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Math Help - Question in rings

  1. #1
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    Question in rings

    Hi All:
    I need the solution of the following question:

    Let (R,+,.) satisfy all conditions of a ring except the condition of a+b=b+a for all a ,b in R ( abelian with +); If 1 in R (R is arig with unity) prove that a+b=b+a for all a,b in R.

    Best wishes
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  2. #2
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    Quote Originally Posted by raed View Post
    Hi All:
    I need the solution of the following question:

    Let (R,+,.) satisfy all conditions of a ring except the condition of a+b=b+a for all a ,b in R ( abelian with +); If 1 in R (R is arig with unity) prove that a+b=b+a for all a,b in R.

    Best wishes

    Existence of additive inverse: (a+b)+(-(a+b))=0

    But also a+b+(-b-a)=a+(b+(-b))+(-a)=a+0+(-a)=a+(-a)=0 , so by uniqueness of inverse

    -b-a = -(a+b)=-a-b .

    Justify each step above and end the argument.

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    a+b+(-b-a)=a+(b+(-b))+(-a)=a+0+(-a)=a+(-a)=0 , so by uniqueness of inverse

    -b-a = -(a+b)=-a-b .


    Tonio
    Thank for your reply but this ring is not a belian under additive is it true that -(a+b)=-a-b=-b-a
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  4. #4
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    Quote Originally Posted by raed View Post
    Thank for your reply but this ring is not a belian under additive is it true that -(a+b)=-a-b=-b-a

    I didn't say it was: that's what you must prove! But you said that "ring" fulfills all the other ring axioms, so there's distributivity:

    -(a+b)=(-1)(a+b)=(-1)a+(-1)b=-a-b . That this equals -b-a follows from uniqueness of the additive inverse.

    Read again my message.

    Tonio
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  5. #5
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    Quote Originally Posted by tonio View Post
    I didn't say it was: that's what you must prove! But you said that "ring" fulfills all the other ring axioms, so there's distributivity:

    -(a+b)=(-1)(a+b)=(-1)a+(-1)b=-a-b . That this equals -b-a follows from uniqueness of the additive inverse.

    Read again my message.

    Tonio
    Thank You Very Much
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