Results 1 to 6 of 6

Math Help - z mod 6

  1. #1
    Newbie
    Joined
    Dec 2006
    Posts
    8

    z mod 6

    why isnt Z mod 6 a field?

    this is an extra credit problem in my pre-calc class...

    I know that it must abide to the field axioms(am i right?)

    but mostly i dont understand the form Z mod 6
    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 tummbler View Post
    why isnt Z mod 6 a field?

    this is an extra credit problem in my pre-calc class...

    I know that it must abide to the field axioms(am i right?)

    but mostly i dont understand the form Z mod 6
    Okay, the basic answer is that it is not a power of a prime. But that is a more advanced answer.

    By you can do it by checking whether it is a field or not.
    The set is,
    \{0,1,2,3,4,5\}
    It froms a group under addition.
    Furthermore, it is a ring (just check the definitions for a ring).
    Furthermore, it is a commutative ring.
    Furthermore, it has unity (1).
    Thus, it is a commutative ring with unity.
    Now, if every non-zero element has an inverse the proof is complete.
    Thus, we need to show every element has a multiplicative inverse.
    The element 2, has no inverse!
    Just check each one and convince thyself.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2006
    Posts
    8
    what parts of the set are a and b..

    im sorry if i sound stupid. ive just neever seen this type of problem before
    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 tummbler View Post
    what parts of the set are a and b..

    im sorry if i sound stupid. ive just neever seen this type of problem before
    I do not understand what you are saying.

    Did you understand anything I said?

    You mentioned it was for a Pre-Calculus class? That make no sense.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2006
    Posts
    8
    this is a extra credit prob for precalc.

    in the definition of the addition field axiom is has the variables a and b. same in alot of the definitions of the terms u used in your original explanation. I used mathworld.com btw..
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by tummbler View Post
    this is a extra credit prob for precalc.

    in the definition of the addition field axiom is has the variables a and b. same in alot of the definitions of the terms u used in your original explanation. I used mathworld.com btw..
    It is not a fair question for a pre-calculus class.
    Just write on your paper, that not all elements (numbers in Z6) has an inverse, for example 2 is in Z6 and it got no inverse.
    Inverse means we need to find a number 'x' (in Z6) such that,
    2x=1
    Now, just check all the possibilities,
    2(0)=0
    2(1)=2
    2(2)=4
    2(3)=6=0 (because we take the mod of 6)
    2(4)=8=2 (same reason)
    2(5)=10=4
    Note, we went through all possibilities because,
    Z6= { 0,1,2,3,4,5 }
    And none of them produce 1.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum