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 $b=0$


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,

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

  • $\begingroup$ 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. $\endgroup$ – symchdmath Jan 12 at 11:49
  • $\begingroup$ 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) $\endgroup$ – Jake Kirsch Jan 12 at 11:58
  • $\begingroup$ and then associativity to get $(a^{-1} a)b = 0$ $\endgroup$ – Jake Kirsch Jan 12 at 11:59
  • $\begingroup$ Yep, you can see it as combining it with transitivity of equality in one line $\endgroup$ – 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.


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