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

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}

• Line (2) uses the associativity of multiplication of the real numbers, that is, for any real numbers $a, b, c$ then $a(bc) = (ab)c$. The reason you can't start with (2) is because when you multiply $a^{-1}$ to both sides of $ab = 0$ then you are implicitly multiplying $a^{-1}$ to $ab$ and so you must use associativity at some point. – symchdmath Jan 12 at 11:49
• ah ok, is it because line two doesn't show $a^{-1}(ab) = a^{-1} \cdot 0 \implies a^{-1}(ab) = 0$ (using $0 \cdot a = a \cdot 0 = 0$ which was proved as theorem 1.6 in the chapter) – Jake Kirsch Jan 12 at 11:58
• and then associativity to get $(a^{-1} a)b = 0$ – Jake Kirsch Jan 12 at 11:59
• Yep, you can see it as combining it with transitivity of equality in one line – symchdmath Jan 12 at 12:19

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.