- Apply the division algorithm for integers by repeatedly dividing by 2 and recording remainders.
- Use the multiplication algorithm for fractions by repeatedly multiplying the fractional part by 2 and extracting the integer parts.
- Determine if the bicimal terminates (denominator is a power of 2) or repeats.
- Results: \(\frac{2}{5} = 0.\overline{0110}_2\), \(\frac{9}{20} = 0.0\overline{11100}_2\), \(\frac{3}{4} = 0.11_2\), \(6 = 110_2\), \(\frac{5}{6} = 0.1\overline{110}_2\). \(\boxed{{0.\overline{0110}_2, 0.0\overline{11100}_2, 0.11_2, 110_2, 0.1\overline{110}_2}}\)
Explanation
- Understanding Bicimal Conversions
A "bicimal" (binary representation of a rational number) is created by converting a number into base 2. For an integer, we repeatedly divide by 2 and record the remainders in reverse order. For a fractional part \(f\), we multiply by 2 repeatedly: the integer part of the product is the next digit, and the new fractional part is used for the next multiplication. A fraction \(\frac{p}{q}\) results in a terminating bicimal if and only if the denominator \(q\) (in simplest form) is a power of 2; otherwise, it results in a repeating sequence.
- Conversion of \(\frac{2}{5}\)
- \(\frac{2}{5} \times 2 = \frac{4}{5} = 0.8 \to\) integer digit 0, remainder 0.8
- \(0.8 \times 2 = 1.6 \to\) integer digit 1, remainder 0.6
- \(0.6 \times 2 = 1.2 \to\) integer digit 1, remainder 0.2
- \(0.2 \times 2 = 0.4 \to\) integer digit 0, remainder 0.4
- \(0.4 \times 2 = 0.8 \dots\) The sequence \(\dots 0.0110 \dots\) repeats.
- Thus, \(\frac{2}{5} = (0.\overline{0110})_2\).
- Conversion of \(\frac{9}{20}\)
- \(\frac{9}{20} \times 2 = \frac{18}{20} = 0.9 \to\) integer digit 0, remainder 0.9
- \(0.9 \times 2 = 1.8 \to\) integer digit 1, remainder 0.8
- \(0.8 \times 2 = 1.6 \to\) integer digit 1, remainder 0.6
- \(0.6 \times 2 = 1.2 \to\) integer digit 1, remainder 0.2
- \(0.2 \times 2 = 0.4 \to\) integer digit 0, remainder 0.4
- \(0.4 \times 2 = 0.8 \to\) integer digit 0, remainder 0.8. The pattern \(011100\) begins to repeat from the second digit.
- Thus, \(\frac{9}{20} = 0.0\overline{11100}_2\).
- Conversion of \(\frac{3}{4}\)
- \(\frac{3}{4} \times 2 = 1.5 \to\) integer digit 1, remainder 0.5
- \(0.5 \times 2 = 1.0 \to\) integer digit 1, remainder 0.0. Since the remainder is 0, the process terminates.
- Thus, \(\frac{3}{4} = (0.11)_2\).
- Conversion of \(6\)
- \(6 \div 2 = 3\) remainder 0
- \(3 \div 2 = 1\) remainder 1
- \(1 \div 2 = 0\) remainder 1
- Reading the remainders from bottom to top gives \(110\).
- Thus, \(6 = (110)_2\).
- Conversion of \(\frac{5}{6}\)
- \(\frac{5}{6} = \frac{4+1}{6} = \frac{2}{3} + \frac{1}{6}\) (Alternative approach: \(5 \div 6 = 0.8333 \dots\))
- \(0.8333 \times 2 = 1.6666 \to\) integer digit 1, remainder 0.6666
- \(0.6666 \times 2 = 1.3333 \to\) integer digit 1, remainder 0.3333
- \(0.3333 \times 2 = 0.6666 \to\) integer digit 0, remainder 0.6666. The pattern \(110\) repeats.
- Thus, \(\frac{5}{6} = 0.1\overline{110}_2\).
- Final Answer
The bicimal conversions are:
A\(\frac{2}{5} = \boxed{0.\overline{0110}_2}\)
B\(\frac{9}{20} = \boxed{0.0\overline{11100}_2}\)
C\(\frac{3}{4} = \boxed{0.11_2}\)
D\(6 = \boxed{110_2}\)
E\(\frac{5}{6} = \boxed{0.1\overline{110}_2}\)
Examples
Converting between bases is a foundational skill in computer science, as computers store all data as binary digits (bits). Understanding bicimals helps you see how numbers that don't fit perfectly into powers of 2, such as \(0.1\) or \(0.6\), are represented in computer memory. This explains why floating-point arithmetic in software can sometimes have small rounding errors—a repeating decimal in base 10 might become a repeating bicimal in base 2, which can only be stored to a limited precision.