0.1 数(NUMBERS)

1.

(a) $\frac{4}{7} = 0.\overline{571428}$となり循環少数である．したがって，有理数

(b) $\frac{\pi}{4}$は無理数．その理由は，もし $\frac{\pi}{4}$が有理数ならば，ある有理数$q$を用いて $\frac{\pi}{4} = q$と表せる．これより，$\pi = 4q$となり，$4q$は有理数の積だから有理数．したがって，$\pi$も有理数となる．これは矛盾である．

2.

(a) $\vert-3.0\vert = 3.0$

(b) $\vert\pi\vert = \pi$

3.

(a) $\vert a+b+c\vert = \vert 2 + (-3) + (-5)\vert = \vert 2 - 3 - 5\vert = \vert-6\vert = 6$

(b) $\vert a-b+c\vert = \vert 2 - (-3) + (-5)\vert = \vert 2 + 3 - 5\vert = \vert\vert = 0$

4.

(a) $9^{\frac{3}{2}} = (9^{\frac{1}{2}})^{3} = 3^{3} = 27$

(b) $27^{-\frac{1}{3}} = (27^{\frac{1}{3}})^{-1} = 3^{-1} = \frac{1}{3}$

5.

(a) $\sqrt[4]{a^{5}} = (a^{5})^{\frac{1}{4}} = a^{\frac{5}{4}}$

(b) $\sqrt{a^{-2}\sqrt[3]{a}} = \sqrt{a^{-2}a^{\frac{1}{3}}} = (a^{-2 + \frac{1}{3}})^{\frac{1}{2}} = (a^{-\frac{5}{3}})^{\frac{1}{2}} = a^{-\frac{5}{6}}$

(c) $\sqrt{a} \times \sqrt[3]{a^{-2}} \div a^{-\frac{1}{6}} = a^{\frac{1}{2}}a^{\fra... ...{\frac{1}{2} - \frac{2}{3} - \frac{1}{6}} = a^{-\frac{2}{6}} = a^{-\frac{1}{3}}$

6.

(a) $\sqrt{18} + \sqrt{50} - \sqrt{72} = \sqrt{2\cdot3^{2}} + \sqrt{2\cdot5^{2}} + \sqrt{2^{3}3^{2}} = 3\sqrt{2} + 5\sqrt{2} - 6\sqrt{2} = 2\sqrt{2}$

(b) $(\sqrt{5} - \sqrt{2})^{2} = (\sqrt{5})^{2} - 2\sqrt{5}\sqrt{2} + (\sqrt{2})^{2} = 5 - 2\sqrt{10} + 2 = 7 - 2\sqrt{10}$

(c) $(4\sqrt{2}+\sqrt{3})(\sqrt{2} + 3\sqrt{3}) = 8 + 12 \sqrt{6} + \sqrt{6} + 9 = 17 + 13\sqrt{6}$ 7.

7.

(a) $\frac{2}{\sqrt{5} - \sqrt{3}} = \frac{2}{\sqrt{5} - \sqrt{3}}\cdot \frac{\sqrt{... ...qrt{5} + \sqrt{3}} = \frac{2(\sqrt{5} + \sqrt{3})}{5 - 3} = \sqrt{5} + \sqrt{3}$

(b) $\frac{\sqrt{3}}{\sqrt{n+1} - \sqrt{n-2}} = \frac{\sqrt{3}}{\sqrt{n+1} - \sqrt{n... ...{n+1} + \sqrt{n-2})}{n+1 - (n-2)} = \frac{\sqrt{3}(\sqrt{n+1} + \sqrt{n-2})}{3}$

(c) $\sqrt{n+2} - \sqrt{n-1} = \frac{(\sqrt{n+2} - \sqrt{n-1})(\sqrt{n+2} + \sqrt{n-... ...\frac{n+2 - (n-1)}{\sqrt{n+2} + \sqrt{n-1}} = \frac{3}{\sqrt{n+2} + \sqrt{n-1}}$