Show in a ring we have a 0a a0 0 b ab ab ab c ab abSolu

Show in a ring we have:

(a) 0a = a0 = 0

(b) (a)b = a(b) = (ab)

(c) (a)(b) = ab

Solution

Let R be a ring and let a and b be elements of R.

then, to prove

(a) a0 = 0a = 0

Let x = a0, We have x = a0 = a(0 + 0) = a0 + a0 = x + x.

Adding x to both sides, we get x = 0, which is (1).

(b) Let y = a(b). We want to show that y is the additive inverse of ab, that is we want to show that y + ab = 0. We have y + ab = a(b) + ab = a(b + b) = a0 = 0, by (1). Hence it is (2).

(c) , (a) is the unique solution of (a)+x = 0R. But x = a is also a solution, so a = (a)

Now, (a)(b) = (a(b)) = ((ab)) = ab Hence Proved.