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What does $\forall x \exists y(x + y = 0)$ mean?

Does it mean "For all x there exists a y for which x + y equals zero"?

Thanks.

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    $\begingroup$ tip: You may accept an answer that you find helpful (you can accept one answer per question) by clicking on the $\checkmark$ to the left of the answer you'd like to accept :-) $\endgroup$ – Namaste Feb 14 '13 at 21:40
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Yes, it does.

In plainer, more succinct words, it means that "every number has an additive inverse".

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    $\begingroup$ Also, be absolutely sure not to confuse this with $\exists y \forall x (x + y = 0)$. The two statements are subtly different, in that in the first one, the $y$ can be different for each $x$, but in the second one, the same $y$ needs to work for all $x$. $\endgroup$ – Joe Z. Feb 14 '13 at 18:26
  • $\begingroup$ Think of it as the "such that" always immediately following the variable that the $\exists$ refers to. "For all $x$ there exists a $y$ such that $x + y = 0$" is not the same as "There exists a $y$ such that for all $x$, $x + y = 0$". And "There exists a $y$ for all $x$ such that $x + y = 0$" would not be a valid transcription of a first-order logic sentence (even though syntactically it should be equivalent to $\forall x \exists y (x + y = 0)$). $\endgroup$ – Joe Z. Feb 14 '13 at 18:43
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Yup, you're correct.

However, it seems a bit clearer if it is written (as suggested in the comments) as

$$\forall x:\exists y : (x + y = 0)$$

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    $\begingroup$ I find this notation unwieldy as well, though I have seen it in Rosen's "Discrete Mathematics and Applications" book. $\endgroup$ – JavaMan Feb 14 '13 at 18:27
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    $\begingroup$ In statements in first-order logic, there is no "such that". It's implicit in the $\exists$. $\endgroup$ – Joe Z. Feb 14 '13 at 18:28
  • $\begingroup$ Such that is denoted by $\mid$, isn't it? $\endgroup$ – saadtaame Feb 14 '13 at 18:28
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    $\begingroup$ I think you just pinged yourself. :P $\endgroup$ – Joe Z. Feb 14 '13 at 18:30
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    $\begingroup$ If at all I'd prefer the colon as a separator,$\forall x\colon\exists y\colon x+y=0$ , but only because its general spacing introduces legibility $\endgroup$ – Hagen von Eitzen Feb 14 '13 at 18:56

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