1. ## Rings

Suppose that $R$ is a set with addition and multiplication defined that satisfy all of the axioms of a ring except possibly for $a+b=b+a$ (commutativity of addition). Prove that $a+b=b+a$, and hence that $R$ is a ring.

Now, we want to show that the group $(R, +)$ is abelian. Since, not every group is abelian, we will have to use the properties of multiplication. The only axiom that links addition to multiplication is the axiom of distributivity so if my reasoning is correct the proof should use that axiom. However, I am a little stuck on this and would really appreciate any hints.

2. Originally Posted by nmatthies1
Suppose that $R$ is a set with addition and multiplication defined that satisfy all of the axioms of a ring except possibly for $a+b=b+a$ (commutativity of addition). Prove that $a+b=b+a$, and hence that $R$ is a ring.

Now, we want to show that the group $(R, +)$ is abelian. Since, not every group is abelian, we will have to use the properties of multiplication. The only axiom that links addition to multiplication is the axiom of distributivity so if my reasoning is correct the proof should use that axiom. However, I am a little stuck on this and would really appreciate any hints.
we do need $R$ to have $1,$ the multiplicative identity, otherwsie your claim would be wrong. the proof now is quite easy:

$(1+1) \cdot (a+b)=1 \cdot (a+b) + 1 \cdot (a+b)=a+b+a+b$ and $(1+1) \cdot (a+b)=(1+1) \cdot a + (1+1) \cdot b=a+a+b+b.$ thus $a+b+a+b=a+a+b+b$ and the result follows.