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we know that "$a$" is a Irrational number .But "$a^2+a$" is Rational. Can You find "$a$"? (more than one answer is available)

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    $\begingroup$ Any irrational number of form $\sqrt n - \frac{1}{2}$, where $n$ is an integer. $\endgroup$ – Ziqian Xie Mar 27 '16 at 15:08
  • $\begingroup$ @vladz no $sqrt(2) $ a bad example $\endgroup$ – openspace Mar 27 '16 at 15:11
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We have $a^2+a=r\in\Bbb Q$ if and only if $$a=\frac{-1\pm\sqrt{1+4r}}{2}.$$ Any $r\in\Bbb Q$ such that $1+4r$ is not a perfect square will do.

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Hint complete the square you get $(a+\frac{1}{2})^2-\frac{1}{4}$ so any number of the form $√n-\frac{1}{2}$ will give it as a cancels out with $-a$ making number rational.

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  • $\begingroup$ though you need $n$ to not be a (rational) square to ensure $\sqrt{n}-\frac{1}{2}$ is irrational $\endgroup$ – Henry Mar 27 '16 at 18:51
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Consider $x^2 + x = \dfrac pq$
for some relatively prime integers $p$ and $q$.

Hence $qx^2 + qx - p = 0$

So, for any relatively prime integers $p$ and $q$ such that $q^2 + 4pq$ is positive and not a perfect square,

$a \in \left\{ \dfrac{-q \pm \sqrt{q^2 + 4pq}}{2q} \right\}$

will have the property that $a$ is irrational and $a^2 + a = \dfrac pq$.

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