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Is this a sufficient way of proving that for for all real numbers $x$ and $y$, if $xy=0$ then $x=0$ or $y=0$?

My attempt:

$$xy=0$$ $$y=\frac{0}{x}$$

$$x=\frac{0}{y}$$

Is this a correct approach using axioms of real numbers?

David G. Stork
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Mark
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    A mathematical proof consists of logical statements, not just symbols. –  Jan 16 '19 at 01:22
  • Perhaps you want to prove that if $x\not=0$, then $y$ must be zero. – Michael Burr Jan 16 '19 at 01:23
  • Well you can only divide by $x$ if you're sure that $x$ isn't zero, and the same goes for $y$, so you need to be a bit more careful about that. But you could combine this with what Michael Burr said. – timtfj Jan 16 '19 at 01:32

1 Answers1

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Let $x$ and $y$ be two real numbers and suppose $xy=0$. There are two cases:

  • $x=0$;
  • $x\ne 0$.

If $x=0$, then "$x=0$ or $y=0$" is true.

If $x\neq 0$, then $x^{-1}$ exists and multiplying this number on both sides of $xy=0$ gives $y=0$, which implies that "$x=0$ or $y=0$" is true.