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Prove (-x)y=-(xy) for rings

  1. Nov 27, 2011 #1
    1. The problem statement, all variables and given/known data

    I need to show that (-x)*y=-(x*y) for a ring. unless it's not true.

    2. Relevant equations

    A ring is a set R and operations +, * such that (R, +) is an Abelian group, * is associative, and a*(b+c)=a*b+a*c and (b+c)*a=b*a+c*a.

    3. The attempt at a solution

    I don't know what the first step of this proof will be, I'm looking for the *trick*, as it were.
     
  2. jcsd
  3. Nov 27, 2011 #2

    I like Serena

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    If we want to know if (-x)*y is the additive inverse (x*y), what is the axiom that we should check?
     
  4. Nov 27, 2011 #3
    I don't know. maybe if I knew what axiom to check I'd have a better idea how to prove it?
     
  5. Nov 27, 2011 #4

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    You wrote that (R,+) is an abelian group.
    What is the definition of an inverse in a group?
     
  6. Nov 27, 2011 #5
    -(a+b)= (-b)+(-a)
    right?
     
  7. Nov 27, 2011 #6

    HallsofIvy

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    That not a definition, is it? I was under the impression that "-x" was the additive inverse of x: x+ (-x)= 0. So to prove that (-x)y= -(xy), you need to show that (-x)y is the additive inverse of xy: that (-x)y+ (xy)= 0.
     
  8. Nov 27, 2011 #7

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    Let's try to find the definition of a ring, and in particular the definition of the additive inverse.

    On wikipedia there is an article on a ring.
    It lists the requirements (aka axioms) of a ring:
    http://en.wikipedia.org/wiki/Ring_math#Formal_definition

    In particular we have:
    Btw, the typical shorthand for the additive inverse of [itex]a[/itex] is [itex]-a[/itex].

    Does this look familiar?
     
  9. Nov 27, 2011 #8
    (-x)y+xy=((-x)+x)y=0y=0
    ta da?
     
  10. Nov 27, 2011 #9

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    Yes! :smile:
    That's basically it.


    Just a couple of things.

    How do you know that 0y=0?

    And the definition requires that a+b=b+a=0.
    Do both equalities hold?
     
  11. Nov 27, 2011 #10
    well I did check that 0y=0, and a+b should equal b+a because its an abelian group. yay :) thanks again!
     
  12. Nov 27, 2011 #11

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    You're making this a bit easy on yourself, aren't you?
    How did you check this?
    If it is so simple, you should be able to easily reproduce the proof...
    (It is not trivial!)


    Yep!
     
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