What is this fallacy called? I tried to show that $\frac{1}{x} \neq 0$ by contradiction:
$$\begin{align} \frac{1}{x} &= 0\ \text{(multiplty by $x$)} \\[6pt]
1 &= 0\end{align}$$
Btw I know other ways to prove this (i.e. proving this is not the goal of my question)
Someone pointed out that you cannot time by $x$. I know why one cannot divide by $x$.
Say $x +1 = 2$. If I times by $x$, I get $x^2 -x = 0$ which leads to the solutions $x = 0,-1$. Obviously $x=0$ is extraneous and can be ruled out, also, this step was unnecessary but this isn't fallacious though- or is it?
 A: Your argument is correct. It is always the case that
$$a=b$$
implies
$$f(a)=f(b)$$
for any function $f$ applied to both sides. In particular,
$$\frac{1}x=0$$
implies
$$1=0$$
since we can take $f$ to be "multiply by $x$" - which is a perfectly good function, even if $x$ were zero (note that the same cannot be said of "divide by $x$"). Of course, $1\neq 0$, so this is a contradiction and thus it cannot be that $\frac{1}x=0$ for any $x$ since that implies a contradiction. (Or, perhaps more precisely we might say that if $y\cdot x=1$ then $y\neq 0$, in order to avoid possibly writing $\frac{1}0$ by accident)
The danger of multiplying by $x$ is that the statements $a=b$ and $f(a)=f(b)$ might not be equivalent - which allows for extra solutions to sneak in. In your example, if you start with
$$x-1=0$$
then multiplying by $x$ gives
$$x^2-x=0.$$
It's certainly true that if $x-1=0$ then $x^2-x=0$. The extraneous solution comes from the fact that $x^2-x=0$ doesn't imply that $x-1=0$ - but your argument isn't trying to use this reverse direction, so it's not an issue that extraneous solutions might pop in.
This is a point that sometimes gets overlooked, since algebra is most often done using functions such as "add $2$" or "subtract $x$" which are invertible (or at least "injective"), which allows us to prove the reverse implication - formally, if for the functions $f$, there's another $g$ so that $g(f(x))=x$, we can find that $f(a)=f(b)$ implies $g(f(a))=g(f(b))$ which then implies $a=b$ - meaning that then $a=b$ and $f(a)=f(b)$ are equivalent. Working with functions that aren't like this - such as squaring both sides or multiplying by some $x$ that might be zero - feels different, but as long as you keep track of the direction of the implication, there is absolutely no problem in doing so.
