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Answer to Question #16529 in Abstract Algebra for Hym@n B@ss

Question #16529
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
Expert's answer
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

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