--- title: "De Morgan's theorem and (-47+27) by 2's complement" subject: "Digital Logic" chapter: 2 key: "semester_3/digital_logic:demorgan-twos-complement-addition" source: https://www.bctengineer.com/q/semester_3/digital_logic/demorgan-twos-complement-addition site: BCT Engineering --- # De Morgan's theorem and (-47+27) by 2's complement Digital Logic · Chapter 2: Digital Logic - Asked in: 2076 Chaitra, 3+3 marks - Video: [De Morgan's Law in Boolean Algebra Explained (with Solved Examples)](https://www.youtube.com/watch?v=W7YTfLaPWRY) — ALL ABOUT ELECTRONICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/digital_logic/demorgan-twos-complement-addition - Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-2 ## Question State and prove the De-Morgan's theorem and perform the addition (-47+27) by using 2's complement method. ## Short answer De Morgan's laws: ¬(A+B)=¬A·¬B, ¬(A·B)=¬A+¬B, proved by truth table. For (-47)+27 using 8-bit 2's complement: -47 is 11010001, +27 is 00011011, sum is 11101100 (no carry), which is -20 in decimal. ## Answer #### De Morgan's theorems **First theorem:** the complement of a sum (OR) of variables equals the product (AND) of their complements. $$ \overline{A + B} = \bar{A} \cdot \bar{B} $$ **Second theorem:** the complement of a product (AND) of variables equals the sum (OR) of their complements. $$ \overline{A \cdot B} = \bar{A} + \bar{B} $$ In words: "break the bar and change the sign". **Proof by truth table** | A | B | $\overline{A+B}$ | $\bar A \cdot \bar B$ | $\overline{AB}$ | $\bar A + \bar B$ | |---|---|---|---|---|---| | 0 | 0 | 1 | 1 | 1 | 1 | | 0 | 1 | 0 | 0 | 1 | 1 | | 1 | 0 | 0 | 0 | 1 | 1 | | 1 | 1 | 0 | 0 | 0 | 0 | For all four input combinations, column 3 equals column 4 and column 5 equals column 6, so both theorems are proved. #### (−47) + (27) using 2's complement Use 8-bit signed numbers (range −128 to +127). **Step 1: +47 and +27 in binary** - $47 = 32 + 8 + 4 + 2 + 1 = 0010\ 1111$ - $27 = 16 + 8 + 2 + 1 = 0001\ 1011$ **Step 2: 2's complement of 47 (to get −47)** ```text +47 = 0010 1111 1's complement = 1101 0000 add 1 = 1101 0001 (= -47) ``` **Step 3: Add** ```text -47 = 1101 0001 + +27 = 0001 1011 ------------------- 1110 1100 ``` No carry out of the MSB, and the sign bit is 1, so the result is negative and is in 2's complement form. **Step 4: Find its magnitude** (take 2's complement) ```text result = 1110 1100 1's complement = 0001 0011 add 1 = 0001 0100 = 20 ``` Check: $-47 + 27 = -20$. **Answer:** $(-47) + (27) = (1110\ 1100)_2$ in 2's complement $= -20_{10}$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. (A+B+C)' = A. A'B'C' B. A'+B'+C' C. ABC D. A+B+C Answer: A. Extends to any number of variables. (Key point: Theorem) 2. How is -47 formed? A. Invert bits of +47 and add 1 B. Invert only C. Add 1 only D. Subtract 1 Answer: A. 47 = 00101111 gives 11010001. (Key point: 2's complement) 3. 8-bit sum of -47 and +27? A. 00010100 B. 11010001 C. 10101100 D. 11101100 Answer: D. 11010001 + 00011011. (Key point: Sum) 4. Sign bit 1 and no carry means? A. Positive 20 B. Overflow C. Negative result in 2's complement; magnitude 20 D. Result is 236 Answer: C. 2's complement of 11101100 is 00010100 = 20, so -20. 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