# Logical proof of the statement $xy = 0 \implies x=0\text{ or } y=0$

Claim:

If $xy=0$, then $x=0$ or $y=0$.

My proof is as follows: case 1: $x=0$, so $0y=0$ case 2: $y=0$, so $x0=0$ Either way, $xy=0$.

I'm very confused by this myself. So if I let $xy=0$ be $P$, and $x=0$ or $y=0$ be $Q$, then the claim "if $xy=0$, then either $x=0$ or $y=0$" is asking me to prove that $P \implies Q$. But I feel as if the proof I gave is $Q \implies P$, which is very different from $P \implies Q$. Can anyone enlighten me on the subject of proving if-then statements?

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You have to specify the domain you are working in. An example: In ${\mathbb Z}_6$ we have $2\ne0$, $3\ne0$, but $2\cdot 3=0$. –  Christian Blatter Oct 18 '13 at 9:17

Your feeling is right: What you have done is the opposite direction. Your argument proves $$(x = 0 \text{ or } y = 0) \implies xy = 0.$$

How can you prove a statement of the form $$A \implies B$$ in general? The direct method is to assume that $A$ is true, and then to conclude that also $B$ is true under this assumption.

Let's apply this to prove the statement $$xy = 0 \implies (x = 0 \text{ or } y = 0).$$ In this case \begin{align*} A & = \text{''}xy = 0\text{''} \\ B & = \text{''}x = 0\text{ or }y = 0\text{''} \end{align*}

So we assume that $xy = 0$ is true. Now we have to show that $(x = 0 \text{ or } y = 0)$ is true. The nature of an "or"-statement often involves a case by case study:

1. If $x = 0$, then of course $(x = 0 \text{ or } y = 0)$ is true.
2. Otherwise, we have $x \neq 0$. Now we may divide our assumption (the equation $xy = 0$) by $x$ to get $y = 0$, so $(x = 0 \text{ or } y = 0)$ is true also in this case.

We have just proven $$xy = 0 \implies (x = 0 \text{ or } y = 0),$$ and the argument in your question proves $$xy = 0 \Longleftarrow (x = 0 \text{ or } y = 0).$$ So in fact, we have an equivalence, which we can write down as $$xy = 0 \iff (x = 0 \text{ or } y = 0).$$

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well, it only happens in an integral domain, or fields as we generally works in which are integral domains automatically. if you are in, say, Z6 , its not true, as take 3*2=6mod6=0 where none of 3 and 2 are zero. but i guess u have already assumed it over reals which is a field, in that case, T.P xy=0⟹x=0 or y=0, you can assume y≠0, and then multiply on both sides by y^-1 which will give you x.1=0*y^-1=0 implies x=0. for integral domains, its basically the definition if an integral domain.

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You should consider using MathJax. –  Student Jul 8 '14 at 18:47

Not quite right, as you have it written you merely show that if either $x$ or $y$ are zero then their product is zero. Not that if the product is zero then either $x$ or $y$ must necessarily be zero.

You can show this by contradiction; assume $xy=0$ but $x\neq0$ and $y\neq 0$ then $xy\neq 0$ which contradicts our original assumption that $xy=0$ so we cannot have that both $x$ and $y$ are non zero. In other words at if $xy=0$ at least one of $x$ and $y$ must be zero.

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‘Not quite right’ is an understatement: it’s simply wrong, I’m afraid. –  Brian M. Scott Oct 18 '13 at 8:16

If $x\ne0$, what happens if you multiply both sides by $\frac1x$?

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Just another way. It might not be a formal method.

The solution satisfying the following equation $$A \times B =0$$ is $A=0$ (for any $B$) or $B=0$ (for any $A$).

You cannot apply the same pattern for the case in which the right hand side is not zero. Why? For example, $$A\times B = 2$$ If you choose $A=2$ then $B$ must be $1$ (rather than for any $B$). If you choose $B=2$ then $A$ must be $1$ (rather than for any $A$).

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