Question

1001010base2-10101

Answer 1

To subtract the binary numbers \(1001010_2\) (which is 74 in decimal) and \(10101_2\) (which is 21 in decimal), we can convert them to decimal, perform the subtraction, and convert the result back to binary.

1. **Convert binary numbers to decimal:**
- \(1001010_2 = 1 \cdot 2^6 + 0 \cdot 2^5 + 0 \cdot 2^4 + 1 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 + 0 \cdot 2^0 = 64 + 0 + 0 + 8 + 0 + 2 + 0 = 74\)
- \(10101_2 = 1 \cdot 2^4 + 0 \cdot 2^3 + 1 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0 = 16 + 0 + 4 + 0 + 1 = 21\)

2. **Perform the subtraction in decimal:**
\[
74 - 21 = 53
\]

3. **Convert the result back to binary:**
To convert \(53\) to binary:
- \(53 / 2 = 26\) remainder \(1\)
- \(26 / 2 = 13\) remainder \(0\)
- \(13 / 2 = 6\) remainder \(1\)
- \(6 / 2 = 3\) remainder \(0\)
- \(3 / 2 = 1\) remainder \(1\)
- \(1 / 2 = 0\) remainder \(1\)

Writing the remainders from bottom to top gives \(110101_2\).

Therefore, \(1001010_2 - 10101_2 = 110101_2\).