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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?

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What you wrote down in the last line can be said in both cases. – DonAntonio Nov 2 '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 2 '13 at 20:10
What does $\forall \in x$ mean? It looks like nonsense to me. – Trevor Wilson Nov 2 '13 at 20:13
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 2 '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 2 '13 at 20:14
up vote 3 down vote accepted

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

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Yes that makes sense. Because squaring makes a number 0 or greater – Fernando Martinez Nov 2 '13 at 20:14
Precisely. $x^2 + y^2 \geq 0 \;\;\forall x, y\in \mathbb R$. – amWhy Nov 2 '13 at 20:16
@amWhy: {}{}{}{}{}{}{}{}{}{}{}{} ->>>>> +1 – Amzoti Nov 2 '13 at 22:14
@amWhy: Just make here +, so + :-) – Babak S. Nov 11 '13 at 11:19

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

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

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