# 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

(

(

Using the additive group laws, we deduce that

(

*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*
## Comments

## Leave a comment