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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:

My idea for (c) (Theorem I.24):
By Axiom 6, we have an x so that ax = 1 and (ax)b = b. If that x is negative, (ax)b = b 0 and both a and b are positive (x is positive, 1 is positive, a must be positive since ax = 1).
Please correct me if I’m wrong.

• Anonymous says:

My comment rendered invalid, there is a problem with this website

2. Anonymous says:

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

3. 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?