Ring/Basic properties/Fact/Proof

Proof

In the noncommutative case we only proof one half of the statements.

1. We have ${}a0=a(0+0)=a0+a0$. Subtracting ${}a0$ (meaning addition with the negative of ${}a0$) on both sides gives the claim.

2. ${}(-a)b+ab=(-a+a)b=0b=0\,$

   due to part (1). Therefore ${}(-a)b$ is the (uniquely determined) negative of ${}ab$.

3. Due to (2), we have ${}(-a)(-b)=(-(-a))b$ and because of ${}-(-a)=a$ (which holds in every group) we get the claim.
4. This follows from the parts proved so far.
5. This follows with a double induction

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