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Ring problem

  1. Mar 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Let R be a ring in which 1_R = 0_R .Show that R has only one element.


    2. Relevant equations



    3. The attempt at a solution
    I'm trying to show that a*0_r=a*1_r implies a*0_r=0_r.
    if 0=0+0=>a*0=a*(0+0)=a*0+a*0=a*1_r+a*1_r=2a=>a*0=2a= >a=0....is this correct? If not
    Is there something I should do?
     
  2. jcsd
  3. Mar 12, 2012 #2

    tiny-tim

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    Hi Stephen88! :smile:
    No, you should be trying to prove that a = 0.

    Hint: what, by definition, is a*0 ?

    what, by definition, is a*1 ? :wink:
     
  4. Mar 12, 2012 #3
    yes a = 0 is something I need to prove that was just I way that I thought it would be possible.
    For instance a*0=a*(0+0)=a*0+a*0=a*1+a*1=2a=>a*0=2a= >a=0..otherwise 2=0 which is a contradiction.Is this correct?
     
  5. Mar 12, 2012 #4

    tiny-tim

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    Sorry, but this is fantasy …

    there is no "2", is there?

    All you're allowed to play with is a 0 and 1.

    what, by definition, is a*0 ?

    what, by definition, is a*1 ?​
     
  6. Mar 12, 2012 #5
    Uh sorry about that...I;m used to integers and real numbers....hmm..given the context both represent the identity of the ring so a*0=a*1=a
     
  7. Mar 12, 2012 #6

    tiny-tim

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    what, by definition, is a*0 ?
     
  8. Mar 12, 2012 #7
    a product of a and the multiplicative identity?
     
  9. Mar 12, 2012 #8

    tiny-tim

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    0 is the additive identity

    (the multiplicative zero)
     
  10. Mar 12, 2012 #9
    Yes normally 0 is the additive identity and 1 the multp. one...but here they are both the same.
    So if 1=0->1*a=0*a=a=>a=1=0?
    If not I'm confused
     
  11. Mar 12, 2012 #10

    tiny-tim

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    I can't tell whether you've got it or not … you need to make it clear. :redface:

    Let's spell it out. :smile:

    a*0 by definition is always 0

    a*1 by definition is always a

    but 0 = 1

    so a*0 = a*1

    so 0 = a :wink:
     
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