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Determine the decimal values of the following 1's complement numbers:

shamieh

Active member
Sep 13, 2013
539
Determine the decimal values of the following 1's complement numbers:

So i understand that if the left most bit number is a 1 it is a negative, and if it is a 0 it is poisitive. But my question is why do they start out with -511 when \(\displaystyle 2^9\) is obviously -512. Why are they adding 1 to it initially?

1011100111 = -511 + 128 + 64 + 32 + 4 + 2 + 1 = -280
 

shamieh

Active member
Sep 13, 2013
539
After looking at it some more, it looks like I would say -511 + [numbers here] and I would + 1 initially, I guess that 's what "1's complacent means", similarly when it 's "2's complacent" it looks like I would not +1, but would still have a negative number if my left most bit began with a 1.

Decimal values (1’s complement)
(a) 0111011110 = 256 + 128 + 64 + 16 + 8 + 4 + 2 = 478
(b) 1011100111 = -511 + 128 + 64 + 32 + 4 + 2 + 1 = -280

Decimal values (2’s complement)
(a) 0111011110 = 256 + 128 + 64 + 16 + 8 + 4 + 2 = 478
(b) 1011100111 = -512 + 128 + 64 + 32 + 4 + 2 + 1 = -281
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The way I look at one's complement is to first find the unsigned value:

\(\displaystyle (1011100111)_2=1+2+4+32+64+128+512=743\)

Now count the number of binary digits, which is 10, and so subtract $2^{10}-1=1023$:

\(\displaystyle 743-1023=-280\)
 

shamieh

Active member
Sep 13, 2013
539
Wow that way is so much easier then what the book teaches.