Восьмиричное число 0,2 ( 1) в восьмиричной системе...

創建於：2025年10月24日

問題

Восьмиричное число 0,2 ( 1) в восьмиричной системе исчисления по основанию 4 равно.”

Python

python
# Let's generate bits for the repeating fractional part 010 001 001 001... up to some length
bits = list("010" + "001"*10)
pairs = ["".join(bits[i:i+2]) for i in range(0, len(bits), 2)]
pairs[:12], len(bits), len(pairs)

(['01', '00', '01', '00', '10', '01', '00', '10', '01', '00', '10', '01'],
 33,
 17)

Python

python
# Attempt to find period after first some initial offset
digits = [int(p, 2) for p in pairs]  # base 4 digits
digits[:20]


[1, 0, 1, 0, 2, 1, 0, 2, 1, 0, 2, 1, 0, 2, 1, 0, 1]

Python

python
bits = list("010" + "001"*50)
pairs = ["".join(bits[i:i+2]) for i in range(0, len(bits), 2)]
digits = [int(p, 2) for p in pairs]
digits[:30]

[1,
 0,
 1,
 0,
 2,
 1,
 0,
 2,
 1,
 0,
 2,
 1,
 0,
 2,
 1,
 0,
 2,
 1,
 0,
 2,
 1,
 0,
 2,
 1,
 0,
 2,
 1,
 0,
 2,
 1]

Python

python
from fractions import Fraction

# value of 0.2\overline{1}_8
# Let's compute as sum: 2*8^-1 + repeating 1 * 8^-2 / (1 - 8^-1)
value_octal = Fraction(2, 8) + Fraction(1, 8**2) * Fraction(1, 1 - Fraction(1, 8))
value_octal_simplified = value_octal
value_octal_simplified, float(value_octal_simplified)

(Fraction(15, 56), 0.26785714285714285)

Python

python
from fractions import Fraction
P = Fraction(1,4) + Fraction(0,16)
B = Fraction(1, 4**3) + Fraction(0, 4**4) + Fraction(2, 4**5)
R = B * Fraction(1, 1 - Fraction(1, 4**3))
val = P + R
val, float(val), Fraction(15,56) == val

(Fraction(15, 56), 0.26785714285714285, True)

思考

Final answer in Russian

The base-4 representation of octal 0.2\overline{1}_8 is 0,10(102)_4. To convert, we first map the digits of octal to binary. For example, 2 becomes 010, and 1 becomes 001. This gives the repeating binary 0.010 001 001..., which matches the repeating pattern in base-4.

Thought for 1m 20s

回答

Ответ: $0{,}10\overline{102}_{(4)}$ .

Кратко: $0{,}2\overline{1}_{(8)} = 0{,}010\overline{001}_{(2)}$ (т.к. $2\to010$ , $1\to001$ ). Группируем по 2 бит → $01,00,01,00,10,01,\ldots$ ⇒ цифры осн. 4: $1,0,1,0,2,1,\ldots$ . Отсюда $0{,}10(102)_{4}$ .

問題

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Python

Python

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Python

思考

Final answer in Russian

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