Explain the four arithmetic systems used in digital circuits or system and convert 0.1001 base 2 to base 10.

Here are the answers for question 1:

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.

b) Convert the following to base 10:

i) $0.1001_2$

Step 1: Expand the binary number using powers of 2.

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

Step 2: Calculate the values.

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

Step 3: Sum the results.

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

ii) $1101.101_2$

Step 1: Expand the binary number using powers of 2.

$$1101.101_2 = (1 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0) + (1 \times 2^{-1}) + (0 \times 2^{-2}) + (1 \times 2^{-3})$$

Step 2: Calculate the values.

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

Step 3: Sum the results.

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

iii) $0.1101_2$

Step 1: Expand the binary number using powers of 2.

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

Step 2: Calculate the values.

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

Step 3: Sum the results.

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

c) Convert the following to base 16:

i) $0.47_{10}$

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

$$0.47 \times 16 = 7.52 \implies \text{integer part } 7$$

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

$$0.52 \times 16 = 8.32 \implies \text{integer part } 8$$

Step 3: Repeat for desired precision.

$$0.32 \times 16 = 5.12 \implies \text{integer part } 5$$ $$0.12 \times 16 = 1.92 \implies \text{integer part } 1$$

Step 4: Read the integer parts from top to bottom.

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

ii) $524_{10}$

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

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

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

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