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3.2 Addition and Subtraction of Signed Numbers We can relate addition and subtraction operations of numbers by the following relationship : (+ A) — (+B) = (+ A) + (-B) and (t A) - (-B) = (+ A) + (+B) iy, +} Therefore, we can change subtraction operation to an addition operation by Chang; the sign of the subtrahend. Let us see how we can represent negative numbers in binay system. 1's Complement Representation The 1's complement of a binary number is the number that results when we change al 1's to zeros and the zeros to ones, Dd Example 3.1: Find 1's complement of (1 1 0 1)>. Solution: 1 1 9 1 © number 9 0 1 0 € 1's complement => Example 3.2: Find 1's complement of 1011 1001. Solution: 101 11 0 0 1 number 01000110 1's complement 2’s Complement Representation The 2's complement is the binary number that results when we add 1 to the 1%, complement. It is given as 2's complement = 1’s complement + 1 The 2’s complement form is used to represent negative numbers. => Example 3.3: Find 2's complement of (1 0 0 1)>. Solution : 1001 number 0110 1’s complement + 1 o1i1i 2's complement wa> Example 3.4: Find 2's complement of (1010 001 1),. Solution : 1010 0011 number 0101 1100 t's complement + 1 0101 1101 2's complement Let us see the subtraction of binary number using 1's complement and 2's complement number representations. Addition of 28 and - 15: + 0 1 4 1 0 0 (28); Sign extension 1 1 0 0 0 0 (-15) 19 a Cany > 1 #0 ) 1 1 0 0 1 Add end-around carry 0 o 1 1 0 1 (13) 46 Case 3 (Greater Negative) : Add (-28)1o and (15)49 We have (011100). — (28);9 and (01111). > (15)o (100011). > (—28),9 Addition of (- 28) and 15 1 0 0 0 1 1 (-28),9 * Sign 4+ 0 0 1 | 1 1 (15)19 extension 1 1 0 0 1 0 (=13)19 Result is in 1's complement form Verification: 10 40 Case 4 (Both Negative) : Add (- 28),) and (- 15)49 We have (011100), —» (28),) and (01111), —> (15),5 (10000), —> 1's complement of 15 (100011), — 1's complement of 28 Addition of (- 28) and (- 15): Sign-extension + 14 1 0 0 0 1 1 + Sign-extension -» 1 1 1 0 0 0 0 Carry 1 1 0 1 0 0 1 1 L 1 Add end-around carry 1 10) 1 0 1 0 0 (~ 43) Result is in 18 complement form Verification: 1 0 1 Le} 4 9 0 sO As BO a4 0 VPA oy MAB) ote : e Here, the magnitude of greater number is 5-bit; however, the magnitude of the result is 6-bit. Therefore, the numbers are sign-extended to 7-bits. e For proper result we suggest to use 1 sign-bit extension to the number having greater magnitude and represent the number having smaller magnitude with extended number of bits. 
