Lecture 2: Signed Numbers  Representation and Arithmetic
We have already noted how in a digital system we can only use 1’s and 0’s to represent
information. For the full implementation of arithmetic functions, we need to be able to
represent both positive and negative numbers and to do so in a way that allows us to
maintain the normal methods of basic arithmetic.
Learning Outcomes:
On completion of this lecture, you will be able
 to define and apply the 2’s complement representation format;
 to demonstrate the addition and subtraction of positive and negative numbers;
 to specify when an overflow occurs;
 to calculate the arithmetic value directly from the 2’s complement representation.
2.1
Background Context
Digital machines are built to handle binary data of some specific maximum wordlength which
we will denote by N. That is, we can assume N-bit words:
X  x N 1 x N 2  x1 x0
For the purposes of numerical examples we will mostly restrict ourselves to the case of
N  4 , but bear in mind that today’s microprocessors typically have N  64 .
One of the important consequences of having finite bit capacity is that when you add two
relatively large numbers, the true sum can fall outside the available bit capacity; an overflow
is generated and you end up with an “incorrect” sum. Consider for example what happens
when adding 9 to 12 with 4-bit capacity:
1 1 0 0
1 0 0 1
(12)
(9)
(1) 0 1 0 1
(5)
With the restriction of N  4 , we lose the most significant carry out which has a weight of
16. Thus rather than getting a result of 21, we get 5. Such finite precision arithmetic is
formally termed addition modulo 2N and is denoted
(12  9) Mod 16  5
A familiar example of modulo addition is the minutes count on a clock:
45 min
 30 min
 15 min
2.2
Sign and Magnitude
Recall that our basic goal is to represent both positive and negative integers using just 1’s
and 0’s. The most simple method, for N-bit capacity, is to use the most significant bit (msb),
xN-1, to designate the sign and the remaining (N-1) bits to designate the magnitude:
x 1 x N 2  x1 x0

N
 

sign
and the usual convention is
magnitude
x N 1  0 
X
positive
x N 1  1 
X
negative
Examples 2.1:
 6  0 11 0
 5  11 0 1
While the Sign and magnitude format is intrinsically simple, it leads to significant
complications if the normal rules of arithmetic are applied:
Examples 2.2:
add
(3)
(2)
0 0 11
0 010
is correct, but
0 1 0 1  (5)
add
(3)
(2)
0 0 11
1010
is incorrect.
1 1 0 1  (5)
Because of the problems arising with addition and subtraction, the sign and magnitude
system is not generally used where arithmetic functions are required; it may, however, be
used for data storage.
2.3
Signed 2’s Complement
This widely used convention is defined as follows:
Let X denote an N-bit number whose most significant bit (msb) is zero,
ie
then
X  0 x N 2  x1 x0
 X 2C
Example 2.3:
X
and
let X  5  0 1 0 1 ,
 X 2C
 2N  X
with as usual
N 4
 52C
 0101
 52C
 24  X  1 0 0 0 0
 52C
ie
01 01
 1 0 11
1 01 1
Clearly this 2’s complement operation, as it is called, is relatively complex and, with
2 4  10 0 0 0 , the subtraction step strictly speaking does not fit within our N  4
constraint. However,
let
X   0 xN 2  x1 x0
where by definition the bit level invert or complement operation is defined by
0  1
1  0
and
By construction, for any bit pattern X
X  X   1 1  1 1
 2N  1
Rearrange to give
 X 2C
 2N  X
 X 1
That is, an algorithm for carrying out the 2’s complement operation is
(i)
invert all the bits, and
(ii)
add a 1 in the least significant bit (lsb) position.
Example 2.4:
then
if
 52C
 52C
 1010
 0101
1
which is the same answer as previous.
1 0 11
Note:
(i)
Similar to the sign and magnitude system, in signed 2’s complement the msb denotes
the polarity of the number:
x N 1  0  X positive;
(ii)
x N 1  1  X negative
If 2C[ ] denotes the 2’s complement operation, then
2C X 2C    X 2C
ie 2’s complementing a negative number yields the corresponding positive number.
For example,
 52C  1 0 1 1
noting that
2C 1 0 1 1  0 1 0 0
1
0 1 0 1  (5) 2C
(iii)
Strictly speaking, the addition of the 1 in the lsb position should be carried out
modulo 2N; consider performing the 2’s complement operation on zero:
2C 0 0 0 0  1 1 1 1
1
0000
In an appendix to this lecture, a table of all possible 4-bit 2’s complement numbers is given.
Note that the range of numbers is not symmetric; the eight positive numbers extend from 0
to +7 while the eight negative numbers range from -8 to -1.
2.4
Addition and Subtraction
We demonstrate the addition and subtraction of numbers in signed 2’s complement format
by means of examples covering all combinations of positive and negative. The basic idea is
that the normal arithmetic rules apply with the exception that the addition is modulo 2N, ie
the most significant carry out is ignored:
010 0
0 0 11
(4)
(3)
010 0
11 0 1
( 4)
( 3)
0 111
(7)
0 0 01
( 1)
11 0 0
0 0 11
(4)
(3)
11 0 0
11 0 1
(4)
(3)
1111
(1)
10 01
(7)
Note that in all four cases the result comes out correct in signed 2’s complement format.
Based on the algebraic rule that ( X )  (Y )  ( X )  (Y ) , we propose to carry out subtraction
by adding the 2’s complement of the subtrahend (Y) to the minuend (X) with, again, the
addition being modulo 2N:
010 0
0 0 11
(4)
(3)

010 0
11 0 0
1
0 0 0 1  (1) 2C
010 0
11 0 1
(4)
(3)

010 0
0 010
1
0 1 1 1  (7) 2C
11 0 0
0 0 11
(4)
(3)

11 0 0
11 0 0
1
1 0 0 1  (7) 2C
11 0 0
11 0 1
(4)
(3)

11 0 0
0 010
1
1 1 1 1  (1) 2C
Again each time the result comes out correct in signed 2’s complement format.
2.5
Overflow
All of the previous results were carefully chosen to the extent that the numbers all fell within
the valid range for 2’s complement representation with N  4 . What happens if the
arithmetic result exceeds the valid range? An incorrect result is produced and the arithmetic
system should have the capability of flagging an overflow. Consider the following two cases:
(6)
(7)
0 11 0
0 111
1 1 0 1  (13) 2C
(6)
(7)
1010
10 01
0 0 1 1  (13) 2C
We observe that overflow conditions are produced when the sign bits of the operands are
the same and differ from the sign bit of the result. Later we will see how to incorporate this
into the hardware of an arithmetic processing unit.
2.6
Arithmetic Value
In the first lecture we saw that, given an unsigned binary integer
A  a N 1 a N 2 a1 a0 r
N 1
we could calculate its arithmetic value from
a
i 0
i
ri
The question now is: can we do something similar for signed 2’s complement
representation?
It may be shown that, given a 2’s complement number X  x N 1 x N 2  x1
x0 , its
arithmetic value may be directly computed from
X   x N 1 2 N 1 
N 2
x
i 0
i
2i
Examples 2.5:
0 1 0 1
  0  2 3  1 2 2  0  21  1 2 0
 0  4  0 1
 5
1 0 1 1
  1 2 3  0  2 2  1 21  1 2 0
 8  0  2 1
 5
Both results are correct.
2.7
Conclusion
At this stage, we have some familiarity with decimal and binary representations, with how to
conveniently represent positive and negative numbers in a digital machine, and with how to
carry out basic arithmetic operations. Having set some relevant application context, we now
proceed to the mathematical framework specific to digital systems.
Appendix: Table of 4-bit 2’s complement numbers
Decimal value
+7
+6
+5
+4
+3
+2
+1
0
-1
-2
-3
-4
-5
-6
-7
-8
X3
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
X2
1
1
1
1
0
0
0
0
1
1
1
1
0
0
0
0
X1
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
X0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0