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How would I solve the following question. And determine if its true or false.

1.$\forall x \in R , \exists y\in R, x^2+y^2=-1$

2: $\exists x\in R,\forall y \in R, x^2+y^2=-1$

For the first one I think I can justify it is false.

As for any arbitrary x must y must be

$y=\sqrt{-x^2-1}$ which would not be real number.

The second one I can say that two numbers squared cannot be a negative.So it would be false?

amWhy
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Fernando Martinez
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    What you wrote down in the last line can be said in both cases. – DonAntonio Nov 02 '13 at 20:07
  • Yes I guess for my second I could see for any y x must be $x=\sqrt{-1-y^2}$ – Fernando Martinez Nov 02 '13 at 20:10
  • What does $\forall \in x$ mean? It looks like nonsense to me. – Trevor Wilson Nov 02 '13 at 20:13
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    I don't understand why people are editing the nonsense "$\forall \in x$" out of the question, when for all we know it might be the entire point of the question. It is better to ask the OP for clarification than to make assumptions about what he meant. – Trevor Wilson Nov 02 '13 at 20:14
  • Just like the question you posed an hour ago, this question is not about discrete mathematics. Please read the tag wiki excerpt next time. – Tomas Nov 02 '13 at 20:14
  • @tomas but I got it out of a discrete math textbook – Fernando Martinez Nov 02 '13 at 20:16
  • @FernandoMartinez. I apologize. It seems that logic (that's what this is about) is sometimes formally viewed as a subtopic of discrete mathematics. So in a broader sense it was a correct tagging, although I find it rather confusing. – Tomas Nov 02 '13 at 20:22

2 Answers2

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Yes, both statements are false because the sum of two squared real numbers, whatever those numbers are, will never be negative.

amWhy
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Another simple way to think about this questions is that for any $$ x ∈ \mathbb{R}, x^2 \geq 0 $$

Then $$x^2 + y^2 \geq 0 $$

meh
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