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Math Help - Prove that for all a,b in a ring R, (-a)b=-ab=a(-b)

  1. #1
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    Prove that for all a,b in a ring R, (-a)b=-ab=a(-b)

    Prove that for all a,b in a ring R, (-a)b=-ab=a(-b)
    The hint says use: a+(-a)=0

    Since any integer has a negative, we know a+(-a)=0

    Multiple by b:

    (a+(-a))b=0(b)\Rightarrow ab+(-a)b=0

    How can I show (-a)b=-ab=a(-b) from the above?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Re: Prove that for all a,b in a ring R, (-a)b=-ab=a(-b)

    Quote Originally Posted by dwsmith View Post
    Prove that for all a,b in a ring R, (-a)b=-ab=a(-b)
    The hint says use: a+(-a)=0

    Since any integer has a negative, we know a+(-a)=0

    Multiple by b:

    (a+(-a))b=0(b)\Rightarrow ab+(-a)b=0

    How can I show (-a)b=-ab=a(-b) from the above?
    Interesting approach (I don't have a problem with it). Note that -ab=-(ab). So the first part follows right away: (-a)b=-ab.

    For the second part, you do pretty much the same thing; however, you start with b+(-b)=0 and then multiply by a on the left:

    a(b+(-b))=0\implies\ldots

    Then because '=' is transitive, it now follows that (-a)b=-ab=a(-b).

    I hope this makes sense.
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  3. #3
    Lord of certain Rings
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    Re: Prove that for all a,b in a ring R, (-a)b=-ab=a(-b)

    Quote Originally Posted by dwsmith View Post
    Prove that for all a,b in a ring R, (-a)b=-ab=a(-b)
    The hint says use: a+(-a)=0

    Since any integer has a negative, we know a+(-a)=0
    integer?
    I think you mean the ring is a group under addition and thus every element in the ring has an additive inverse.

    Multiple by b:

    (a+(-a))b=0(b)\Rightarrow ab+(-a)b=0
    Have you proved that b.0 = 0 for all elements b in the ring?

    How can I show (-a)b=-ab=a(-b) from the above?
    Your approach is right.
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