4. Subtract the following binary numbers (convert these binary numbers to decimal numbers at the end and check to see if the subtraction is correct). a) \( 1101_{2}-101_{2} \) b) \( 10000_{2}-1010_{2} \) (b)

To subtract the binary numbers \( 10000_2 \) and \( 1010_2 \), follow these steps: ### **Step 1: Align the Numbers** First, ensure both binary numbers have the same number of digits by adding leading zeros to the shorter number if necessary. \[ \begin{array}{c} \phantom{0}10000 \\ - \;01010 \\ \hline \end{array} \] ### **Step 2: Perform Binary Subtraction** We'll subtract each bit starting from the rightmost (least significant bit) to the leftmost (most significant bit), borrowing as needed. \[ \begin{array}{c@{}c@{}c@{}c@{}c@{}c} & 1 & 0 & 0 & 0 & 0 \\ - & 0 & 1 & 0 & 1 & 0 \\ \hline & ? & ? & ? & ? & ? \\ \end{array} \] Let's subtract bit by bit: 1. **Position 0 (Rightmost Bit):** - **Minuend:** 0 - **Subtrahend:** 0 - **Result:** \( 0 - 0 = 0 \) \[ \text{Result Bit 0} = 0 \] 2. **Position 1:** - **Minuend:** 0 - **Subtrahend:** 1 - **Since 0 < 1, we need to borrow from the next higher bit.** **Borrowing Process:** - **Position 2:** 0 → Cannot borrow; move to Position 3. - **Position 3:** 0 → Cannot borrow; move to Position 4. - **Position 4:** 1 → Borrow 1 from Position 4. - **Position 4 becomes:** 0 - **Position 3 becomes:** \( 0 + 2 = 2 \) (binary '10') - **Position 3:** Borrow 1 - **Position 3 becomes:** \( 2 - 1 = 1 \) - **Position 2 becomes:** \( 0 + 2 = 2 \) - **Position 2:** Borrow 1 - **Position 2 becomes:** \( 2 - 1 = 1 \) - **Position 1 becomes:** \( 0 + 2 = 2 \) - **Now, subtract Position 1:** \( 2 - 1 = 1 \) \[ \text{Result Bit 1} = 1 \] 3. **Position 2:** - **Minuend After Borrowing:** 1 - **Subtrahend:** 0 - **Result:** \( 1 - 0 = 1 \) \[ \text{Result Bit 2} = 1 \] 4. **Position 3:** - **Minuend After Borrowing:** 1 - **Subtrahend:** 0 - **Result:** \( 1 - 0 = 1 \) \[ \text{Result Bit 3} = 1 \] 5. **Position 4 (Leftmost Bit):** - **Minuend After Borrowing:** 0 - **Subtrahend:** 0 - **Result:** \( 0 - 0 = 0 \) \[ \text{Result Bit 4} = 0 \] **Final Binary Result:** \[ \begin{array}{c@{}c@{}c@{}c@{}c@{}c} & 0 & 1 & 1 & 1 & 0 \\ \end{array} \] So, the result of \( 10000_2 - 01010_2 \) is \( 00110_2 \). Removing leading zeros, we get: \[ 00110_2 = 110_2 \] ### **Step 3: Convert to Decimal for Verification** Let's convert the binary numbers and the result to decimal to ensure the subtraction is correct. 1. **Convert \( 10000_2 \) to Decimal:** \[ 1 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 0 \times 2^0 = 16 + 0 + 0 + 0 + 0 = 16 \] 2. **Convert \( 1010_2 \) to Decimal:** \[ 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 0 \times 2^0 = 8 + 0 + 2 + 0 = 10 \] 3. **Convert the Result \( 110_2 \) to Decimal:** \[ 1 \times 2^2 + 1 \times 2^1 + 0 \times 2^0 = 4 + 2 + 0 = 6 \] 4. **Verify the Subtraction in Decimal:** \[ 16 - 10 = 6 \] Since the decimal result of the binary subtraction (\(6\)) matches the direct decimal subtraction, the binary subtraction is **correct**. ### **Final Answer:** \[ 10000_2 - 1010_2 = 110_2 \quad \text{(which is } 6 \text{ in decimal)} \]