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# Prove some consequences of the order axioms

Prove the following consequences of the order axioms.

1. If and , then .
2. If , then .
3. If , then and are both both positive or and are both negative.
4. If and , then .

1. 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., 2. Proof. First, implies , and then we have, Thus, ; hence, 3. 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 4. Proof. Since we have , and since we have . Then, Hence, 1. Anonymous says:

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

• Yes indeed, that’s what i though

2. Anonymous says:

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.

• Anonymous says:

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

• Anonymous says:

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?