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~(∃ x ∈ Z | ∀ y ∈ Z, y/x ∈ Z)

= ∀ x ∈ Z | ∀ y ∈ Z, y/x ∈ Z

I'm just not 100% sure. I'm under the impression that if expressions following "such that" are not included in the quantifier they do not get negated. Is that correct?

And if that is true, how does one read the bottom expression? It seems bizarre to me.

2 Answers 2

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Hint:

Assuming Z refers to the integers, your original un-negated statement reads:

There exists an integer $x$ so that for any integer $y$, the value of $\frac{y}{x}$ is an integer.

(It's referring to the integers $1$ and $−1$.)

Your (incorrectly) negated statement is read:

For any integers $x$ and $y$, the value of $\frac{y}{x}$ is an integer.

Correctly negated, it should read:

For any integer $x$, there is an integer $y$ so that $\frac{y}{x}$ is not an integer.

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    Thanks for the beautiful explanation. I understand it conceptually (like an "English" translation of negation, if you will) but mathematically converting is where I'm having trouble. Could you see the comment on the other answer?2019-02-22
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The negation is

$ \forall x \in \mathbb{Z}, \exists y \in \mathbb{Z} \,\,\,\,\,\, \frac{y}{x} \notin \mathbb{Z}$

When you have a statement of the form

$\forall \exists \exists \forall \exists \cdots P$

that is, a bunch of quantifiers before a quantifier-free formula, you negate by turning $\forall$ into $\exists$ and vice versa, and then negating $P$ to get

$\neg (\forall \exists \exists \forall \exists \cdots P) = \exists \forall \forall \exists \forall \cdots \neg P$

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    @DougSmith: The "such that" is just there to make it read nicely in English. Most people leave it out, I believe. For example, I would have written $\exists x \in \mathbb{Z} \; \forall y \in \mathbb{Z} \; \frac{y}{x} \in \mathbb{Z}$ for the original expression.2012-09-25