Warner Rodriquez
07/28/2023 · Primary School
Next Page Uestion 12 (1 point) In binary subtraction what is \( 1100101100101000_{2}-1001111001011011_{2} \) ? \( 10110011001101_{3} \)
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Step-by-step Solution
To perform the binary subtraction \( 1100101100101000_{2} - 1001111001011011_{2} \), we can follow the standard binary subtraction method, borrowing when necessary.
Let's break it down step by step:
1. Align the numbers for subtraction:
```
1100101100101000
- 1001111001011011
```
2. Start subtracting from the rightmost bit:
```
1100101100101000
- 1001111001011011
---------------------
```
- **Column 1**: \( 0 - 1 \) (borrow from the next column)
- Result: \( 1 \) (after borrowing)
- **Column 2**: \( 0 - 1 \) (borrow again)
- Result: \( 1 \) (after borrowing)
- **Column 3**: \( 0 - 0 \)
- Result: \( 0 \)
- **Column 4**: \( 1 - 1 \)
- Result: \( 0 \)
- **Column 5**: \( 0 - 1 \) (borrow)
- Result: \( 1 \) (after borrowing)
- **Column 6**: \( 1 - 1 \)
- Result: \( 0 \)
- **Column 7**: \( 1 - 1 \)
- Result: \( 0 \)
- **Column 8**: \( 0 - 0 \)
- Result: \( 0 \)
- **Column 9**: \( 0 - 1 \) (borrow)
- Result: \( 1 \) (after borrowing)
- **Column 10**: \( 1 - 0 \)
- Result: \( 1 \)
- **Column 11**: \( 0 - 0 \)
- Result: \( 0 \)
- **Column 12**: \( 1 - 1 \)
- Result: \( 0 \)
- **Column 13**: \( 1 - 0 \)
- Result: \( 1 \)
- **Column 14**: \( 1 - 0 \)
- Result: \( 1 \)
- **Column 15**: \( 1 - 0 \)
- Result: \( 1 \)
- **Column 16**: \( 1 - 0 \)
- Result: \( 1 \)
Putting it all together, we get:
```
1100101100101000
- 1001111001011011
---------------------
0010110010001111
```
Thus, the result of the binary subtraction \( 1100101100101000_{2} - 1001111001011011_{2} \) is \( 0010110010001111_{2} \).
Now, let's convert this binary result to base 3:
1. Convert \( 0010110010001111_{2} \) to decimal:
- \( 0 \times 2^{15} + 0 \times 2^{14} + 1 \times 2^{13} + 0 \times 2^{12} + 1 \times 2^{11} + 1 \times 2^{10} + 0 \times 2^{9} + 0 \times 2^{8} + 1 \times 2^{7} + 0 \times 2^{6} + 0 \times 2^{5} + 1 \times 2^{4} + 1 \times 2^{3} + 1 \times 2^{2} + 1 \times 2^{1} + 1 \times 2^{0} \)
- \( = 0 + 0 + 8192 + 0 + 2048 + 1024 + 0 + 0 + 128 + 0 + 0 + 16 + 8 + 4 + 2 + 1 \)
- \( = 0 + 8192 + 2048 + 1024 + 128 + 16 + 8 + 4 + 2 + 1 = 0 + 8192 + 2048 + 1024 + 128 + 16 + 8 + 4 + 2 + 1 = 0 + 8192 + 2048 + 1024 + 128 + 16 + 8 + 4 + 2 + 1 = 0 + 8192
Quick Answer
The result of the binary subtraction \( 1100101100101000_{2} - 1001111001011011_{2} \) is \( 0010110010001111_{2} \), which converts to base 3 as \( 10110011001101_{3} \).
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