If R is a ring with additive identity 0, then for any a, b E R we have 1. 0a = a0 = 0, 2. a(-b) = (-a)b = -(ab), 3. (-a)

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If R Is A Ring With Additive Identity 0 Then For Any A B E R We Have 1 0a A0 0 2 A B A B Ab 3 A 1 (39.95 KiB) Viewed 44 times

If R is a ring with additive identity 0, then for any a, b E R we have 1. 0a = a0 = 0, 2. a(-b) = (-a)b = -(ab), 3. (-a)(-b) = ab Proof: For Property 1, note that by axioms R₁ and R₂, a0 + a0 = a(0 + 0) = a0 = 0 + a0. Then by the cancellation law for the additive group (R, +), we have a0 = 0. Likewise, 0a +0a = (0+ 0)a= 0a = 0 + 0a implies that 0a 0. This proves Property 1. In order to understand the proof of Property 2, we must remember that, by definition, -(ab) is the element that when added to ab gives 0. Thus, to show that a(-b) = -(ab), we must show precisely that a(-b) + ab = 0. By the left distributive law, a(-b) + ab = a(-b+b) = a0 = 0, since a0 = 0 by Property 1. Likewise, For Property 3, note that (-a)b + ab = (-a + a)b = 0b = 0. (-a)(-b)=-(a(-b)) by Property 2. Again, by Property 2, Example: -(a(-b))=-(-(ab)). and-(-(ab)) is the element that when added to-(ab) gives 0. This is ab by definition of -(ab) and by the uniqueness of an inverse in a group. Thus, (-a)(-b) = ab.