Hello anderson Originally Posted by

**anderson** Hi everyone

Need help to verify my working,especially for Qiii, the extra 9th bit is not present.

Perform the arithmetic using 2’s complement notation. The numbers are given in decimal, so first convert them to appropriate form. Make suitable assumptions.

(i) -29 – 37

-29 =00011101

nvert = 11100010+1

= 11100011

-37 =00100101

invert = 11011010+1

= 11011011

11100011+11011011=110111110

=-66

(ii) 95 – 28

95=01011111

-28=00011100

INVERT=11100011+1=11100100

01011111+11100100=01000011

=67

(iii) 28 – 78

28=00011100

-78=01001110=10110001+1=10110010

00011100+10110010=11001110

-50

Thank you for all your kind help & support.

The assumption we need to make is that this is 8-bit 2's complement. So the valid range of integers is -128 to +127 (decimal).

Provided we don't overflow this range, we can work to 8 bits and ignore any 9th bit that may arise.

So in (i) when you add the 2's complement numbers you get:

...11100011 +

...11011011

...---------

(1)10111110

Ignoring the overflow 9th bit (1), converting the remaining 8 bits back to decimal gives the answer -66.

This situation with the overflow bit doesn't arise in (ii) and (iii), so there's no problem with your working.

Grandad