Show that there is no order relation on K

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In summary, we discussed the impossibility of finding an order relation on $K$ that would make it an ordered field given the existence of an element $u\in K$ such that $u^2+1=0$. We analyzed different cases and concluded that no such relation can exist.
  • #1
mathmari
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Hey! :eek:

Let $K$ be a field. Suppose that there is an element $u\in K$ such that $u^2+1=0$. Show that there is no order relation on $K$ that would make $K$ an ordered field. We have to show that no relation $<$ can exist that satisfies the order axioms, i.e.:
  1. Only one of $a < b$, $a = b$, or $a > b$ is true
  2. If $a < b$ and $b < c$, then $a < c$
  3. If $a < b$ and $c < d$, then $a + c < b + d$
  4. If $a < b$ and $c < d$, then $a c < b d$
Suppose that there is an order relation on $K$ that would make $K$ an ordered field.

In $K$ there is an element $u$ such that $u^2+1=0\Rightarrow u^2=-1<0$.

According to the $4$th axiom we have that $u^2<0$ and $u^2<0$ then $u^2\cdot u^2=\left (u^2\right )^2=\left (-1\right )^2=1>0$, which is a contradiction since it should be $u^2\cdot u^2 <0$.

That means that there cannot be an order relation on $K$ that would make $K$ an ordered field. Is the proof correct and complete? Could I improve something? (Wondering)
 
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  • #2
mathmari said:
If $a < b$ and $c < d$, then $a c < b d$
Then real numbers don't form an ordered field...
 
  • #3
Hey mathmari!

See Evgeny.Makarov's comment. There must be something wrong with that 4th axiom. (Worried)

mathmari said:
In $K$ there is an element $u$ such that $u^2+1=0\Rightarrow u^2=-1<0$.

According to the $4$th axiom we have that $u^2<0$ and $u^2<0$ then $u^2\cdot u^2=\left (u^2\right )^2=\left (-1\right )^2=1>0$, which is a contradiction since it should be $u^2\cdot u^2 <0$.

Additionally, it is not given that $-1 < 0$ is it? Nor that $0 < 1$. (Worried)
We only know that either $-1 < 0$ or $-1 > 0$ according to the first axiom (they cannot be equal in a field).
 
  • #4
Ahh ok! Could you give me a hint what I am supposed to do? (Wondering)
 
  • #5
mathmari said:
Could you give me a hint what I am supposed to do?
Find a correct axiomatization of an ordered field in a textbook or in Wikipedia.

Additionally, it is not given that $-1<0$ is it?
It is not given, but this is a separate problem. If we just want to show that assuming that $K$ is an ordered field leads to a contradiction, we can use theorems of ordered fields.
 
  • #6
Evgeny.Makarov said:
Find a correct axiomatization of an ordered field in a textbook or in Wikipedia.

The Order Axioms are the following:
  • (Trichotemy) Either $a = b$, $a < b$ or $b < a$;
  • (Addition Law) $a < b$ if and only if $a + c < b + c$;
  • (Multiplication Law) If $c > 0$, then $ac < bc$ if and only if $a < b$. If $c < 0$, then $ac < bc$ if and only if $b < a$;
  • (Transitivity) If $a < b$ and $b < c$, then $a < c$.

Correct? We suppose that $K$ is an ordered field.

We have that $u^2=-1$. Do we have to show first that $-1<0$ or do we have to do something else? (Wondering)
 
  • #7
Yes. Any field necessarily contains an "additive identity", 0. The requirement is that "if a< b and 0< c then ac< bc".

By the way, we can also define F to be an "ordered field" if and only if there exist a set, P, such that
1) P is closed under addition.
2) P is closed under multiplication.
3) Given any x in F, one and only one is true:
x= 0, x is in P, or -x is in P.

Given these, you can prove your requirements and vice versa. Of course, given your conditions for an ordered field, you would take "P" to be the set of "positive" elements, x such that 0< x.
 
  • #8
mathmari said:
We suppose that $K$ is an ordered field.

We have that $u^2=-1$. Do we have to show first that $-1<0$ or do we have to do something else? (Wondering)

We could.
As Evgeny pointed out, we can actually already use that $-1 < 0$, since that is an existing proposition that follows from the axioms.
But suppose we start with distinguishing cases for $u$?
Suppose $u>0$. Then what can we say about $u^2$ and about $u^4$? (Wondering)
 
  • #9
Suppose that $u>0$.
Then from the multiplicative rule with $c=b=u$ and $a=0$ we get $ac=0<u^2=bc$.
But it holds that $u^2=-1<0$, a contradiction.

Therefore it cannot hold that $u>0$. Suppose that $u<0$.
Then from the multiplicative rule with $c=b=u$ and $a=0$ we get $ac=0<u^2=bc$.
But it holds that $u^2=-1<0$, a contradiction.

Therefore it cannot hold that $u<0$. That would mean that it must hold that $u=0$. Is that correct? (Wondering)
 
  • #10
mathmari said:
That would mean that it must hold that $u=0$.
And this is not possible either (why?).

mathmari said:
Is that correct?
Yes. For further practice, it is recommended proving that in an ordered field $-1<0$ and that your multiplication law in post 6 follows from HallsofIvy's in post 7.
 

Related to Show that there is no order relation on K

What is an order relation?

An order relation is a mathematical concept that defines the relationship between two elements in a set. It establishes whether one element is greater than, less than, or equal to the other element.

Why is it important to show that there is no order relation on K?

It is important because it helps us understand the properties and limitations of the set K. It also allows us to explore other ways of organizing the elements in K, which may be useful in certain applications.

Can an order relation exist on any set?

No, not all sets have an order relation. For example, sets that contain only abstract concepts or undefined elements cannot have an order relation.

How can we prove that there is no order relation on K?

We can prove that there is no order relation on K by showing that the properties of an order relation, such as reflexivity, transitivity, and antisymmetry, cannot be satisfied by the elements in K.

What are some examples of sets without an order relation?

Some examples of sets without an order relation include the set of all colors, the set of all emotions, and the set of all mathematical expressions. These sets do not have a natural way of ordering their elements.

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