Tomi's Challenges - #24

Feel free to write a report if you don't agree with me, with your own reasoning.

This is a very common proof, but I think you can't really do it this way. Here is why. Multiplication is indeed an inverse operation of division and vice versa, but ONLY for non-zero numbers. Multiplication is NOT an inverse operation of division by zero. To see why, let's look at the formal definition of inverse elements:

$b$ is called an inverse of $a$ if: $a\circ b=b\circ a=e$ , where $e$ is the identity element.

$\mathbb{R}$ is very commonly defined as an ordered field. Let's look at what a field really is. The formal definition of a field is that it's a ring with operations $+$ and $\times$ such that all elements which are not equal to the identity element in terms of $+$ , which is $0$ , form a group in terms of $\times$ . Since the group of elements doesn't even contain $0$ in it, it's tempting to think that $x\cdot 0$ also isn't defined in fields. But $x\cdot 0=0$ follows from the ring axioms instead. Given our axioms, there is no inverse element in terms of multiplication in the case of $0$ , which is why we cannot say that multiplication is an inverse operation of division.

The error in the reasoning is introduced after multiplication by $0$ on both sides: That step leads to the true statement $"0=0"$ (from which nothing follows at all!!) instead of the statement $"0\cdot x=6"$ . I'll try my hand at a correct proof below using field axioms only.

---

Let $r'$ denote the additive inverse of any $r\in\mathbb{R}$ . Let $0=0+0$ and $1=1\cdot 1$ denote the additive and muliplicative neutral elements in $\mathbb{R}$ . Suppose $0$ had a multiplicative inverse $x\in\mathbb{R}$ . Then $0\cdot x$ has the additive inverse $(0\cdot x)'$ with $\begin{aligned} 0&=0\cdot x + (0\cdot x)' = (0+0)\cdot x + (0\cdot x)' \underset{(*)}{=} 0\cdot x + \red{\Bigl(0\cdot x + (0\cdot x)'\Bigr)}=0\cdot x+\red{0}=0\cdot x=1 \end{aligned}$ In step $(*)$ we use the associative property of addition and distributive property. As $0\neq 1$ in $\mathbb{R}$ , we have a contradiction.