

Thus, by Axiom 7 , and then using the field properties (Section I.3.3) we have

Hence, , i.e.,

* Proof. * First, implies , and then we have,

Thus, ; hence,

* Proof. * First, we cannot have or since by Theorem I.11 this would mean ; hence, we could not have .

Assume then that and . By assumption . Now, if is positive then

On the other hand, if then

Thus, implies as well. Hence, if is positive then so is , and if is negative then so is , i.e., either and are both positive or both negative

* Proof. * Since we have , and since we have . Then,

Hence,

*Related*

for b) you can really just apply a) for c=-1

Yes indeed, that’s what i though

For the problem C, i’ve came to this brief solution:

Let there be a positive real number “x” such that a.x = 1

Then, for a>0, we have that:

a.b>0 (a.b).x> 0, using Axiom 7 (because “x” is a positive real number)

(a.x).b > 0, commutativity and associativity

1.b > 0, definition of a.x

b > 0, Axiom 4.

Now, let there be a negative real number “x” such that a.x=1

Then, for a 0 (a.b).x< 0, because of the Theorem I.23 (multiplying by a negative number changes the sign)

(a.x).b < 0, commutativity and associativity

1.b < 0, definition of a.x

b<0, Axiom 4.

Forgot to point out that we use Axiom 7 in the first step for when a<0

I have a concern: How do we know that if a > 0, then there exists a “positive” real number x so that a.x = 1?