The document discusses various methods for representing signed integers in binary, including signed magnitude, 1's complement, 2's complement, and excess binary. It provides examples of adding, subtracting, and multiplying numbers in binary using these different representations. 2's complement is described as the most common method used today due to its simplicity. The key aspects of 2's complement include representing negative numbers by flipping all bits and adding 1, and performing subtraction by adding the 2's complement. Overflow conditions for addition are also explained.

1. MBA admission in India
By:
Admission.edhole.com

2. Binary numbers and arithmetic
Admission.edhole.com

3. addition
Admission.edhole.com

4. Addition (decimal)
1
+
4
5
5
+
14
19
15
+
5
10
16
+
5
11
1 1
+
12
99
111
Admission.edhole.com

5. Addition (binary)
0
+
0
0
1
+
0
1
0
+
1
1
11
+
1
10
Admission.edhole.com

6. Addition (binary)
1 1 1 1
+
01101
01011
11000
Admission.edhole.com

7. Addition (binary)
0
+
0
1
+
0
So can w0e count in binary?
1
0
+
1
1
11
+
1
10
Admission.edhole.com

8. Counting in binary (4 bits)
0
0000
1
0001
2
…
3
4
5
6
7
8
9
10
11
12
13
14
15
Admission.edhole.com

9. multiplication
Admission.edhole.com

10. Multiplication (decimal)
13
11
13
130
143
+
´
Admission.edhole.com

11. Multiplication (binary)
1101
1011
1101
´
11010
1101000
+
10001111
Admission.edhole.com

12. Multiplication (binary)
1101
1011
1101
´
11010
1101000
+
10001111
It’s interesting to note
that binary multiplication
is a sequence of shifts
and adds of the first
term (depending on the
bits in the second term.
110100 is missing here
because the
corresponding bit in the
second terms is 0.
Admission.edhole.com

13. Representing signed (positive and
negative) numbers

14. Representing numbers (ints)
Fixed, finite number of bits.
bits bytes C/C++ Intel
Sun
8 1 char [s]byte byte
16 2 short [s]word half
32 4 int or long [s]dword word
64 8 long long [s]qword xword

15. Representing numbers (ints)
Fixed, finite number of bits.
bits Intel signed unsigned
8 [s]byte -27..+27-1 0..+28-1
16 [s]word -215..+215-1 0..+216-1
32 [s]dword -231..+231-1 0..+232-1
64 [s]qword -263..+263-1 0..+264-1
In general, for k bits, the unsigned range is [0..+2k-1] and the signed
range is [-2k-1..+2k-1-1].

16. Methods for representing signed ints.
1. signed magnitude
2. 1’s complement (diminished radix complement)
3. 2’s complement (radix complement)
4. excess bD-1

17. Signed magnitude
Ex. 4-bit signed magnitude
1 bit for sign
3 bits for magnitude
+ N - N
0 0000 1000
1 0001 1001
2 0010 1010
3 0011 1011
4 0100 1100
5 0101 1101
6 0110 1110
7 0111 1111

18. Signed magnitude
Ex. 4-bit signed magnitude
1 bit for sign
3 bits for magnitude
+ N - N
0 0000 1000
1 0001 1001
2 0010 1010
3 0011 1011
4 0100 1100
5 0101 1101
6 0110 1110
7 0111 1111

19. 1’s complement
(diminished radix complement)
Let x be a non-negative number.
Then –x is represented by bD-1+(-x) where
b = base
D = (total) # of bits (including the sign bit)
Ex. Let b=2 and D=4.
Then -1 is represented by 24-1-1 = 1410 or 11102.

20. 1’s complement
(diminished radix complement)
Let x be a non-negative number.
Then –x is represented by bD-1+(-x) where
b = base & D = (total) # of bits (including the sign bit)
Ex. What is the 9’s complement of 1238910?
Given b=10 and D=5. Then the 9’s complement of 12389
= 105 – 1 – 12389
= 100000 – 1 – 12389
= 99999 – 12389
= 87610

21. 1’s complement
(diminished radix complement)
Let x be a non-negative number.
Then –x is represented by bD-1+(-x) where
b = base
D = (total) # of bits (including the sign bit)
Shortcut for base 2?
All combinations used, but 2 zeros!
+ N - N
0 0000 1111
1 0001 1110
2 0010 1101
3 0011 1100
4 0100 1011
5 0101 1010
6 0110 1001
7 0111 1000

22. 2’s complement
(radix complement)
Let x be a non-negative number.
Then –x is represented by bD+(-x).
Ex. Let b=2 and D=4. Then -1 is represented by 24-1 = 15 or
11112.
Ex. Let b=2 and D=4. Then -5 is represented by 24 – 5 = 11
or 10112.
Ex. Let b=10 and D=5. Then the 10’s complement of 12389
= 105 – 12389 = 100000 – 12389 = 87611.

23. 2’s complement
(radix complement)
Let x be a non-negative number.
Then –x is represented by bD+(-x).
Ex. Let b=2 and D=4. Then -1 is
represented by 24-1 = 15 or 11112.
Ex. Let b=2 and D=4. Then -5 is
represented by 24 – 5 = 11 or 10112.
Shortcut for base 2?
+ N - N
0 0000 0000
1 0001 1111
2 0010 1110
3 0011 1101
4 0100 1100
5 0101 1011
6 0110 1010
7 0111 1001

24. 2’s complement
(radix complement)
Shortcut for base 2?
Yes! Flip the bits and add 1.
+ N - N
0 0000 0000
1 0001 1111
2 0010 1110
3 0011 1101
4 0100 1100
5 0101 1011
6 0110 1010
7 0111 1001

25. 2’s complement
(radix complement)
Are all combinations of 4 bits used?
No. (Now we only have one zero.)
1000 is missing!
What is 1000?
Is it positive or negative?
Does -8 + 1 = -7 work in 2’s complement?
+ N - N
0 0000 0000
1 0001 1111
2 0010 1110
3 0011 1101
4 0100 1100
5 0101 1011
6 0110 1010
7 0111 1001

26. excess bD-1 (biased representation)
For pos, neg, and 0, x is represented by
bD-1 + x
Ex. Let b=2 and D=4. Then the excess 8 (24-1)
representation for 0 is 8+0 = 8 or 10002.
Ex. Let b=2 and D=4. Then excess 8 for -1 is 8 – 1 = 7 or
01112.

27. excess bD-1
For pos, neg, and 0, x is represented by
bD-1 + x.
Ex. Let b=2 and D=4. Then the excess 8
(24-1) representation for 0 is 8+0 = 8 or
10002.
Ex. Let b=2 and D=4. Then excess 8 for -1
is 8 – 1 = 7 or 01112.
+ N - N
0 1000 1000
1 1001 0111
2 1010 0110
3 1011 0101
4 1100 0100
5 1101 0011
6 1110 0010
7 1111 0001

28. 2’s complement vs. excess bD-1
In 2’s, positives start with 0; in excess,
positives start with 1.
Both have one zero (positive).
Remaining bits are the same.
+ N - N
0 1000 1000
1 1001 0111
2 1010 0110
3 1011 0101
4 1100 0100
5 1101 0011
6 1110 0010
7 1111 0001

29. Summary of methods for
representing signed ints.
signedMag sComp sComp excess
1 2 8
N + n - n - n - n - n +
n
0 0000 1000 1111 0000 1000 1000
1 0001 1001 1110 1111 0111 1001
2 0010 1010 1101 1110 0110 1010
3 0011 1011 1100 1101 0101 1011
4 0100 1100 1011 1100 0100 1100
5 0101 1101 1010 1011 0011 1101
6 0110 1110 1001 1010 0010 1110
7 0111 1111 1000 1001 0001 1111
1000=-8| 0000 unused

30. Binary arithmetic
Signed magnitude
1’s complement
2’s complement
Excess K (biased)

31. Binary Arithmetic
Signed magnitude

32. Addition w/ signed magnitude
algorithm
For A - B, change the sign of B and perform addition of A + (-B) (as
in the next step)
For A + B:
if (Asign==Bsign) then { R = |A| + |B|; Rsign = Asign; }
else if (|A|>|B|) then { R = |A| - |B|; Rsign = Asign; }
else if (|A|==|B|) then { R = 0; Rsign = 0; }
else { R = |B| - |A|; Rsign = Bsign; }
Complicated?

33. Binary Arithmetic
2’s complement

34. Representing numbers (ints) using
2’s complement
Fixed, finite number of bits.
bits Intel signed
8 sbyte -27..+27-1
16 sword -215..+215-1
32 sdword -231..+231-1
64 sqword -263..+263-1
In general, for k bits, the signed range is [-2k-1..+2k-1-1].
So where does the extra negative value come from?

35. Representing numbers (ints)
Fixed, finite number of bits.
bits Intel signed
8 sbyte -27..+27-1
16 sword -215..+215-1
32 sdword -231..+231-1
64 sqword -263..+263-1
+ n - n
0 0000 0000
1 0001 1111
2 0010 1110
3 0011 1101
4 0100 1100
5 0101 1011
6 0110 1010
In general, for k bits, the signed range is
[-2k-1..+2k-1-1].
So where does the extra negative value come
from? 8 1000
7 0111 1001

36. Addition of 2’s complement binary
numbers
Consider 8-bit 2’s complement binary numbers.
Then the msb (bit 7) is the sign bit. If this bit is 0, then this is a
positive number; if this bit is 1, then this is a negative number.
Addition of 2 positive numbers.
Ex. 40 + 58 = 98
1 1 1
00101000
00111010
01100010
+

37. Addition of 2’s complement binary
numbers
Consider 8-bit 2’s complement binary
numbers.
Addition of a negative to a positive.
What are the values of these 2 terms?
-88 and 122
-88 + 122 = 34
1 1 1 1
+
10101000
01111010
1 00100010

38. So how can we perform subtraction?

39. Addition of 2’s complement binary
numbers
Consider 8-bit 2’s complement binary numbers.
Subtraction is nothing but addition of the 2’s complement.
Ex. 58 – 40 = 58 + (-40) = 18
1 1 1 1
+
00111010
11011000
1 00010010
discard carry

40. Carry vs. overflow

41. Addition of 2’s complement binary
numbers
Carry vs. overflow when adding A + B
If A and B are of opposite sign, then overflow cannot occur.
If A and B are of the same sign but the result is of the opposite
sign, then overflow has occurred (and the answer is therefore
incorrect).
Overflow occurs iff the carry into the sign bit differs from the carry out
of the sign bit.

42. Addition of 2’s complement binary
numbers
class test {
public static void main ( String args[] )
{
byte A = 127;
byte B = 127;
byte result = (byte)(A + B);
System.out.println( "A + B = "
+ result );
}
}
#include <stdio.h>
int main ( int argc, char* argv[] )
{
char A = 127;
char B = 127;
char result = (char)(A + B);
printf( "A + B = %d n", result );
return 0;
} Result = -2 in both
Java (left) and C++
(right). Why?

43. Addition of 2’s complement binary
numbers
class test {
public static void main ( String args[] )
{
byte A = 127;
byte B = 127;
byte result = (byte)(A + B);
System.out.println( "A + B = "
+ result );
}
}
Result = -2 in both Java and C++.
Why?
What’s 127 as a 2’s complement
binary number?
01111111
+
01111111
11111110
What is 111111102?
Flip the bits: 00000001.
Then add 1: 00000010.
This is -2.

44. Binary Arithmetic
1’s complement

45. Addition with 1’s complement
Note: 1’s complement has two 0’s!
1’s complement addition is tricky (end-around-
carry).
+ N - N
0 0000 1111
1 0001 1110
2 0010 1101
3 0011 1100
4 0100 1011
5 0101 1010
6 0110 1001
7 0111 1000

46. 8-bit 1’s complement addition
Ex. Let X = A816 and Y = 8616.
Calculate Y - X using 1’s complement.

47. 8-bit 1’s complement addition
Ex. Let X = A816 and Y = 8616.
Calculate Y - X using 1’s complement.
Y = 1000 01102 = -12110
X = 1010 10002 = -8710
~X = 0101 01112
1000 0110
+
0101 0111
(Note: C=0 out of msb.) 1101 1101
Y - X = -121 + 87 = -34 (base 10)

48. 8-bit 1’s complement addition
Ex. Let X = A816 and Y = 8616.
Calculate X - Y using 1’s complement.

49. 8-bit 1’s complement addition
Ex. Let X = A816 and Y = 8616.
Calculate X - Y using 1’s complement.
X = 1010 10002 = -8710
Y = 1000 01102 = -12110
~Y = 0111 10012
(Note: C=1 out of msb.)
1010 1000
0111 1001
+
1 0010 0001
1
+
end around
carry
0010 0010
X - Y = -87 + 121 = 34 (base 10)

50. Binary Arithmetic
Excess K (biased)

51. Binary arithmetic and Excess K
(biased)
Method: Simply add and then flip the sign bit.
-1 0111
+5 1101
-- ----
+4 0100 -> flip sign -> 1100
+1 1001
-5 0011
-- ----
-4 1100 -> flip sign -> 0100
+1 1001
+5 1101
-- ----
+6 0110 -> flip sign -> 1110
-1 0111
-5 0011
-- ----
-6 1010 -> toggle sign -> 0010
+ N - N
0 1000 1000
1 1001 0111
2 1010 0110
3 1011 0101
4 1100 0100
5 1101 0011
6 1110 0010
7 1111 0001
(Not used for integer arithmetic but employed
in IEEE 754 floating point standard.)