# ring axiom

• Mar 10th 2009, 04:08 AM
rajr
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..

• Mar 10th 2009, 07:15 AM
red_dog
b) $0=0\cdot b=(a+(-a))b=ab+(-a)b\Rightarrow (-a)b=-ab$

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