Results 1 to 4 of 4

Math Help - no set that makes

  1. #1
    Member
    Joined
    Jun 2008
    Posts
    170

    no set that makes

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

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

    How would I obtain a contradiction. Is this the right way to go about proving it?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by particlejohn View Post
    Prove that there is no set  P \subset \mathbb{Z}_{3} (integers modulo  3 ) which makes  \mathbb{Z}_{3} into an ordered field.

    Assume for contradiction that there is a set  P \subset \mathbb{Z}_{3} which makes  \mathbb{Z}_{3} into an ordered field. Then there is a set  P \subset \mathbb{Z}_{3} such that if  a,b \in P, \ \text{then} \ a+b, ab \in P and if  a \in \mathbb{Z}_{3} , then exactly one of the following is true:  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: \{ 0\}, \{1\},\{2\},\{0,1\},\{0,2\},\{1,2\}. Now show each one does not satisfy the condition of being a ring ordering.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2008
    Posts
    170
    And from a real analysis viewpoint, this would amount to showing that the above properties described above are violated?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by particlejohn View Post
    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 \{0,1\}. We require that it is closed. Is it? Well 1+1 = 2 is not in the set. So it cannot represent the "positives".
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. the interval for which the solution to DE makes sense
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: June 8th 2010, 09:32 AM
  2. What makes this matrix true?
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: April 26th 2010, 05:44 PM
  3. Current Makes Magnetic Field
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 13th 2010, 05:58 PM
  4. Does this equation makes sense? If so, how to solve it?
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: December 2nd 2009, 02:47 PM
  5. Replies: 3
    Last Post: June 15th 2008, 07:08 AM

Search Tags


/mathhelpforum @mathhelpforum