Question

Let (R, +, ×) be a system satisfying all axioms of a ring with identity, except possibly a + b = b + a. Show that a + b = b + a for all a, b ∈ R, so R is indeed a ring

Answer 1

By the two distributive laws, we have
(a + b)(1 + 1) = a(1 + 1) + b(1 + 1) = a + a + b + b,
(a + b)(1 + 1) = (a + b)1 + (a + b)1 = a + b + a + b.
Using the additive group laws, we deduce that a + b = b + a for all a, b ∈ R