This was just a good excuse to use LaTeX.

# Introduction

Okay, what’s all the hubbub about dividing by 0? Why can’t you do it? Well, the issue runs deeper than just your garden variety numbers you see…

If you know what “abelian groups” are, you can skip this. Otherwise, forget everything you learned about numbers and basic arithmetic.

Let’s suppose we have a well-defined collection of unique items, also known as a set. We’ll name this $\mathbb{F}$. Let’s also suppose we invent 2 binary operations to perform on the elements of this set, one named “addition” and one named “multiplication.” Note that binary operations map two values of $\mathbb{F}$ to a unique value in $\mathbb{F}$: $\mathbb{F}\times\mathbb{F}\to\mathbb{F}$.

“Addition,” denoted $+$ , has the following properties by definition. Note that $a$, $b$, $c$ are arbitrary elements of $\mathbb{F}$.

1. Commutativity: $a+b=b+a$
2. Associativity: $(a+b)+c=a+(b+c)$
3. Additive Identity: There exists an element, $0$, in $\mathbb{F}$ such that $a+0=a$
4. Additive Inverse: For every $a$ in $\mathbb{F}$ there exists a unique element, $-a$ (the additive inverse of $a$), such that $a+(-a)=0$

And “multiplication,” denoted $\cdot$ , has the following properties by definition:

1. Commutativity: $a\cdot b = b\cdot a$
2. Associativity: $(a\cdot b) \cdot c = a \cdot (b \cdot c)$
3. Multiplicative Identity: There exists an element, $1$, in $\mathbb{F}$ such that $a\cdot 1 = a$
4. Multiplicative Inverse: For every $a$ in $\mathbb{F}$ there exists a unique element, $a^{-1}$ (the multiplicative inverse of $a$), such that $a\cdot a^{-1}=1$

And finally, the distributive property: $a\cdot(b+c) = a\cdot b + a\cdot c$. Cool, if you noticed, we just (classically) defined $\mathbb{F}$ as a field, dodging abelian groups and abstract algebra.

# The Proof

First, let’s see how the additive identity behaves under multiplication. Suppose we have

\begin{aligned} &a \cdot 0 = a\cdot (0+0) = a \cdot 0 + a\cdot 0 \\ \implies &a\cdot 0 + (-a\cdot 0) = a \cdot 0 + a\cdot 0 + (-a\cdot 0) \\ \implies &a\cdot 0 = 0 \end{aligned}

Therefore, the additive identity also behaves as the multiplicative absorbing element (such a delightful name).

By rule 4 of multiplication, $0$ should have an inverse, $0^{-1}$. First consider $$0 \cdot (0 \cdot 0^{-1}) = 0 \cdot (1) = 0$$

Then consider

$$(0 \cdot 0) \cdot 0^{-1} = (0) \cdot 0^{-1} = 1$$

By the associativity of multiplication, we expect $$0 \cdot (0 \cdot 0^{-1}) = (0 \cdot 0) \cdot 0^{-1} \implies 0 = 1$$

However, this is nonsense! Suppose we have a field with an element $a \neq 0$, then obviously $a \cdot 0 = 0$ and $a\cdot 1 = a$. If $0=1$, then $a = 0$ but $a\neq 0$.

You see, $0$ is multiplication’s problem child. Notice how we didn’t even mention numbers, repeated addition or anything of that sort; yet still, $0$ manages to break multiplication.