Prove that $(-a)b = -ab$ can you please tell me if my proof is right or if I'm taking some steps for granted? Thanks  $$\forall a, b\in \Bbb R \Rightarrow (-a)b=-ab$$  $$(-a)b\cdot (-a)^{-1} = -ab\cdot (-a)^{-1} \\ 1\cdot b = \frac{-ab}{-a} \\ b=b $$
Rings and some of their properties So I am studying for an exam and was wondering if anyone could explain some of the reasoning being the proofs of the following properties:   Let R be a ring with $a,b\in R$. Then    $a0=0a=0$  $a(-b)=(-a)b=-ab$ $(-a)(-b)=ab$   Proof 1:   $a0=a(0+0)=a0+a0$ hence $a0=0$   I don't really understand why $a0+a0$ proves that $a0=0$
Why is $\sqrt{-x}*\sqrt{-x}=-x?$ Q1 - Why is $\sqrt{-x}*\sqrt{-x}=-x?$  Q2 - I was thinking it would be:  $\sqrt{-x}*\sqrt{-x}=\sqrt{-x*-x}=\sqrt{x^2}$ but apparently not (why not?)  Q3 - What are the formal algebra rules to use? Can I calculate this without using i such as in: $\sqrt{-x}*\sqrt{-x}=i\sqrt{x}*i\sqrt{x}=-\sqrt{x^2}=-x$.