# Is $\forall\space x \in \mathbb{R} \space \exists \space y \in \mathbb{R} \space(x+y^2=10)$ true or false?

Given the expression $\forall\space x \in \mathbb{R} \space \exists \space y \in \mathbb{R}\space(x+y^2=10)$ tell what is its logical value?

When I look to an expression with quantifiers, I try to translator into current language. In this case my atempt was "for each real number $x$ there is a real number $y$ that satisfy $x+y^2=10$". Or " for a given real number $x$ there is a real number $y$ that satisfy $x+y^2=10$".

Can I take from the previous thought that $y=\sqrt{10-x} \space$ ?(for a given $x$...there is a $y$). Thanks

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false, because when x=-11, there does not exist real $y$ such that it satisfies the equation – Idonknow Jan 8 '13 at 9:51
No, if you chose x to be greater then 10, no real number y will satisfy the equation. – Ethan Jan 8 '13 at 9:51
@NgHongWai: How about $\sqrt{21}$? – Clive Newstead Jan 8 '13 at 10:00

The statement is false, as you won't find an appropriate $y$ for $x>10$. If, however, you would allow $y$ to be a complex number, the statement would be true.

Concerning your translation. I agree with the first one, but

"for a given real number $x$ there is a real number $y$ that satisfy $x+y^2=10$"

should rather be

"for ANY given real number $x$ there is a real number $y$ that satisfy $x+y^2=10$"

even though your expression isn't wrong, but using "any" makes it more clear in my opinion.

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I chose this, because it answer to my doubt. That was if my "translation" was the right one.Thanks – João Jan 8 '13 at 21:32

As you can see, there is no $y \in \mathbb{R}$ for $x>10$ which satisfies the equation

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Your answer gave me a geometric poin of view. Thanks – João Jan 8 '13 at 21:35

Suppose this is true. Then $\ \ \ y^2 =10 - x \iff y = \sqrt{10-x} \iff 10-x\ge0 \iff x \le 10$ but that is a contradiction to $x$ can be all real numbers. $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \boxed{}$

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