[tex]\it N=\dfrac{\sqrt3-1}{\sqrt3+1}:\dfrac{2-\sqrt3}{2}=\dfrac{\sqrt3-1}{\sqrt3+1}\cdot\dfrac{2}{2-\sqrt3}=\dfrac{(\sqrt3-1)\cdot2}{(\sqrt3+1)(2-\sqrt3)}=\\ \\ \\ =\dfrac{(\sqrt3-1)\cdot2}{2\sqrt3-3+2-\sqrt3}= \dfrac{(\sqrt3-1)\cdot2}{\sqrt3-1}= 2\in\mathbb{N}[/tex]
