Ring/Basic properties/Fact/Proof

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

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

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

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

3. Due to (2), we have ${\displaystyle {}(-a)(-b)=(-(-a))b}$, and because of ${\displaystyle {}-(-a)=a}$ (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.