Tag: Ordered Fields

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

by RoRi

February 25, 2016

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

The first axiom fails. Take $z = 2 + i$ is positive since $2 > 1$ but $z^2 = 3 + 4i$ is not positive since $3 \not > 4$ .

The second axiom fails since for any $z \neq 0$ such that $x = y$ then neither $z$ nor $-z$ is positive.

The third axiom holds since $0 = 0 + 0i$ implies 0 is not positive.

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

by RoRi

February 25, 2016

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

The final axiom holds since $|z| > 0$ is true for all $z \neq 0$ . Hence, $0$ is not positive so the axiom holds.

The first axiom fails since under this ordering both $1$ and $-1$ are positive, but $1 + (-1) = 0$ is not positive.

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

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

by RoRi

February 25, 2016

Consider a “pseudo-ordering” on the complex field defined by saying a complex number $z = x +iy$ is positive if and only if $x$ 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 $z$ with $x \neq 0$ either $z$ is positive or $-z$ is positive, but not both. By our definition, $z$ positive implies $x >0$ . Therefore, since $-z = -x-iy$ we have $-x < 0$ . Hence, $-z$ is not positive.

The third axiom is satisfied since by our definition of positive if $z = x+iy$ and $x = 0$ , then $z$ is not positive.

Finally, the first axiom fails since if we take $z = 1 + 2i$ , then $z$ is positive, but

$z^2 = (1+2i)(1+2i) = 1 - 4 + 4i = -3 + 4i$

is not positive.

Prove that the complex numbers cannot be ordered

by RoRi

February 25, 2016

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

If $x$ and $y$ are in $\mathbb{R}^+$ , so are $x+y$ and $xy$ .
For every real $x \neq 0$ , either $x \in \mathbb{R}^+$ or $-x \in \mathbb{R}^+$ , but not both.
$0 \notin \mathbb{R}^+$ .

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 $\mathbb{C}^+$ . Then, since $i \neq 0$ , we must have either $i \in \mathbb{C}^+$ or $i \in \mathbb{C}^-$ , but not both by the second axiom.

But, if $i \in \mathbb{C}^+$ then $i^3 \in \mathbb{C}^+$ (since the product of positive elements must also be positive by applying the first axiom twice). Since $i^3 = -i$ this is a contradiction; therefore, $i \notin \mathbb{C}^+$ .

Similarly, if $-i \in \mathbb{C}^+$ then $(-i)^3 \in \mathbb{C}^+$ which is also a contradiction since $(-i)^3 = i$ . Hence, there can be no such ordering $. \qquad \blacksquare$

Prove a squeeze-like property for reals.

by RoRi

June 30, 2015

If $x \in \mathbb{R}$ and $0 \leq x < h$ for all $h \in \mathbb{R}^+$ , then $x = 0$ .

Proof. Suppose otherwise, that $0 \leq x < h$ for all $h \in \mathbb{R}^+$ , but $x \neq 0$ . Then, we have $0 \leq x$ and $0 \neq x$ implies $0 < x$ , so $x$ is a positive real number. Then let $h = x$ (since $x \in \mathbb{R}^+$ this is a valid choice of $h$ ). By assumption we then have $x < h = x$ (since $x < h$ for all $h \in \mathbb{R}^+$ ), but this is a contradiction since $x = x$ (equality is symmetric) implies $x \not< x$ . Therefore, we must have $x = 0. \qquad \blacksquare$

Sum of squares is nonnegative.

by RoRi

June 30, 2015

Prove that $a^2 + b^2 \geq 0$ with equality if and only if $a = b = 0$ .

Proof. First, by Theorem I.20 we know that for any $x \in \mathbb{R}$ we have if $x \neq 0$ , then $x^2 > 0$ . Further, since $0 \cdot 0 = 0$ , we have that if $x = 0$ , then $x^2 = 0$ . Thus, $a^2 \geq 0$ and $b^2 \geq 0$ and so $a^2 + b^2 \geq 0$ .
Now, if $a = b = 0$ , then $a^2 + b^2 = 0 + 0 = 0$ . Conversely, if $a^2 + b^2 = 0$ , then if $a^2 > 0$ we must have $b^2 < 0$ (otherwise the sum would be greater than 0). However, we know $b^2 \geq 0$ , so this is a contradiction. Therefore, $a^2 = 0$ , and hence, $a^2 + b^2 = 0 \implies b^2 = 0$ , as well. By Theorem I.20 again, this must mean $a = b =0$ (since if $a \neq 0$ or $b \neq 0$ then $a^2$ or $b^2$ is greater than 0).
Hence, we indeed have $a^2 + b^2 \geq 0$ with equality if and only if $a = b = 0. \qquad \blacksquare$

Another proof on the less than or equal relation.

by RoRi

June 30, 2015

Prove that if $a \leq b$ and $b \leq c$ and $a = c$ , then we also must have $b = c$ .

Proof.
Since $b \leq c$ and $a = c$ we have $b \leq a$ , by substitution. But this means $a \not< b$ . Since $a \leq b$ by assumption, and $a \not< b$ we must have $a = b. \qquad \blacksquare$

Less than or equal relation is transitive.

by RoRi

June 30, 2015

Prove that if $a \leq b$ and $b \leq c$ , then $a \leq c$ .

Proof.
By the transitivity of $<$ (Theorem I.17), we have that if $a < b$ and $b < c$ , then $a < c$ .
Then, if $a = b$ and $b \leq c$ we have $a \leq c$ by substitution.
If $b = c$ and $a \leq b$ , then $a \leq c$ by substitution.
If $a = b$ and $b = c$ , then $a = c$ by transitivity of the $=$ relation. Hence, $a \leq c$ by definition of $\leq$ .
Thus, in all cases $a \leq b$ and $b \leq c$ implies $a \leq c. \qquad \blacksquare$

Taking reciprocals is order-reversing

by RoRi

June 30, 2015

Prove that if $a$ and $b$ are positive reals with $a < b$ , then $0 < b^{-1} < a^{-1}$ .

Proof. Since $a$ and $b$ are positive we have $a \neq 0$ and $b \neq 0$ , so $a^{-1}$ and $b^{-1}$ exist. Then, $a > 0$ and $b > 0$ implies $a^{-1} > 0$ and $b^{-1} > 0$ (by I.3.5, Exercise #4). Hence,

$a^{-1} b^{-1} > 0$

So, by Theorem I.19, we have

$\begin{align*} &&a(a^{-1} b^{-1}) &< b(a^{-1}b^{-1}) \\ \implies&& b^{-1} &< a^{-1}. \qquad \blacksquare \end{align*}$

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

by RoRi

June 30, 2015

If $a \neq 0$ , then $a$ and $1/a$ have the same sign.

Proof.
First, since $a \neq 0$ , we know $1/a$ exists and $a(1/a) = 1 > 0$ . By Theorem I.24, this means $a$ and $1/a$ are both positive or both negative $. \qquad \blacksquare$