講義（山本裕樹）

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====== 【数値情報処理a】第2回の答え ======
===== 課題 =====

  * (1) $x$ の整数部を2進数に変換しなさい。
  * (2) $x$ の小数部を2進数に変換しなさい（2進数で小数点第10位まで書けば十分）。
  * (3) $x$ を2進数の浮動小数点数にしなさい（倍精度で全ての桁を書くこと）。

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==== 解答例 ====

$x=2021.2$ の場合
|  (1)  | $(111 1011 1100)_{2}$                                                                                                                                                                                                                                                                                                   |
|  (2)  | $(0.0110 0110 0110 \cdots)_{2}$                                                                                                                                                                                                                                                                                         |
|  (3)  | $(11110111100.0110011001100110011001100110011001100110011\cdots)_{2}$\\ $=(1.11101111000110011001100110011001100110011001100110011\cdots)_{2}\times 2^{10}$\\ $\simeq(1.1110111100011001100110011001100110011001100110011010)_{2}\times 2^{10}$\\ $\simeq(11110111100.011001100110011001100110011001100110011010)_{2}$  |

$x=1980.4$ の場合
|  (1)  | $(111 1011 1100)_{2}$                                                                                                                                                                                                                                                                                                   |
|  (2)  | $(0.0110 0110 0110 \cdots)_{2}$                                                                                                                                                                                                                                                                                         |
|  (3)  | $(11110111100.0110011001100110011001100110011001100110011\cdots)_{2}$\\ $=(1.11101111000110011001100110011001100110011001100110011\cdots)_{2}\times 2^{10}$\\ $\simeq(1.1110111100011001100110011001100110011001100110011010)_{2}\times 2^{10}$\\ $\simeq(11110111100.011001100110011001100110011001100110011010)_{2}$  |

$x=2901.6$ の場合
|  (1)  | $(1011 0101 0101)_{2}$                                                                                                                                                                                                                                                                                                   |
|  (2)  | $(0.1001 1001 1001 \cdots)_{2}$                                                                                                                                                                                                                                                                                         |
|  (3)  | $(101101010101.100110011001100110011001100110011001100110\cdots)_{2}$\\ $=(1.01101010101100110011001100110011001100110011001100110\cdots)_{2}\times 2^{11}$\\ $\simeq(1.0110101010110011001100110011001100110011001100110011)_{2}\times 2^{11}$\\ $\simeq(101101010101.10011001100110011001100110011001100110011)_{2}$  |

$x=1527.7$ の場合
|  (1)  | $(101 1111 0111)_{2}$                                                                                                                                                                                                                                                                                                   |
|  (2)  | $(0.1011 0011 0011 0011 \cdots)_{2}$                                                                                                                                                                                                                                                                                         |
|  (3)  | $(10111110111.1011001100110011001100110011001100110011001\cdots)_{2}$\\ $=(1.01111101111011001100110011001100110011001100110011001\cdots)_{2}\times 2^{10}$\\ $\simeq(1.0111110111101100110011001100110011001100110011001101)_{2}\times 2^{10}$\\ $\simeq(10111110111.101100110011001100110011001100110011001101)_{2}$  |

$x=2333.9$ の場合
|  (1)  | $(1001 0001 1101)_{2}$                                                                                                                                                                                                                                                                                                   |
|  (2)  | $(0.1110 0110 0110 \cdots)_{2}$                                                                                                                                                                                                                                                                                         |
|  (3)  | $(100100011101.111001100110011001100110011001100110011001\cdots)_{2}$\\ $=(1.00100011101111001100110011001100110011001100110011001\cdots)_{2}\times 2^{11}$\\ $\simeq(1.0010001110111100110011001100110011001100110011001101)_{2}\times 2^{11}$\\ $\simeq(100100011101.11100110011001100110011001100110011001101)_{2}$  |