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Show that there is no order relation on K

mathmari

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Apr 14, 2013
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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)
 

Evgeny.Makarov

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MHB Math Scholar
Jan 30, 2012
2,488

Klaas van Aarsen

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Mar 5, 2012
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Hey mathmari !!

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

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).
 

mathmari

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Apr 14, 2013
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Ahh ok! Could you give me a hint what I am supposed to do? (Wondering)
 

Evgeny.Makarov

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MHB Math Scholar
Jan 30, 2012
2,488
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.
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,009
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)
 

HallsofIvy

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Jan 29, 2012
1,151
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.
 

Klaas van Aarsen

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Mar 5, 2012
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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)
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,009
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)
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,488
That would mean that it must hold that $u=0$.
And this is not possible either (why?).

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.