Prove that $(-a)b = -ab$

Question

can you please tell me if my proof is right or if I'm taking some steps for granted? Thanks

$$\forall a, b\in \Bbb R \Rightarrow (-a)b=-ab$$

$$(-a)b\cdot (-a)^{-1} = -ab\cdot (-a)^{-1} \\ 1\cdot b = \frac{-ab}{-a} \\ b=b $$

Answer 1

The whole approach doesn't make sense. You deduce that $b=b$. So what? What can you deduce from that? And how do you know that $a^{-1}$ exists?

You can observe that$$(-a)b+ab=\bigl((-a)+a\bigr)b=0\times b=0.$$Therefore, $(-ab)=-(ab)$. Of course, this assumes that you already know that $(\forall x\in\mathbb{R}):0\times x=0$.

Answer 2

You have $(-a)b + ab = (-a+a)b = 0\cdot b = 0 \implies (-a)b = -ab$

Answer 3

Hint:

What is meant by $-ab$ is the additive inverse of $ab$, that is, the unique element such that $$(-ab)+ab=0$$

To show that $(-a)b=-ab$, you need to show that $$(-a)b+ab=0$$

Answer 4

You've started your proof by assuming that which you want to prove. Then you derived a tautology ($b=b$). There are several issues with this.

The first is that this is regarded as a sloppy form of proof. The only way that assuming what you want to show and deriving a tautology produces a logically valid argument, is if the steps can be reversed to go from the tautology to that which you want to show. And even in this case, people who read a lot of proofs are going to regard it as fishy.

Now in this proof specifically, certain steps are not reversible. When you multiply by $(-a)^{-1}$, for instance, you have added in the assumption that $a \neq 0$. That's not part of the original proposition.

Also, buried in the algebra, you have once again assumed that which you want to prove. When you go from $\frac{-ab}{-a}$ to $b$, you are factoring the numerator as $\frac{(-a)b}{-a}$. So the entire argument is circular.

You've already got some hints on how to proceed. But essentially, because you have a statement about additive inverses, you should use addition to prove it, not multiplication or division.