Braila 2008

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alex2008
Leibniz
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Location: Tulcea

Braila 2008

Post by alex2008 »

Demonstrati ca \( \sqrt{\frac{a^2+2008^2}{3}} \in \mathb{R} \)\\( \mathb{Q} \) \( (\forall)a\in\mathb{Z}. \)








... acolo e R minus Q ...
. A snake that slithers on the ground can only dream of flying through the air.
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Marius Mainea
Gauss
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Location: Gaesti (Dambovita)

Post by Marius Mainea »

\( a^2+2008^2=\mathcal{M}3+1 \) sau \( a^2+2008^2=\mathcal{M}3+2 \)
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