Proof clarification - If $ab = 0$ then $a = 0$ or $b =0$

Question

I came across a proof for the following theorem in Apostol Calculus 1. My question is regarding (1) in the proof, why is this part necessary? I don't see why you can't begin with (2)

Theorem 1.11

If $ab = 0$ then $a = 0$ or $b=0$

Proof

Let $a, b \in \mathbb{R}$ with $ab =0$

Then, if $a \neq 0$ we know there exists $a^{-1} \in \mathbb{R}$ such that $a * a^{-1} = 1$

Thus, $$\begin{align} ab = 0 &\implies a^{-1}(ab) = a^{-1} \cdot 0 = 0\tag{1}\label{1} \\ \end{align}$$ But,

$$\begin{align} a^{-1}(ab) = 0 &\implies (a^{-1}a)b = 0\tag{2}\\ &\implies 1\cdot b = 0\tag{3}\\ &\implies b = 0 \tag{4} \end{align}$$

Answer 1

Looks a bit weird. If $a\ne 0$, then $a$ is invertible and so $$b = 1b = (a^{-1}a)b = a^{-1}(ab) = a^{-1}0 = 0.$$ In the last step, $b0=0$, I used the fact that $0$ is absorbing.

In the general case, if you have a ring $R$ and a unit $a\in R$, then the above proof shows that units are not zero divisors.

Answer 2

Suppose $ab=0$ and $a\neq 0$. Then $a^{-1}(ab)=(a^{-1}a)b=b=0$