Confusion in how to handle the "or both" part of If $x, y \in \mathbb{F}$, then if $x\cdot y = 0$, then $x=0$ or $y = 0$ (or both).

Question

I am proving that if $x, y \in \mathbb{F}$ (an arbitrary field), then if $x\cdot y = 0$, then $x=0$ or $y = 0$ (or both).

What I am a bit confused about is the "(or both)" statement at the end.

I can think of several ways to prove this. The first is to do a direct proof, case by case. We have two cases: $x=0$ and $x\neq 0$. In the $x=0$ case, we trivially obtain the result. In the $x\neq 0$ case, $y$ is forced to be zero by multiplicative inverses of fields. However, I am not sure here where the "(or both)" statement comes in. It obviously seems to appear in the $x=0$ case, but that is a bit unclear to me. If $x=0$, then $y$ can be anything, but how can we justify this rigorously?

Additionally, I have tried to do the contradiction proof. That is, I am assuming the negation of "$x=0$ or $y = 0$ (or both)". The negation of this appears to be:

$$ x \neq 0 \ \text{AND} \ y \neq 0 \ \text{AND} \ NOT \ BOTH $$

Would this then turn into:

$$ x \neq 0 \ \text{AND} \ y \neq 0 \ \text{AND} \ \{(x=0, y \neq 0)\ or \ (x\neq0, y=0) \ or \ (x\neq0, y \neq 0)\} $$

and so because of the "and" conditions, we can just take the overlap of $ \{(x=0, y \neq 0)\ or \ (x\neq0, y=0) \ or \ (x\neq0, y \neq 0)\}$ with $x \neq 0 $ and $ y \neq 0 $, which is just $x \neq 0 \ \text{AND} \ y \neq 0 $.

In other words (apologies for being long-winded):

$$ x \neq 0 \ \text{AND} \ y \neq 0 \ \text{AND} \ NOT \ BOTH $$

$$ \implies $$

$$ x \neq 0 \ \text{AND} \ y \neq 0 \ \text{AND} \ \{(x=0, y \neq 0)\ or \ (x\neq0, y=0) \ or \ (x\neq0, y \neq 0)\} $$

$$ \implies $$

$$ x \neq 0 \ \text{AND} \ y \neq 0 $$

It is obviously easy to obtain the contradiction from this, but I am not sure if the "(or both)" statement is addressed fully here. Does anyone have any ideas? Thanks.

Answer 1

It's usually superfluous to use "or both", even in English, unless you want to make very sure that the other party does not misinterpret it. This is why the phrase was placed in parentheses. It also means that you can simply ignore it. Also, $P \lor Q \lor ( P \land Q ) \equiv P \lor Q$ for any assertions $P,Q$.