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I'm trying to prove this following theorem:

If $x, y \in \mathbb{Z}$, use the cancellation law for $\mathbb{Z}$ to demonstrate that $xy = 0 \implies$ $x = 0$ or $y = 0$

The proof I came up with doesn't quite seem definitive enough. I know how to prove this without using the cancellation law, but this requirement seems to make things much more difficult.

So, clearly $0 = 0 \cdot x = 0 \cdot y$ $\forall x, y \in \mathbb{Z}$. So, we can clearly say that this holds for $x \neq 0$ and $y \neq 0$.

So, first we can write that $xy = 0 \cdot x$ for $x \neq 0$. So, by the cancellation law, $y = 0$. Similarily, we can write that $xy =0 \cdot y$ for $y \neq 0$, so $x = 0$ by the cancellation law.

It seems to me that we can write $xy = 0 \cdot a$ for any integer, so this doesn't quite "prove" anything, though in such a case we wouldn't be able to use the cancellation law, so that wouldn't at all be a pertinent fact.

How does this sound?

  • Which is the "cancellation law" (for $\Bbb Z$) ? – dan_fulea Jul 25 '18 at 17:53
  • I believe it is that, for $x, y, z \in \mathbb{Z}$, $x \cdot y = y \cdot z \wedge z \neq 0 \implies x = y$. At least, this is the theorem I should be using, though it's possible that this is a more colloquial name for it. –  Jul 25 '18 at 17:55
  • Then assume the contrary, that both $x,y$ are not zero and get the contradiction from $x0=xy=0y$. – dan_fulea Jul 25 '18 at 17:57

  • "So, first we can write that $xy=0⋅x$ for $x≠0$. So, by the cancellation law, $y=0$. Similarily, we can write that $xy=0⋅y$ for $y≠0$, so $x=0$ by the cancellation law"

    But wait, if $x\ne 0$ then $x=0$, no need the second part.

     – ℋolo Jul 25 '18 at 18:03
  • "It seems to me that we can write xy=0⋅a for any integer, so this doesn't quite "prove" anything" What do you mean. You just proved either $y=0$ or $x=0$. So why does $xy = 0a$ cancel that? If you have to prove George was an American and you proved that George was born in Maine and said "Yeah, but this would be true of everybody* born in Maine so it doesn't prove anything" um.... why not? – fleablood Jul 25 '18 at 18:23
  • If $xy = 0a$ then $xyy^{-1} = 0(ay^{-1}) $ so $x = 0$. No, difference. That still* proves .... everything. – fleablood Jul 25 '18 at 18:24

2 Answers2

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The cancellation law says that if $c\ne 0$ and $ac = bc$ then $a=b$.

If $xy=0$ and $x\ne 0$, then since $xy = x0$, $y=0$.

John Douma
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It seems to me that we can write $xy=0⋅a$ for any integer, so this doesn't quite "prove" anything,

And if $y\ne 0$ you can write $x = x*y*y^{-1} = 0*a*y^{-1} = 0*(a*y^{-1}) = 0$ for any integer. So that proves everything.

fleablood
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