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How many binary bits are needed to represent decimal number 231? - Answers

At least 8 bits are needed to represent the number 231. 231 ÷ 2 = 115 r 1 115 ÷ 2 = 57 r 1 57 ÷ 2 = 28 r 1 28 ÷ 2 = 14 r 0 14 ÷ 2 = 7 r 0 7 ÷ 2 = 3 r 1 3 ÷ 2 = 1 r 1 1 ÷ 2 = 0 r 1 → 231 is 1110 0111 in binary and has 8 binary digits. Thus 231 can be represented in 8 bits, but if more are provided, eg 16, it can still be represented (in 16 bits it would be 0000 0000 1110 0111, unless there is a binary point, with say 8 bits after it, then 231 would be 1110 0111 . 0000 0000).



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How many binary bits are needed to represent decimal number 231? - Answers

https://math.answers.com/other-math/How_many_binary_bits_are_needed_to_represent_decimal_number_231

At least 8 bits are needed to represent the number 231. 231 ÷ 2 = 115 r 1 115 ÷ 2 = 57 r 1 57 ÷ 2 = 28 r 1 28 ÷ 2 = 14 r 0 14 ÷ 2 = 7 r 0 7 ÷ 2 = 3 r 1 3 ÷ 2 = 1 r 1 1 ÷ 2 = 0 r 1 → 231 is 1110 0111 in binary and has 8 binary digits. Thus 231 can be represented in 8 bits, but if more are provided, eg 16, it can still be represented (in 16 bits it would be 0000 0000 1110 0111, unless there is a binary point, with say 8 bits after it, then 231 would be 1110 0111 . 0000 0000).



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https://math.answers.com/other-math/How_many_binary_bits_are_needed_to_represent_decimal_number_231

How many binary bits are needed to represent decimal number 231? - Answers

At least 8 bits are needed to represent the number 231. 231 ÷ 2 = 115 r 1 115 ÷ 2 = 57 r 1 57 ÷ 2 = 28 r 1 28 ÷ 2 = 14 r 0 14 ÷ 2 = 7 r 0 7 ÷ 2 = 3 r 1 3 ÷ 2 = 1 r 1 1 ÷ 2 = 0 r 1 → 231 is 1110 0111 in binary and has 8 binary digits. Thus 231 can be represented in 8 bits, but if more are provided, eg 16, it can still be represented (in 16 bits it would be 0000 0000 1110 0111, unless there is a binary point, with say 8 bits after it, then 231 would be 1110 0111 . 0000 0000).

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      At least 8 bits are needed to represent the number 231. 231 ÷ 2 = 115 r 1 115 ÷ 2 = 57 r 1 57 ÷ 2 = 28 r 1 28 ÷ 2 = 14 r 0 14 ÷ 2 = 7 r 0 7 ÷ 2 = 3 r 1 3 ÷ 2 = 1 r 1 1 ÷ 2 = 0 r 1 → 231 is 1110 0111 in binary and has 8 binary digits. Thus 231 can be represented in 8 bits, but if more are provided, eg 16, it can still be represented (in 16 bits it would be 0000 0000 1110 0111, unless there is a binary point, with say 8 bits after it, then 231 would be 1110 0111 . 0000 0000).

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