De Morgan's theorem and (-47+27) by 2's complement

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

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)

 +47            = 0010 1111
 1's complement = 1101 0000
 add 1          = 1101 0001  (= -47)

Step 3: Add

   -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)

 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 $= -20_{10}$