Proof of a ring property

Question

Let $R$ be a ring. Prove that, for every $a, b$ in $R$, $(-a)b = -(ab) = a(-b)$

Now what I did was:

1) $ab + (-a)b = (a-a)b$ (distributivity) $= 0b$ (additive inverse) $= 0$ (can easily be proved by ring axioms) and similarly, $(-a)b + ab = 0$.

2) $ab + (-ab) = 0 = (-ab) + ab$ (additive inverse of $ab$)

3) $ab + a(-b) = a(b-b) = a0 = 0$ (similar to the first)

And it can be proven that the additive inverse is unique for each element in a ring, therefore $(-a)b = -(ab) = a(-b)$.

Are there any mistakes with this?

Answer 1

It's good. The proof works. I have no criticisms, but I do have a couple of small, largely stylistic pointers:

• You can use the commutativity of the additive group to show that $(-a)b + ab = 0$ instead of saying "similarly".
• The second line of logic is not really necessary here; you showed in 1) and 3) that $(-a)b$ and $a(-b)$ both satisfy the definition of an additive inverse of $ab$, which means they are equal by the uniqueness of inverses, and we name such an element $-(ab)$.