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Explain the four arithmetic systems used in digital circuits or systems and convert binary to decimal.

Computer Science

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6 parts
b(i)

0.5625100.5625_{10}

b(ii)

13.6251013.625_{10}

b(iii)

0.8125100.8125_{10}

c(i)

0.7851...160.7851..._{16}

c(ii)

20C1620C_{16}

c(iii)

17616176_{16}

Step-by-step solution

21 steps

Here are the answers for question 1:

Part a

Explain the four arithmetic systems used in digital circuits or system.

  1. Binary (Base 2): This system uses only two digits, 0 and 1. It is the fundamental language of digital computers and electronic circuits because it can be easily represented by two distinct states, such as on/off or high/low voltage.
  2. Octal (Base 8): This system uses eight digits (0-7). It is often used as a compact way to represent binary numbers, as each octal digit corresponds to exactly three binary digits.
  3. Decimal (Base 10): This is the standard number system used by humans, employing ten digits (0-9). Digital systems often convert inputs and outputs to decimal for user interaction.
  4. Hexadecimal (Base 16): This system uses sixteen symbols (0-9 and A-F, where A represents 10, B is 11, and so on up to F for 15). It is widely used in computing as a shorthand for binary numbers, with each hexadecimal digit representing four binary digits, making it efficient for representing memory addresses and data.

Part b(i)

0.5625100.5625_{10}

0.100120.1001_2

Convert the following to base 10

  1. Expand the binary number using powers of 2.

    0.10012=(1×2−1)+(0×2−2)+(0×2−3)+(1×2−4)0.1001_2 = (1 \times 2^{-1}) + (0 \times 2^{-2}) + (0 \times 2^{-3}) + (1 \times 2^{-4})

  2. Calculate the values.

    =(1×0.5)+(0×0.25)+(0×0.125)+(1×0.0625)= (1 \times 0.5) + (0 \times 0.25) + (0 \times 0.125) + (1 \times 0.0625)

  3. Sum the results.

    =0.5+0+0+0.0625=0.562510= 0.5 + 0 + 0 + 0.0625 = \boxed{0.5625_{10}}

Part b(ii)

13.6251013.625_{10}

1101.10121101.101_2

  1. Expand the binary number using powers of 2.

    1101.1012=(1×23)+(1×22)+(0×21)+(1×20)+(1×2−1)+(0×2−2)+(1×2−3)1101.101_2 = (1 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0) + (1 \times 2^{-1}) + (0 \times 2^{-2}) + (1 \times 2^{-3})

  2. Calculate the values.

    =(1×8)+(1×4)+(0×2)+(1×1)+(1×0.5)+(0×0.25)+(1×0.125)= (1 \times 8) + (1 \times 4) + (0 \times 2) + (1 \times 1) + (1 \times 0.5) + (0 \times 0.25) + (1 \times 0.125)

  3. Sum the results.

    =8+4+0+1+0.5+0+0.125=13+0.625=13.62510= 8 + 4 + 0 + 1 + 0.5 + 0 + 0.125 = 13 + 0.625 = \boxed{13.625_{10}}

Part b(iii)

0.8125100.8125_{10}

0.110120.1101_2

  1. Expand the binary number using powers of 2.

    0.11012=(1×2−1)+(1×2−2)+(0×2−3)+(1×2−4)0.1101_2 = (1 \times 2^{-1}) + (1 \times 2^{-2}) + (0 \times 2^{-3}) + (1 \times 2^{-4})

  2. Calculate the values.

    =(1×0.5)+(1×0.25)+(0×0.125)+(1×0.0625)= (1 \times 0.5) + (1 \times 0.25) + (0 \times 0.125) + (1 \times 0.0625)

  3. Sum the results.

    =0.5+0.25+0+0.0625=0.812510= 0.5 + 0.25 + 0 + 0.0625 = \boxed{0.8125_{10}}

Part c(i)

0.7851...160.7851..._{16}

0.47100.47_{10}

Convert the following to base 16

  1. Multiply the fractional part by 16 and record the integer part.

    0.47×16=7.52  ⟹  integer part 70.47 \times 16 = 7.52 \implies \text{integer part } 7

  2. Take the new fractional part and repeat the multiplication.

    0.52×16=8.32  ⟹  integer part 80.52 \times 16 = 8.32 \implies \text{integer part } 8

  3. Repeat for desired precision.

    0.32×16=5.12  ⟹  integer part 50.32 \times 16 = 5.12 \implies \text{integer part } 5 0.12×16=1.92  ⟹  integer part 10.12 \times 16 = 1.92 \implies \text{integer part } 1

  4. Read the integer parts from top to bottom.

    0.7851...16\boxed{0.7851..._{16}}

Part c(ii)

20C1620C_{16}

52410524_{10}

  1. Divide the integer by 16 and record the remainder.

    524÷16=32 remainder 12 (C in hex)524 \div 16 = 32 \text{ remainder } 12 \text{ (C in hex)}

  2. Divide the quotient by 16 and record the remainder.

    32÷16=2 remainder 032 \div 16 = 2 \text{ remainder } 0

  3. Divide the quotient by 16 and record the remainder.

    2÷16=0 remainder 22 \div 16 = 0 \text{ remainder } 2

  4. Read the remainders from bottom to top.

    20C16\boxed{20C_{16}}

Part c(iii)

17616176_{16}

37410374_{10}

  1. Divide the integer by 16 and record the remainder.

    374÷16=23 remainder 6374 \div 16 = 23 \text{ remainder } 6

  2. Divide the quotient by 16 and record the remainder.

    23÷16=1 remainder 723 \div 16 = 1 \text{ remainder } 7

  3. Divide the quotient by 16 and record the remainder.

    1÷16=0 remainder 11 \div 16 = 0 \text{ remainder } 1

  4. Read the remainders from bottom to top.

    17616\boxed{176_{16}}
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