2
$\begingroup$

I have to show: $xy\neq0 \Leftrightarrow x\neq0 \wedge y \neq0 $

I think I can "simplify" it to this: $xy=0 \Leftrightarrow x=0 \vee y=0 $

Since $a\cdot0=0$ is an proven theorem, I can show: $x=0 \vee y=0 \rightarrow xy=0 $

But that is just one of the directions. How can I use the field axioms to show this theorem?

  • 0
    @queuoverflow: No, it doesn't necessarily hurt, nor is it a problem. I'm just saying that calling it a "simplification" is a misnomer. You aren't simplifying (you are not reducing the question to something 'smaller'), you are replacing the question with something which is logically equivalent to it.2011-10-21

3 Answers 3

0

This is much the same as the other answers, but you do not need to consider cases. If you want to show that $p \implies q$, you can assume $p \wedge \neg q$ is true and derive a contradiction.

Suppose $x \neq 0$, $y \neq 0$, and assume $xy = 0$. Then, $x$ and $y$ have inverses. Hence, $xy$ does too (this needs formal justification). But,...(this is where you come in).

  • 0
    Okay, and the other is easy to show since one of $x$ and $y$ being 0, $0a=0$ immediately shows this.2011-10-21
5

Hint: Suppose that $xy=0$. If $x=0$, then certainly $x=0\vee y=0$. Otherwise, $x\neq0$. What do we know about non-zero elements of a field that would let us conclude that $y=0$?

  • 1
    @queueoverflow If you assume $x\neq 0$, then you know $x^{-1}$ exists. Try multiplying $xy=0$ through by that. What can you conclude about $y$?2011-10-21
1

It helps to spell out what you're trying to show.

Claim: If $xy = 0$, then $x = 0$ or $y = 0$ (or both).

Proof: We'll consider two cases.

If $x = 0$ , then there is nothing to show (we already have one of $x$ and $y$ equal to 0, which is what we want).

If $x \neq 0$ , we hope to establish that $y = 0$ (since the conclusion we desire is that at least one of the two is equal to 0). Since $x \neq 0$ and we are in a field, the multiplicative inverse of $x$ exists.

With $x^{-1}$ in hand, how can you isolate $y$ in the equation $xy=0$ ? What do you learn about $y$ when you do so?

  • 2
    I think I see your confusion now. You are wondering whether "multiply both sides by something" is a legal maneuver when trying to prove something from axioms, correct? If so, think of the operation "multiply by $x^{-1}$" as a function $f$ (it is indeed a function). Since $xy = 0$, we know that $f(xy) = f(0)$, but this is just $x^{-1}xy = x^{-1}0$.2011-10-21