# Proving that for every real $x$ there exists $y$ with $x+y^2\in\mathbb{Q}$

I'm having trouble with proving this question on my assignment:

For all real numbers $x$, there exists a real number $y$ so that $x + y^2$ is rational.

I'm not sure exactly how to prove or disprove this. I proved earlier that for all real numbers $x$, there exists some $y$ such that $x+y$ is rational by cases and i'm assuming this is similar but I'm having trouble starting.

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Please do not use homework as the only tag for your question. You should a tag that is appropriate to the subject in which this question came up (e.g. algebra-precalculus). –  J. M. is back. Jul 12 '12 at 15:28

Why isn't this trivial? For take $z$ to be any positive rational you like greater than $x$, put $y = \sqrt{(z-x)}$ ...

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There is an integer $n$ bigger than $x$. Let $y=\sqrt{n-x}$.

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Hint $\rm\ x\! +\! y^2 = q\in\mathbb Q\iff y^2 = q\!-\!x\:$ so choose a rational $\rm\:q\ge x\:$ then let $\rm\:y = \sqrt{q-x}.$

Remark $\$ Note how writing it out equationally makes it clearer what we need to do to solve it. Just as for "word problems", the key is learning how to translate the problem into the correct algebraic (equational) representation - the rest is easy.

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Hint They are basically hiding a simpler fact by stating the problem as find a real number y such that $x+y^2$ is rational. So basically; look for how are x and $x+y^2$ related.

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