ガウス記号

(1)$\small{ \ \lbrack 2x+1 \rbrack=4 \ }$より$\small{ \ 4\leqq 2x+1 \lt 5 \ }$
これを解いて$\small{ \ \displaystyle\frac{3}{2}\leqq x \lt 2 \ }$
(2)$\small{ \ \lbrack x+1 \rbrack \ }$は$\small{ \ x \lt \lbrack x+1 \rbrack \leqq x+1 \ }$より
$\small{ \ x \lt \displaystyle\frac{1}{2}x+3 \leqq x+1 \ }$
これを解くと$\small{ \ 4\leqq x \lt 6 \ }$
(i)$\small{ \ 4\leqq x \lt 5 \ }$のとき$\small{ \ \lbrack x+1 \rbrack=5 \ }$
$\small{ \ \displaystyle\frac{1}{2}x+3=5 \ }$
$\small{ \ \therefore x=4 \ }$
(ii)$\small{ \ 5\leqq x \lt 6 \ }$のとき$\small{ \ \lbrack x+1 \rbrack=6 \ }$
$\small{ \ \displaystyle\frac{1}{2}x+3=6 \ }$
$\small{ \ x=6 \ }$　$\small{ \ 5\leqq x \lt 6 \ }$より不適
(i)(ii)より$\small{ \ x=4 \ }$

(1)
$\small{ \ x^2-nx-1=0 \ }$
$\small{ \ \alpha\gt0 \ }$より$\small{ \ x=\displaystyle\frac{n+\sqrt{n^2+4}}{2} \ }$

(2)
$\small{ \ n^2\lt n^2+4 \lt n^2+4n+4 \ }$
$\small{ \ n\lt \sqrt{n^2+4}\lt n+2 \ }$
$\small{ \ n\lt \displaystyle\frac{n+\sqrt{n^2+4}}{2}\lt n+1 \ }$
$\small{\begin{eqnarray} \ \lbrack \alpha \rbrack&=&\lbrack \displaystyle\frac{n+\sqrt{n^2+4}}{2} \rbrack\\[3pt] &=&n \ \end{eqnarray}}$

(3)
$\small{ \ \langle \alpha \rangle=\alpha-\lbrack \alpha \rbrack \ }$より
$\small{\begin{eqnarray} \ \langle \alpha \rangle&=&\displaystyle\frac{n+\sqrt{n^2+4}}{2}-n\\[3pt] &=&\displaystyle\frac{\sqrt{n^2+4}-n}{2} \ \end{eqnarray}}$

(4)
$\small{\begin{eqnarray} \ \displaystyle\frac{1}{\alpha}&=&\displaystyle\frac{2}{n+\sqrt{n^2+4}}\\[3pt] &=&\displaystyle\frac{2(n-\sqrt{n^2+4})}{(n+\sqrt{n^2+4})(n-\sqrt{n^2+4})}\\[3pt] &=&\displaystyle\frac{\sqrt{n^2+4}-n}{2} \ \end{eqnarray}}$
ここで
$\small{ \ 0\lt \sqrt{n^2+4}-n \lt 2 \ }$より
$\small{ \ 0\lt \displaystyle\frac{\sqrt{n^2+4}-n}{2}\lt 1 \ }$
$\small{\begin{eqnarray} \ \lbrack \displaystyle\frac{1}{\alpha} \rbrack&=&\lbrack \displaystyle\frac{\sqrt{n^2+4}-n}{2} \rbrack\\[3pt] &=&0 \ \end{eqnarray}}$

(5)
$\small{\begin{eqnarray} \ \langle \displaystyle\frac{1}{\alpha} \rangle&=&\displaystyle\frac{1}{\alpha}-\lbrack \displaystyle\frac{1}{\alpha} \rbrack\\[3pt] &=&\displaystyle\frac{\sqrt{n^2+4}-n}{2} \ \end{eqnarray}}$

(6)
$\small{\begin{eqnarray} \ \lbrack \displaystyle\frac{1}{\langle \alpha \rangle}\rbrack&=&\lbrack \displaystyle\frac{2}{\sqrt{n^2+4}- n}\rbrack\\[3pt] &=&\lbrack \displaystyle\frac{\sqrt{n^2+4}+n}{2}\rbrack\\[3pt] &=&n \ \end{eqnarray}}$

(7)
$\small{\begin{eqnarray} \ \langle \displaystyle\frac{1}{\langle \alpha \rangle}\rangle&=&\displaystyle\frac{1}{\langle \alpha \rangle}-\lbrack \displaystyle\frac{1}{\langle \alpha \rangle}\rbrack\\[3pt] &=&\displaystyle\frac{\sqrt{n^2+4}+n}{2}-n\\[3pt] &=&\displaystyle\frac{\sqrt{n^2+4}-n}{2} \ \end{eqnarray}}$