# no set that makes

• Jul 3rd 2008, 03:46 PM
particlejohn
no set that makes
Prove that there is no set $\displaystyle P \subset \mathbb{Z}_{3}$ (integers modulo $\displaystyle 3$) which makes $\displaystyle \mathbb{Z}_{3}$ into an ordered field.

Assume for contradiction that there is a set $\displaystyle P \subset \mathbb{Z}_{3}$ which makes $\displaystyle \mathbb{Z}_{3}$ into an ordered field. Then there is a set $\displaystyle P \subset \mathbb{Z}_{3}$ such that if $\displaystyle a,b \in P, \ \text{then} \ a+b, ab \in P$ and if $\displaystyle a \in \mathbb{Z}_{3}$, then exactly one of the following is true: $\displaystyle a \in P, -a \in P, a = 0$.

How would I obtain a contradiction. Is this the right way to go about proving it?
• Jul 3rd 2008, 04:05 PM
ThePerfectHacker
Quote:

Originally Posted by particlejohn
Prove that there is no set $\displaystyle P \subset \mathbb{Z}_{3}$ (integers modulo $\displaystyle 3$) which makes $\displaystyle \mathbb{Z}_{3}$ into an ordered field.

Assume for contradiction that there is a set $\displaystyle P \subset \mathbb{Z}_{3}$ which makes $\displaystyle \mathbb{Z}_{3}$ into an ordered field. Then there is a set $\displaystyle P \subset \mathbb{Z}_{3}$ such that if $\displaystyle a,b \in P, \ \text{then} \ a+b, ab \in P$ and if $\displaystyle a \in \mathbb{Z}_{3}$, then exactly one of the following is true: $\displaystyle a \in P, -a \in P, a = 0$.

How would I obtain a contradiction. Is this the right way to go about proving it?

The only possibly non-empty subsets are: $\displaystyle \{ 0\}, \{1\},\{2\},\{0,1\},\{0,2\},\{1,2\}$. Now show each one does not satisfy the condition of being a ring ordering.
• Jul 3rd 2008, 06:04 PM
particlejohn
And from a real analysis viewpoint, this would amount to showing that the above properties described above are violated?
• Jul 3rd 2008, 06:33 PM
ThePerfectHacker
Quote:

Originally Posted by particlejohn
And from a real analysis viewpoint, this would amount to showing that the above properties described above are violated?

It makes no difference what viewpoint you have it is always violated. Take for example $\displaystyle \{0,1\}$. We require that it is closed. Is it? Well $\displaystyle 1+1 = 2$ is not in the set. So it cannot represent the "positives".