Let's solve the problem step by step: **Given:** - \( X = 2245.75 \) - \( Y = 0.625 \) - **Operation:** \( Z = X + Y \) - **Task:** Convert both numbers to IEEE 754 single precision (32-bit), perform the addition in floating point, and give the result. --- ## **Step 1: Convert X and Y to IEEE 754 Single Precision Format** ### **IEEE 754 Single Precision Structure** - **1 bit**: Sign (S) - **8 bits**: Exponent (E) (bias = 127) - **23 bits**: Mantissa (M) #### **A. Convert X = 2245.75** 1. **Determine the sign bit:** - \( X > 0 \) ⇒ Sign bit = **0** 2. **Convert integer part (2245) to binary:** - 2245 / 2 = 1122, remainder 1 - 1122 / 2 = 561, remainder 0 - 561 / 2 = 280, remainder 1 - 280 / 2 = 140, remainder 0 - 140 / 2 = 70, remainder 0 - 70 / 2 = 35, remainder 0 - 35 / 2 = 17, remainder 1 - 17 / 2 = 8, remainder 1 - 8 / 2 = 4, remainder 0 - 4 / 2 = 2, remainder 0 - 2 / 2 = 1, remainder 0 - 1 / 2 = 0, remainder 1 Now, write the remainders in reverse: - 1 0 0 0 0 1 1 0 0 1 0 1 So, **2245 in binary = 100011001101** 3. **Convert fractional part (0.75) to binary:** - 0.75 × 2 = 1.5 → 1 - 0.5 × 2 = 1.0 → 1 - 0.0 (done) - So, **0.75 in binary = .11** So, **2245.75 in binary = 100011001101.11** 4. **Normalize the binary number:** - Move the decimal after the first 1: - \( 1.0001100110111 \times 2^{11} \) - (We keep more digits for mantissa calculation.) 5. **Exponent:** - \( 2^{11} \) → Exponent = 11 - Exponent field = 11 + bias = 11 + 127 = **138** - 138 in binary = 10001010 6. **Mantissa:** - Drop the leading 1, take the next 23 bits: - 00011001101111000000000 (pad with zeros as needed) - Let's get the full mantissa from the binary: - 0001100110111...(rest are zeros) - So, mantissa = **00011001101110000000000** 7. **IEEE 754 Representation of X = 2245.75:** - **Sign:** 0 - **Exponent:** 10001010 - **Mantissa:** 00011001101110000000000 **X (binary):** `0 10001010 00011001101110000000000` --- #### **B. Convert Y = 0.625** 1. **Sign bit:** 0 (since Y is positive) 2. **0.625 in binary:** - 0.625 × 2 = 1.25 → 1 - 0.25 × 2 = 0.5 → 0 - 0.5 × 2 = 1.0 → 1 - So, **0.625 in binary = 0.101** 3. **Normalize:** - \( 1.01 \times 2^{-1} \) 4. **Exponent:** - Exponent = -1 - Exponent field = -1 + 127 = **126** - 126 in binary = 01111110 5. **Mantissa:** - Remove the leading 1, take next 23 bits: - 01000000000000000000000 6. **IEEE 754 Representation of Y = 0.625:** - **Sign:** 0 - **Exponent:** 01111110 - **Mantissa:** 01000000000000000000000 **Y (binary):** `0 01111110 01000000000000000000000` --- ## **Step 2: Perform Floating Point Addition** ### **Step A: Align Exponents** - X exponent (E_X) = 138 - Y exponent (E_Y) = 126 **Difference = 12** - Shift Y's mantissa right by 12 bits to align with X's exponent. ### **Step B: Add Mantissas** #### **X:** - Mantissa: 1.0001100110111 (binary), pad with zeros: 1.00011001101110000000000 #### **Y:** - Mantissa: 1.01000000000000000000000 (remember the implicit leading 1) - After right shift by 12: 0.00000000000101000000000 #### **Add:** ``` X: 1.00011001101110000000000 Y: + 0.00000000000101000000000 ------------------------------- 1.00011001110011000000000 ``` - The result is approximately: 1.00011001110011000000000 in binary. ### **Step C: Normalize Result (if needed)** - No need to normalize, as the sum is still of the form 1.xxxxx... - Exponent remains 138 (binary: 10001010). ### **Step D: Final IEEE 754 Representation of Z** - **Sign bit:** 0 - **Exponent:** 10001010 (138) - **Mantissa:** 00011001110011000000000 --- ## **Step 3: Final Answer** ### **A. IEEE 754 Single Precision Format for Z** **Z = 2245.75 + 0.625 = 2246.375** - **Sign bit:** 0 - **Exponent:** 10001010 - **Mantissa:** 00011001110011000000000 **IEEE 754 (binary):** ``` 0 10001010 00011001110011000000000 ``` **In Hexadecimal:** - Sign: 0 - Exponent: 10001010 = 8A - Mantissa: 00011001110011000000000 - Mantissa in hex: 00011001110011000000000 = 0x19CC00 So, the hexadecimal representation: - 0 8A 19 CC 00 Or, as 32 bits: **0x48A19CC0** --- ## **Step 4: Final Numeric Answer** **Z = X + Y = 2246.375** --- ## **Summary Table** | Number | IEEE 754 Binary | IEEE 754 Hex | |-----------|-------------------------------|--------------| | 2245.75 | 0 10001010 00011001101110000000000 | 0x48A19B80 | | 0.625 | 0 01111110 01000000000000000000000 | 0x3F200000 | | **2246.375** | 0 10001010 00011001110011000000000 | **0x48A19CC0** | --- ## **Final Answers** **(a) X in IEEE 754 single precision:** `0 10001010 00011001101110000000000` **(b) Y in IEEE 754 single precision:** `0 01111110 01000000000000000000000` **(c) Z = X + Y in IEEE 754 single precision:** `0 10001010 00011001110011000000000` **Hexadecimal:** `0x48A19CC0` **(d) Z (decimal):** `2246.375`