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