【数値情報処理a】第2回の答え

課題

$x=2021.2$ 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)	$(100100011101.111001100110011001100110011001100110011001\cdots)_{2}$ $=(1.00100011101111001100110011001100110011001100110011001\cdots)_{2}\times 2^{11}$ $\simeq(1.0010001110111100110011001100110011001100110011001101)_{2}\times 2^{11}$ $\simeq(100100011101.11100110011001100110011001100110011001101)_{2}$