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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?

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In any way of axiomatizing the concept of field I'm familiar with, $a\cdot0=0$ is not an axiom, it is a theorem that one proves using the axiom that multiplication distributes over addition. It might help, therefore, to actually list what you have been given as the axioms for a field. –  Zev Chonoles Oct 21 '11 at 19:18
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(Off-topic: That's an interesting distinction Mathworld draws between additive and multiplicative distributivity ...) –  Henning Makholm Oct 21 '11 at 19:27
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Hint: Suppose that $xy=0$. If $x\ne 0$, then $x$ has a multiplicative inverse. –  André Nicolas Oct 21 '11 at 19:31
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@queueoverflow: Note that you aren't "simplifying" the original question. Rather, you are replacing the statement you want to prove (which is of the form $\neg p\Longleftrightarrow \neg q$) with a statement which is logically equivalent to it (namely $p\Longleftrightarrow q$). –  Arturo Magidin Oct 21 '11 at 19:45
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Depends on your specific list of axioms, and previously proved theorems that you can quote, but probably yes. Out of habit, though it makes no real difference, I would multiply by $x^{-1}$ on the left. Of course, this does one direction only, but it is the "harder" direction. –  André Nicolas Oct 21 '11 at 19:54

3 Answers 3

up vote 1 down vote accepted

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?

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In case I can "multiply both sides", then I can simply do $x^{-1}xy = x^{-1}0$, obtaining $y=0$. –  queueoverflow Oct 21 '11 at 19:58
    
Exactly what you wanted, right? –  Austin Mohr Oct 21 '11 at 19:59
    
Absolutely. I was just not sure, whether I am allowed do "divide by x" since it is not on the axiom list explicitly. But now I have both sides of my problem done. Thanks! –  queueoverflow Oct 21 '11 at 20:04
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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 $x^{-1}xy = x^{-1}0$. –  Austin Mohr Oct 21 '11 at 20:05

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 $y=0$?

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I know that $0a=0$ for any $a$. Therefore, if I start with $y=0$ ($0=0$) and use this axiom to make it $xy=0$, then a non-zero $x$ would make the $y$ $0$. Is that a proof? –  queueoverflow Oct 21 '11 at 19:20
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@queue, in the last "then" it looks like you're using the property you're supposed to be proving. Or do you have any particular reason in mind why $xy=0$ with a nonzero $x$ would imply $y=0$? You're supposed to end with $y=0$, not start with it. –  Henning Makholm Oct 21 '11 at 19:25
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@queue: That's a bunch of formulas. It's not clear to me how you think they relate to each other. What do the square brackets around $y0=0$ signify? Are you attempting to conclude $x=0$ from $x0=0$ (which won't work at all)? You need to be much more generous with the connecting prose in your reasoning. –  Henning Makholm Oct 21 '11 at 19:41
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@queue, multiplying everything by zero just loses every information the equation contains. Hint: you need to use the assumption that $x$ is nonzero, and the knowledge that you're working in a field rather than a ring. –  Henning Makholm Oct 21 '11 at 19:47
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@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 $y$? –  yunone Oct 21 '11 at 19:54

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).

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I have $p \Longleftrightarrow q$ in this case. Isn't that a different story? –  queueoverflow Oct 21 '11 at 20:17
    
But $p \Longleftrightarrow q$ is the same as both $p \implies q$ and $q \implies p$. You have already shown that $q \implies p$. –  JavaMan Oct 21 '11 at 20:44
    
Okay, and the other is easy to show since one of $x$ and $y$ being 0, $0a=0$ immediately shows this. –  queueoverflow Oct 21 '11 at 20:46

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