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i. The 2's complement of the number 111000111 is:

ii. Perform the conversion [tex]\((124.025)_{10} = X_{16}\)[/tex]. [tex]\(X\)[/tex] is equivalent to:


Sagot :

Sure, let's break down each part of the question with a detailed, step-by-step solution.

### Part i: Finding the 2's complement of the binary number 111000111

1. Starting Binary Number: 111000111

2. Step 1: Inversion of the digits (1's Complement)
- Inversion means swapping all `0`s to `1`s and all `1`s to `0`s.
- The 1's complement of 111000111 is 000111000.

3. Step 2: Add 1 to the inverted number (1's Complement) to get 2's Complement
- 1's complement: 000111000
- Adding 1: 000111000 + 1 = 000111001

4. Result: Ensure the result is in 9 bits
- The result already fits 9 bits, so no padding is needed.

Therefore, the 2's complement of 111000111 is 000111001.

### Part ii: Converting [tex]\(124.025\)[/tex] to base 16 (hexadecimal)

Step 1: Convert the integer part (124) to hexadecimal

1. Divide 124 by 16:
- 124 ÷ 16 = 7 remainder 12
- The remainder 12 corresponds to the hexadecimal digit 'C'.

Therefore, the integer part in hexadecimal is 7C.

Step 2: Convert the fractional part (0.025) to hexadecimal

1. Multiply the fractional part by 16 repeatedly and take the integer part of the result:
- [tex]\(0.025 \times 16 = 0.4\)[/tex]
- Integer part: 0
- Fractional part: 0.4
- [tex]\(0.4 \times 16 = 6.4\)[/tex]
- Integer part: 6
- Fractional part: 0.4
- [tex]\(0.4 \times 16 = 6.4\)[/tex]
- Integer part: 6
- Fractional part: 0.4
- [tex]\(0.4 \times 16 = 6.4\)[/tex]
- Integer part: 6
- Fractional part: 0.4
- Repeat for as many digits as desired (for illustrative purposes, we continue this pattern).

After eight iterations, the sequence of hexadecimal digits forms a repeating pattern of '66'.

Therefore, the fractional part in hexadecimal is 0.06666666.

Combining the results:

- Integer part: 7C
- Fractional part: 0.06666666

Thus, [tex]\(124.025\)[/tex] in base 16 is: 7C.06666666

### Final Answer:
i. The 2's complement of the binary number 111000111 is 000111001.
ii. The conversion [tex]\(124.025\)[/tex] to base 16 is 7C.06666666.
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