Given , 2\'s complement of A= 10101011 (here sign bit = 1 => negative) A= -128*1

Question

Given , 2's complement of A= 10101011 (here sign bit = 1 => negative) A= -128*1+64*0+32*1+16*0+8*1+4*0+2*1+1*1 = -128+32+8+2+1 = -128 + 43 = -85 A= -85 2's complement of B = 01101111 (here sign bit = 0=> positive) B = 128*0+64*1+32*1+16*0+8*1+4*1+2*1+1*1 = 64+32+8+4+2+1 = 111 B=111 A+B = 10101011 01101111 ------------ 00011010 , carry = 1 overflow occurs in A+B A-B = 00111100 If A=-85 and B=111, shouldn't A-B= -196? but this A-b=00111100 doesn't equal -196, since it starts with a 0. can someone explain why A-B=00111100?

Explanation / Answer

A= 10101011

B=01101111

To find A-B , 1st convert B into 2's complement

2's complement of B= 10010001

A-B=

10101011

10010001

------------

00111100 , carry =1

=> 100111100

overflow occurs

A-B = 100111100

-256*1+128*0+64*0+32*1+16*1+8*1+4*1+2*0+1*0

= -256+32+16+8+4

=-196

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