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# Determine which order axioms are satisfied for a given “pseudo” ordering on the complex numbers

Consider a “pseudo-ordering” on the complex field defined by saying a complex number is positive if and only if . Which of the three order axioms (listed in the previous exercise) are satisfied.

The first axiom fails. Take is positive since but is not positive since .

The second axiom fails since for any such that then neither nor is positive.

The third axiom holds since implies 0 is not positive.

# Determine which order axioms are satisfied for a given “pseudo” ordering on the complex numbers

Consider a “pseudo-ordering” on the complex field defined by saying a complex number is positive if and only if is positive. Which of the three order axioms (listed in the previous exercise) are satisfied.

The final axiom holds since is true for all . Hence, is not positive so the axiom holds.

The first axiom fails since under this ordering both and are positive, but is not positive.

The second axiom fails since for any complex number we have which implies that if is positive then so is .

# Determine which order axioms are satisfied for a given “pseudo” ordering on the complex numbers

Consider a “pseudo-ordering” on the complex field defined by saying a complex number is positive if and only if is positive. Which of the three order axioms (listed in the previous exercise) are satisfied.

The second and third axioms are satisfied, but not the first.

The second axiom states that for every with either is positive or is positive, but not both. By our definition, positive implies . Therefore, since we have . Hence, is not positive.

The third axiom is satisfied since by our definition of positive if and , then is not positive.

Finally, the first axiom fails since if we take , then is positive, but

is not positive.

# Prove that the complex numbers cannot be ordered

The three axioms for an order relation are (see page 20 of Apostol):

1. If and are in , so are and .
2. For every real , either or , but not both.
3. .

Prove that the complex number system cannot be equipped with an ordering relation satisfying all of these axioms.

Proof. Suppose otherwise, that such an ordering exists. Denote the positive complex numbers by . Then, since , we must have either or , but not both by the second axiom.

But, if then (since the product of positive elements must also be positive by applying the first axiom twice). Since this is a contradiction; therefore, .

Similarly, if then which is also a contradiction since . Hence, there can be no such ordering

# Prove a squeeze-like property for reals.

If and for all , then .

Proof. Suppose otherwise, that for all , but . Then, we have and implies , so is a positive real number. Then let (since this is a valid choice of ). By assumption we then have (since for all ), but this is a contradiction since (equality is symmetric) implies . Therefore, we must have

# Sum of squares is nonnegative.

Prove that with equality if and only if .

Proof. First, by Theorem I.20 we know that for any we have if , then . Further, since , we have that if , then . Thus, and and so .
Now, if , then . Conversely, if , then if we must have (otherwise the sum would be greater than 0). However, we know , so this is a contradiction. Therefore, , and hence, , as well. By Theorem I.20 again, this must mean (since if or then or is greater than 0).
Hence, we indeed have with equality if and only if

# Another proof on the less than or equal relation.

Prove that if and and , then we also must have .

Proof.
Since and we have , by substitution. But this means . Since by assumption, and we must have

# Less than or equal relation is transitive.

Prove that if and , then .

Proof.
By the transitivity of (Theorem I.17), we have that if and , then .
Then, if and we have by substitution.
If and , then by substitution.
If and , then by transitivity of the relation. Hence, by definition of .
Thus, in all cases and implies

# Taking reciprocals is order-reversing

Prove that if and are positive reals with , then .

Proof. Since and are positive we have and , so and exist. Then, and implies and (by I.3.5, Exercise #4). Hence,

So, by Theorem I.19, we have

# Prove that a nonzero number and its reciprocal have the same sign.

If , then and have the same sign.

Proof.
First, since , we know exists and . By Theorem I.24, this means and are both positive or both negative