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Math Help - ring axiom

  1. #1
    Junior Member
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    Exclamation ring axiom

    Let R be a ring.
    (a) Show that
    a.0 = 0.a = 0 for all a belongs toR.
    (b) Using (a), show that for any
    a, b in R, (-a)b = -(ab):

    (c) Using (b), show that for any a, b in R, (-a)(-b) = ab:

    (a) so far I have done part (a) and showed that a.0 = 0.a = 0
    and the steps are as follows:

    We have 0+0 = 0, since 0 is the zero element. Multiply both sides by a:

    a0+a0 = a(0+0) = a0 = a0+0;

    where the last equality uses the zero law again. Now from
    a0+a0 = a0+0, we get a0 = 0 by the cancellation law. The other part 0a = 0 is proved similarly.

    But I couldnt do part (b) and (c) help would be appreciated thank you..

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  2. #2
    MHF Contributor red_dog's Avatar
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    b) 0=0\cdot b=(a+(-a))b=ab+(-a)b\Rightarrow (-a)b=-ab

    c) (-a)(-b)=-a(-b)=-(-ab)=ab
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