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Suppose that we have rational numbers $q_1$, $q_2$ such that $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{q_1}e^{-\frac{t^2}{2}} \,\mathrm{d}t=q_2.$$ Does this imply that $q_1=0$ and $q_2=\dfrac{1}{2}$?

Ѕᴀᴀᴅ
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Jaakko Seppälä
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    Two different questions here: (a) $\Phi(\Phi^{-1}(1/4)) = 1/4$ and $q = \Phi^{-1}(1/4) \approx -0.6744898$ is a number with rational $\Phi(q),$ with $q \ne 0$, so Yes to the title. (b) However, I would not want to claim that $q$ is rational, so the main question remains unanswered. – BruceET Jan 10 '18 at 23:29
  • I suspect $q_1$, $q_2$ can be even algebraic instead of rational (like in Lindemann-Weierstrass theorem showing that $\sin x$ is transcendental) – sdcvvc Feb 11 '18 at 01:48
  • We know that $ q_{2} = \frac{1}{2} \Big( 1 + \text {erf} \Big( \frac{ q_{1} }{ \sqrt{2} } \Big) \Big)$ and $ \text{erf} (x) = 1 - \frac{1}{ \sqrt{ \pi }} \Gamma \Big( \frac{1}{2} , x^2 \Big)$, so for $ q_{2} $ to be rational $ \text {erf} \Big( \frac{ q_{1} }{ \sqrt{2} } \Big) $ must be zero. Therefore $ q_{1} = 0$ and $q_{2} = \frac{1}{2}$. –  Jun 10 '20 at 07:58

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