# lithium barbie doll

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 Mar1 revised Is the set of integers with respect to the p-adic metric compact?added 4 characters in body Mar1 asked Is the set of integers with respect to the p-adic metric compact? Feb22 accepted Prove that $\displaystyle\prod_{q\in \mathbb{Q}^{\times}}|q|=1$ Feb22 comment Prove that $\displaystyle\prod_{q\in \mathbb{Q}^{\times}}|q|=1$I do! =] ok thanks everyone for the quick responses. Feb22 comment Prove that $\displaystyle\prod_{q\in \mathbb{Q}^{\times}}|q|=1$Ohh ok! I get it, so it's for a fixed rational, and it's the absolute value that is running through all primes, is that correct? Feb22 revised Prove that $\displaystyle\prod_{q\in \mathbb{Q}^{\times}}|q|=1$added 170 characters in body Feb22 comment Prove that $\displaystyle\prod_{q\in \mathbb{Q}^{\times}}|q|=1$@git gud thanks good tip Feb22 asked Prove that $\displaystyle\prod_{q\in \mathbb{Q}^{\times}}|q|=1$ Feb19 accepted Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$ Feb17 answered Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$ Feb17 comment Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$@WillJagy That's probably the same contour I'm using, it is a parallelogram(rhombus), not a rectangle. It's definitely doable anyways, since the contour is given to me explicitly in my book. Feb17 revised Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$deleted 2 characters in body Feb17 comment Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$Ok that makes sense Feb17 revised Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$edited title Feb17 comment Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$Something I'm not completely confident about is are these poles in fact of order 1? Feb17 revised Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$added 1 characters in body Feb17 comment Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$Yes but $a=\sqrt{\pi}e^{\frac{\pi i}{4}}$, so bringing it up to the numerator and making the powers negative simplifies to $a$ up above, since $i=e^{\frac{\pi i}{2}}$. Or I guess simpler would be just to refer to my note that $\pi i=a^2$. Feb17 comment Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$The only thing different between mine and yours is you have an extra factor of 2 in the denominator, and I think that might have been a mistake on your part. Feb17 asked Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$ Feb15 accepted The Group of points on the Elliptic curve $y^2=x^3+1$ over $\mathbb{F}_5$