- Digital computers perform arithmetic operations like addition, subtraction, multiplication and division on binary numbers. - Signed binary numbers use the most significant bit as the sign bit to represent positive and negative values. Common representations are sign-magnitude, one's complement, and two's complement. - Subtraction is performed using the two's complement method by taking the two's complement of the subtrahend and adding it to the minuend. Overflow needs to be handled for accurate results.

1. COMPUTER ARITHMETIC

2. • Addition and Subtraction
• Two’s Compliment Representation
• Signed Addition and Subtraction
• Multiplication and division
• Division Operation

3. • Data is manipulated by using the arithmetic instructions in
digital computers.
• Data is manipulated to produce results necessary to give solution for
the computation problems.
• The Addition, subtraction, multiplication and division are the four
basic arithmetic operations.

4. • Inside a computer system, all operations are carried out on fixed-
length binary values
• The unsigned (binary number system) and signed (2’s complement)
representations have the advantage that addition and subtraction
operations have simple
• and that the same algorithm can be used for both representations.

5. ADDITION AND
SUBTRACTION

6. General concept
• Decimal addition
(carry) 1_
19
+ 7
26
• Binary addition
( carry) 111_
10011
+ 111
11010
• 16+8+2 = 26

7. • Signed Binary Numbers
• Signed Binary Numbers use the
MSB as a sign bit to display a
range of either positive numbers
or negative numbers

9. • In mathematics, positive numbers (including zero) are represented as
unsigned numbers. That is we do not put the +ve sign in front of them
to show that they are positive numbers.
• However, when dealing with negative numbers we do use a -ve sign in
front of the number to show that the number is negative in value and
different from a positive unsigned value, and the same is true
with signed binary numbers.

10. • However, in digital circuits there is no provision made to put a plus or
even a minus sign to a number,
• since digital systems operate with binary numbers that are
represented in terms of “0’s” and “1’s”.
• When used together in microelectronics, these “1’s” and “0’s”, called
a bit (being a contraction of BInary digiT), fall into several range sizes
of numbers which are referred to by common names, such as
a byte or a word.

11. • Mathematical numbers are generally made up of a sign and a value
(magnitude) in which the sign indicates whether the number is
positive, ( + ) or negative, ( – ) with the value indicating the size of the
number,
• for example 23, +156 or -274. Presenting numbers is this fashion is
called “sign-magnitude” representation since the left most digit can
be used to indicate the sign and the remaining digits the magnitude
or value of the number.

12. • But how do we represent signed binary numbers if all we have is just
one’s and zero’s.
• We know that binary digits, or bits only have two values, either a “1”
or a “0” and conveniently for us, a sign also has only two values, being
a “+” or a “–“.

13. signed binary numbers
• For signed binary numbers the most significant bit (MSB) is used as
the sign bit.
• If the sign bit is “0”, this means the number is positive in value.
• If the sign bit is “1”, then the number is negative in value.
• The remaining bits in the number are used to represent the
magnitude of the binary number in the usual unsigned binary number
format way.

14. signed binary numbers
• Then we can see that the Sign-and-Magnitude (SM) notation stores
positive and negative values by dividing the “n” total bits into two
parts:
• 1 bit for the sign and n–1 bits for the value which is a pure binary
number.

15. Positive Signed Binary Numbers

16. Negative Signed Binary Numbers

17. Signed and unsigned additions
• Unsigned addition in 4-
bit arithmetic
( carry) 11_
1011
+ 0011
1110
• 11 + 3 = 14
(8 + 4 + 2)
• Signed addition in
4-bit arithmetic
( carry) 11_
1011
+ 0011
1110
• -5 + 3 = -2

18. • 1 0
• 0+0=0
• 1+0=1
• 1+1=0 carry =1
• 01+01=10
• 1+1+1=11
• 10+1=11

19. • 0-0=0
• 1-0=1
• 0-1=-1
• 1-1=0
• 10-11

20. Signed and unsigned additions
• Same rules apply even though bit strings represent different values
• Sole difference is overflow handling

21. One’s Complement of a Signed Binary Number
• One’s Complement or 1’s Complement is another method which we
can use to represent negative binary numbers in a signed binary
number system.
• In one’s complement, positive numbers remain unchanged as before
with the sign-magnitude numbers.

22. One’s Complement of a Signed Binary Number
• Negative numbers are represented by taking the one’s complement of the
unsigned positive number.
• Since positive numbers always start with a “0”, the complement will always
start with a “1” to indicate a negative number.
• The one’s complement of a negative binary number is the complement of
its positive counterpart, so to take the one’s complement of a binary
number, all we need to do is change each bit in turn.
• Thus the one’s complement of “1” is “0” and vice versa,
• then the one’s complement of 100101002 is simply 011010112 as all the 1’s
are changed to 0’s and the 0’s to 1’s.

23. 2’s complement
• The 2’s complement is defined as 2n-N
• Can be done by subtraction of N from 2n or adding 1 to the 1’s
complement of a number.
• For 6 = 0110
• The 1’s complement is 1001
• The 2’s complement is 1010

24. operation is carried out by means of the following steps:
• (i) At first, 2’s complement of the subtrahend is found.
• (ii) Then it is added to the minuend.
• (iii) If the final carry over of the sum is 1, it is dropped and the result
is positive.
• (iv) If there is no carry over, the two’s complement of the sum will be
the result and it is negative.

25. (i) 110110 - 10110
• The numbers of bits in the subtrahend is 5 while that of minuend is 6.
• We make the number of bits in the subtrahend equal to that of minuend by taking a `0’ in the
sixth place of the subtrahend.
• Now, 2’s complement of 010110 is (101101 + 1) i.e.101010.
• Adding this with the minuend.
•
1 1 0 1 1 0 Minuend
1 0 1 0 1 0 2’s complement of subtrahend
Carry over 1 1 0 0 0 0 0 Result of addition
• After dropping the carry over we get the result of subtraction to be 100000.

26. (ii) 10110 – 11010
• 2’s complement of 11010 is (00101 + 1) i.e. 00110.
• Hence
• Minued - 1 0 1 1 0
2’s complement of subtrahend - 0 0 1 1 0
Result of addition - 1 1 1 0 0
• As there is no carry over, the result of subtraction is negative and is
obtained by writing the 2’s complement of 11100 i.e.(00011 + 1) or 00100.
• Hence the difference is – 100.

27. (iii) 1010.11 – 1001.01
• 2’s complement of 1001.01 is 0110.11. Hence
• Minued - 1 0 1 0 . 1 1
2’s complement of subtrahend - 0 1 1 0 . 1 1
Carry over 1 0 0 0 1 . 1 0
• After dropping the carry over we get the result of subtraction as 1.10.

28. Subtract the following numbers:
• i) 101 from 1001
Solution:
101 from 1001
1 Borrow
1 0 0 1
1 0 1
1 0 0

29. ii) 1010101.10 from 1111011.11
• Solution:
1010101.10 from 1111011.11
1 Borrow
1 1 1 1 0 1 1 . 1 1
1 0 1 0 1 0 1 . 1 0
1 0 0 1 1 0 . 0 1

30. • The operation is carried out by means of the following steps:
• (i) At first, 2’s complement of the subtrahend is found.
• (ii) Then it is added to the minuend.
• (iii) If the final carry over of the sum is 1, it is dropped and the result
is positive.
• (iv) If there is no carry over, the two’s complement of the sum will be
the result and it is negative.
Subtraction by 2’s Complement

31. (i) 110110 - 10110
• The numbers of bits in the subtrahend is 5 while that of minuend is 6. We make the
number of bits in the subtrahend equal to that of minuend by taking a `0’ in the sixth
place of the subtrahend.
• Now, 2’s complement of 010110 is (101101 + 1) i.e.101010. Adding this with the
minuend.
•
1 1 0 1 1 0 Minuend
1 0 1 0 1 0 2’s complement of subtrahend
Carry over 1 1 0 0 0 0 0 Result of addition
• After dropping the carry over we get the result of subtraction to be 100000.
•

32. • (iii) 1010.11 – 1001.01
• Solution:
• 2’s complement of 1001.01 is 0110.11. Hence
• Minued - 1 0 1 0 . 1 1
2’s complement of subtrahend - 0 1 1 0 . 1 1
Carry over 1 0 0 0 1 . 1 0
• After dropping the carry over we get the result of subtraction as 1.10.
•

33. Subtraction by 1’s Complement
• The steps to be followed in subtraction by 1’s complement are:
• i) To write down 1’s complement of the subtrahend.
• ii) To add this with the minuend.
• iii) If the result of addition has a carry over then it is dropped and an 1
is added in the last bit.
• iv) If there is no carry over, then 1’s complement of the result of
addition is obtained to get the final result and it is negative.

34. • 1’s complement of 10011 is 011010. Hence
• Minued - 1 1 0 1 0 1
1’s complement of subtrahend - 0 1 1 0 1 0
Carry over - 1 0 0 1 1 1 1
1
0 1 0 0 0 0
• The required difference is 10000
Evaluate:
(i) 110101 – 100101

35. • (iii) 1011.001 – 110.10
• Solution:
• 1’s complement of 0110.100 is 1001.011 Hence
• Minued - 1 0 1 1 . 0 0 1
1’s complement of subtrahend - 1 0 0 1 . 0 1 1
Carry over - 1 0 1 0 0 . 1 0 0
1
0 1 0 0 . 1 0 1

36. Binary Multiplication
• (i) 10111 by 1101
• Solution:
• 1 0 1 1 1
1 1 0 1
1 0 1 1 1 ← First partial product
1 0 1 1 1
1 1 1 0 0 1 1 ← First intermediate sum
1 0 1 1 1
1 0 0 1 0 1 0 1 1 ← Final sum.
• Hence the required product is 100101011.
• 1x0=0
• 1x1=1
•

37. • 12x11
• 12
• X11
• 11x2 22
• +11x1 11
• 132

38. • (i) 11001 ÷ 101
• 101) 11001 (101
101
101
101
• Hence the quotient is 101
Binary Division

39. • (ii) 11101.01 ÷ 1100
• Solution:
• 1100) 11101.01 (10.0111
1100
10101
1100
0010
1100
1100
1100
• Hence the quotient is 10.0111

41. Operation with 2’s comp
• Add 4 and -6
• Will use the 2’s complement of -6 or 1010
• 4 0100
• -6 1010 1010
• 1110
• And taking the 2’s complement of 1110 get 0001 + 1 = 0010

42. Overflow
• When adding 2 n-bit numbers it is possible to get a n+1 bit result if
there is a carry out.
• On paper it is easy just add another bit.
• In 2’s complement add a msb 0 for a positive or a msb 1 for a
negative.
• In a computer the number of bits that can be used is fixed

43. Overflow handling
• No overflow in signed
arithmetic
( carry) 111_
1110
+ 0011
0001
• -2 + 3 = 1
(correct)
• Signed addition in
4-bit arithmetic
( carry) 1__
0110
+ 0011
1001
• 6 + 3  -7
(false)

44. Overflow handling
• In signed arithmetic an overflow happens when
• The sum of two positive numbers exceeds the maximum positive value that
can be represented using n bits: 2n – 1 – 1
• The sum of two negative numbers falls below the minimum negative value
that can be represented using n bits: – 2n – 1

45. MULTIPLICATION

46. Decimal multiplication
(carry) 1_
37
x 12
74
370
444
• What are the rules?
• Successively multiply the multiplicand
by each digit of the multiplier starting
at the right shifting the result left by
an extra left position each time each
time but the first
• Sum all partial results

47. Binary multiplication
(carry)
111 _
1101
x 101
1101
00
110100
1000001
• What are the rules?
• Successively multiply the
multiplicand by each digit of the
multiplier starting at the right
shifting the result left by an extra
left position each time each time
but the first
• Sum all partial results
• Binary multiplication is easy!

48. Binary multiplication table
X 0 1
0 0 0
1 0 1

49. Example (I)
• Multiply 0011 by 0011
• Start
Multiplicand Multiplier Product
0011 0011 0000
• First addition
Multiplicand Multiplier Product
0011 0011 0011

50. DIVISION

51. Division
• Implemented by successive subtractions
• Result must verify the equality
Dividend = Multiplier× Quotient + Remainder

52. Decimal division (long division
• What are the rules?
• Repeatedly try to subtract smaller
multiple of divisor from dividend
• Record multiple (or zero)
• At each step, repeat with a lower
power of ten
• Stop when remainder is smaller
than divisor
303
7 2126
-210
26
-21
5

53. Binary division
• What are the rules?
• Repeatedly try to subtract powers of
two of divisor from dividend
• Mark 1 for success, 0 for failure
• At each step, shift divisor one position
to the right
• Stop when remainder is smaller than
divisor
011
11 1011
-11
1011
>-11
101
>>-11
10
X
X

54. Same division in decimal
• What are the rules?
• Repeatedly try to subtract powers of
two of divisor from dividend
• Mark 1 for success, 0 for failure
• At each step, shift divisor one position
to the right
• Stop when remainder is smaller than
divisor
2+1=3
3 11
-12
11
>-6
5
>-3
2
X
X

55. Normalizing binary numbers
• 0.1 becomes 1.0×2-1
• 0.01 becomes 1.0×2-2
• 0.11 becomes 1.1×2-1
• 1.1 is already normalized and equal to1.0×20
• 10.01 becomes 1.001×21
• 11.11 becomes 1______×2_____