Sure, let’s dive into the problem step by step. We will perform subtraction using 2's complement for both decimal and binary numbers.
### Part (i) Decimal subtraction using 2's complement
We need to subtract [tex]\(5_{10}\)[/tex] from [tex]\(52_{10}\)[/tex].
1. Step 1: Convert both numbers to the same bit-width of binary form.
- [tex]\(52_{10}\)[/tex] in binary is [tex]\(110100_2\)[/tex].
- [tex]\(5_{10}\)[/tex] in binary is [tex]\(000101_2\)[/tex] using 6 bits to match the bit-width of 52.
2. Step 2: Find the 2's complement of the second number (000101).
- Find the 1's complement by flipping the bits: [tex]\(000101_2 \rightarrow 111010_2\)[/tex].
- Add 1 to the 1's complement: [tex]\(111010_2 + 1 = 111011_2\)[/tex].
- Thus, the 2's complement of [tex]\(5_{10}\)[/tex] is [tex]\(111011_2\)[/tex].
3. Step 3: Add the original number to the 2's complement of the second number.
- [tex]\(110100_2\)[/tex] (52 in binary)
- [tex]\(+ 111011_2\)[/tex] (2's complement of 5)
- [tex]\(\hline\)[/tex]
- [tex]\( 110100_2\)[/tex]
- [tex]\(+111011_2\)[/tex]
- [tex]\(\hline\)[/tex]
- [tex]\( 1100111_2 \)[/tex] (with carry)
4. Step 4: Discard any carry beyond the original bit-width.
- We initially used 6 bits.
- So, the resultant binary is [tex]\(100111_2\)[/tex], discarding the leftmost bit.
5. Step 5: Convert the result back to decimal.
- [tex]\(100111_2\)[/tex] equals [tex]\(39_{10}\)[/tex].
Hence, the result is [tex]\( 52_{10} - 5_{10} = 47_{10} \)[/tex].
### Part (ii) Binary subtraction using 2's complement
We need to subtract [tex]\(11010_2\)[/tex] from [tex]\(101011_2\)[/tex].
1. Step 1: Convert both numbers to the same bit-width if necessary.
- [tex]\(101011_2\)[/tex] is already 6 bits.
- [tex]\(11010_2\)[/tex] is 5 bits, so we pad it to 6 bits: [tex]\(011010_2\)[/tex].
2. Step 2: Find the 2's complement of the second number (011010).
- Find the 1's complement by flipping the bits: [tex]\(011010_2 \rightarrow 100101_2\)[/tex].
- Add 1 to the 1's complement: [tex]\(100101_2 + 1 = 100110_2\)[/tex].
- Thus, the 2's complement of [tex]\(011010_2\)[/tex] is [tex]\(100110_2\)[/tex].
3. Step 3: Add the original number to the 2's complement of the second number.
- [tex]\(101011_2\)[/tex] (original number)
- [tex]\(+ 100110_2\)[/tex] (2's complement of 011010)
- [tex]\(\hline\)[/tex]
- [tex]\(101011_2\)[/tex]
- [tex]\(+100110_2\)[/tex]
- [tex]\(\hline\)[/tex]
- [tex]\(1010001_2\)[/tex] (with carry)
4. Step 4: Discard any carry beyond the original bit-width.
- We initially used 6 bits.
- So, the resultant binary is [tex]\(010001_2\)[/tex], discarding the leftmost bit.
Hence, the result is [tex]\( 101011_2 - 11010_2 = 010001_2 \)[/tex].
Summary:
- Part (i): [tex]\( 52_{10} - 5_{10} = 47_{10} \)[/tex].
- Part (ii): [tex]\( 101011_2 - 11010_2 = 010001_2 \)[/tex].
