Where is the error in this "proof" that $1=0$?

Question

Suppose

$$x-2z = 0$$

Here is what I did to the equation, by dividing $x-2z$ on both sides,

$$\frac{x-2z}{x-2z} = \frac{0}{x-2z} \implies 1 = 0$$

I don't understand where I am wrong.

Answer 1

Since $x-2z=0$, dividing by $x-2z$ is dividing by zero; such a thing cannot be done (outside of Wheel theory).

Answer 2

Well, suppose $x \ne 2z$. Then, sure, you can divide like this, but you would be starting on a false premise, since if $x \ne 2z$, then $x - 2z \ne 0$, so you would be starting with the claim that $a = 0$ for some $a \ne 0$. Obviously nonsensical; if we start with a falsehood, we can prove all sorts of nonsense.

On the other hand, suppose $x = 2z$. Then sure, you start with the true statement $0=0$. But then you are dividing by $x-2z$, which equals $0$. You can't divide by zero though, so you can't do that either.

So in either case, your argument is either starting from a falsehood, or outright doing illegal things.

Answer 3

See if division by $0$ were permissible, we would have

$$7\times 0=9\times 0\Rightarrow 7=9$$ In general, $$a\times 0=b\times 0\Rightarrow a=b, \forall a,b\in \mathbb{C}$$