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Consider the statement

$$ \exists x\in\mathbb{R} \; \forall y\in\mathbb{R}, \; x^{2}+y^{2}\leq 1 \Rightarrow xy\neq 0 $$

The negation of it (i.e.) $$\lnot(\exists x\in\mathbb{R} \forall y\in\mathbb{R}, \; x^{2}+y^{2}\leq 1 \Rightarrow xy\neq 0)$$

should be (not sure if this is correct)$$\forall x\in\mathbb{R} \; \exists y\in\mathbb{R}, \; x^{2}+y^{2}\leq 1 \land xy=0$$

Which of these statements is true? I'm having trouble figuring it out.

The first one should be false because if $x=0$ and $y<1$ then $x^{2}+y^{2}$ will be true but $xy$ will still equal $0$ and be false.

In the second one if $x>1$ and $y=0$ then $x^{2}+y^{2}$ will be greater than $1$ (be false) and $xy$ will equal $0$ (true), so all together it's false (for conjunction to be true both have to be true, right?)

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    $\begingroup$ Welcome to MSE. What are your thoughts on the question? Please edit the post and share your thoughts so that it becomes clearer where exactly you have a problem. $\endgroup$ Commented Sep 20, 2021 at 19:47
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    $\begingroup$ "The first one should be false because if x=0..." There only has to exist some x that works for that formula. See if something other than x=0 works. $\endgroup$
    – TomKern
    Commented Sep 20, 2021 at 20:03

1 Answer 1

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For $\forall x\in\mathbb{R} \; \exists y\in\mathbb{R}, \; x^{2}+y^{2}\leq 1 \land xy=0$, consider $x = 2$. As $\forall y \in \mathbb{R}, 4 + y^2 \gt 1$, this statement is false. Thus its negation (the first statement) is true.

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  • $\begingroup$ This is correct, but a little unsatisfying. The first statement is true, but why? Well, any $x$ with $|x| > 1$ is a witness: for any $y$, $x^2 + y^2 > 1$, so the statement is vacuously true. $\endgroup$
    – BrianO
    Commented Sep 20, 2021 at 21:28

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