The document discusses computer arithmetic, focusing on number representation, arithmetic operations, and the components such as the Arithmetic Logic Unit (ALU). It covers various numbering systems including binary, octal, and hexadecimal, and explains conversion techniques between these systems and decimal. Additionally, it elucidates on signed numbers using methods like signed magnitude and two's complement, detailing binary addition and subtraction operations.

1. Computer Arithmetic
By
Roll No. 2,Roll No.10
Department of CS & IT
University of Azad Jammu & Kashmir, MZD
To
Prof. Dr. Wajid Aziz Loun
Chairman, DCS & IT/Prof. of Subject
Advance Computer Architecture
University of Azad Jammu & Kashmir, MZD

2. Presentation Topics
Part I: Number Representation
Part II: Addition / Subtraction
Part III: Multiplication
Part IV: Division

3. Arithmetic & Logic Unit
• Does the calculations
• Handles integers
• May handle floating point (real) numbers

4. Integer Representation
• In binary number system, an arbitrary number can be represented with
– Zero and one
– The minus sign
– And period, or radix point
– e.g.
• For computer storage and processing
– Only have 0 & 1 to represent everything
– Positive numbers stored in binary. (For example 41=00101001)
– No minus sign
– No period
• In general, for n-bit sequence of binary digits an-1 an-2 … a1, ao interpreted
as unsigned integer A will have value
å-
=
=
1
0
2
n
i
i
A i a

5. What’s ALU?
1. ALU stands for: Arithmetic Logic Unit
2. ALU is a digital circuit that performs
Arithmetic (Add, Sub, . . .) and Logical (AND,
OR, NOT) operations.
3. John Von Neumann proposed the ALU in
1945 when he was working on EDVAC.
Computer Arithmetic, Number
Representation

6. Number Systems
and Binary Arithmetic

7. Introduction to Numbering Systems
• We are all familiar with the decimal number
system (Base 10). Some other number
systems that we will work with are:
– Binary ® Base 2
– Octal ® Base 8
– Hexadecimal ® Base 16

8. Characteristics of Numbering Systems
1) The digits are consecutive.
2) The number of digits is equal to the size of the
base.
3) Zero is always the first digit.
4) When 1 is added to the largest digit, a sum of zero
and a carry of one results.
5) Numeric values determined by the have implicit
positional values of the digits.

9. Significant Digits
Binary: 11101101
Most significant digit Least significant digit
Hexadecimal: 1D63A7A
Most significant digit Least significant digit

10. Binary Number System
• Also called the “Base 2 system”
• The binary number system is used to model the
series of electrical signals computers use to
represent information
• 0 represents the no voltage or an off state
• 1 represents the presence of voltage or an
on state

11. Binary Numbering Scale
Base 2
Number
Base 10
Equivalent Power Positional
Value
000 0 20 1
001 1 21 2
010 2 22 4
011 3 23 8
100 4 24 16
101 5 25 32
110 6 26 64
111 7 27 128

12. Binary Addition
4 Possible Binary Addition Combinations:
(1) 0 (2) 0
+0 +1
00 01
Carry Sum
(3) 1 (4) 1
+0 +1
01 10
Note that leading
zeroes are frequently
dropped.

13. Decimal to Binary Conversion
• The easiest way to convert a decimal number to its
binary equivalent is to use the Division Algorithm
• This method repeatedly divides a decimal number by
2 and records the quotient and remainder
– The remainder digits (a sequence of zeros and ones) form
the binary equivalent in least significant to most
significant digit sequence

14. Division Algorithm
Convert 67 to its binary equivalent:
6710 = x2
Step 1: 67 / 2 = 33 R 1 Divide 67 by 2. Record quotient in next row
Step 2: 33 / 2 = 16 R 1 Again divide by 2; record quotient in next row
Step 3: 16 / 2 = 8 R 0 Repeat again
Step 4: 8 / 2 = 4 R 0 Repeat again
Step 5: 4 / 2 = 2 R 0 Repeat again
Step 6: 2 / 2 = 1 R 0 Repeat again
Step 7: 1 / 2 = 0 R 1 STOP when quotient equals 0
1 0 0 0 0 1 12

15. Binary to Decimal Conversion
• The easiest method for converting a binary
number to its decimal equivalent is to use the
Multiplication Algorithm
• Multiply the binary digits by increasing powers
of two, starting from the right
• Then, to find the decimal number equivalent,
sum those products

16. Multiplication Algorithm
Convert (10101101)2 to its decimal equivalent:
Binary 1 0 1 0 1 1 0 1
Positional Values x x x x x x x x
27 26 25 24 23 22 21 20
Products 128 + 32 + 8 + 4 + 1
17310

17. Octal Number System
• Also known as the Base 8 System
• Uses digits 0 - 7
• Readily converts to binary
• Groups of three (binary) digits can be used to
represent each octal digit
• Also uses multiplication and division
algorithms for conversion to and from base 10

18. Decimal to Octal Conversion
Convert 42710 to its octal equivalent:
427 / 8 = 53 R3 Divide by 8; R is LSD
53 / 8 = 6 R5 Divide Q by 8; R is next digit
6 / 8 = 0 R6 Repeat until Q = 0
6538

19. Octal to Decimal Conversion
Convert 6538 to its decimal equivalent:
6 5 3
x x 2 81 80
x8
384 + 40 + 3
42710
Octal Digits
Positional Values
Products

20. Octal to Binary Conversion
Each octal number converts to 3 binary digits
To convert 6538 to binary, just
substitute code:
6 5 3
110 101 011

21. Hexadecimal Number System
• Base 16 system
• Uses digits 0-9 &
letters A,B,C,D,E,F
• Groups of four bits
represent each
base 16 digit

22. Decimal to Hexadecimal Conversion
Convert 83010 to its hexadecimal equivalent:
830 / 16 = 51 R14
51 / 16 = 3 R3
3 / 16 = 0 R3
33E16
= E in Hex

23. Hexadecimal to Decimal Conversion
Convert 3B4F16 to its decimal equivalent:
Hex Digits
3 B 4 Fx
x x
163 162 161 160
12288 +2816 + 64 +15
15,18310
Positional Values
Products
x

24. Binary to Hexadecimal Conversion
• The easiest method for converting binary to
hexadecimal is to use a substitution code
• Each hex number converts to 4 binary digits

25. Substitution Code
Convert 0101011010101110011010102 to hex using the
4-bit substitution code :
5 6 A E 6 A
0101 0110 1010 1110 0110 1010
56AE6A16

26. Substitution Code
Substitution code can also be used to convert binary to
octal by using 3-bit groupings:
2 5 5 2 7 1 5 2
010 101 101 010 111 001 101 010
255271528

27. Complementary Arithmetic
• 1’s complement
– Switch all 0’s to 1’s and 1’s to 0’s
Binary # 10110011
1’s complement 01001100

28. Complementary Arithmetic
• 2’s complement
– Step 1: Find 1’s complement of the number
Binary # 11000110
1’s complement 00111001
– Step 2: Add 1 to the 1’s complement
00111001
+ 00000001
00111010

29. Signed Magnitude Numbers
110010.. …00101110010101
Sign bit
0 = positive
1 = negative
31 bits for magnitude
This is your basic
Integer format

30. Sign-Magnitude
• Left most bit is sign bit
– 0 means positive
– 1 means negative
• +18 = 00010010
• -18 = 10010010
• The general case can be expressed as follows
• Problems
ü
=
å
a if a
2 0
-
– Need to consider both sign and magnitude in arithmetic
– Two representations of zero (+0 and -0)
+010 = 00000000
-010 = 10000000
ï ïþ
ï ïý
ï ïî
ï ïí ì
- =
=
å
=
-
-
=
-
2
0
1
2
0
1
2 1
n
i
i n
i
n
i
i n
i
a if a
A

31. Negation (Sign Magnitude Representation)
• In sign magnitude representation, the sign bit
is inverted for forming negative of an integer.
– For 8 bit sign magnitude representation
+1810 = 00010010
-1810 = 10010010
– For 16 bit sign magnitude representation
+1810 =0000000000010010
-1810 =1000000000010010

32. Two’s Compliment
• The general expression for two’s complement
representation is
A n a a
• +3 = 00000011
• +2 = 00000010
• +1 = 00000001
• +0 = 00000000
• -1 = 11111111
• -2 = 11111110
• -3 = 11111101
å-
2 1 2
=
= - - +
-
2
0
1
n
i
i
i
n

33. Negation (Twos Complement Representation)
• In twos complement representation, negation of
an integer can be formed with following rules.
– Take the Boolean complement of each bir of integer
(including sign bit).
– Add 1 treating the result as an unsigned integer.
• The two step process is referred to as twos
complement operation.
+18=00010010
Bitwise Complement =11101101
+ 1
11101110= -18

34. Negation Special Case 1
• The negation of 0 is 0.
0 = 00000000
Bitwise not 11111111
Add 1 to LSB +1
Result 100000000=0
• Overflow is ignored, so:
• The result of negation of 0 is 0.

35. Negation Special Case 2
• The negation of the bit pattern followed by n-1
zeros results back the same number.
-128 = 10000000
Bitwise not 01111111
Add 1 to LSB +1
Result 10000000=-128
• Monitor MSB (sign bit)
• It should change during negation
• Such anomaly is unavoidable.

36. Value Box for Conversion between Twos Complement Binary
and Decimal
-128 64 32 16 8 4 2 1
An Eight Position Twos Complement Value Box
-128 64 32 16 8 4 2 1
1 0 0 0 0 0 1 1
-128 +2 +1 =-125
Converting Binary 10000011 to Decimal
-128 64 32 16 8 4 2 1
1 0 0 0 1 0 0 0
-128 +8 =-120
Converting Decimal -120 to Binary

37. Conversion Between Lengths
• Positive number pack with leading zeros
+18 = 00010010
+18 = 00000000 00010010
• Negative numbers pack with leading ones
-18 = 10010010
-18 = 11111111 10010010
• i.e. pack with MSB (sign bit)

38. Addition and Subtraction
• Normal binary addition
• Monitor sign bit for overflow
• Take twos compliment of subtrahend and add
to minuend
– i.e. a - b = a + (-b)
• So we only need addition and complement
circuits

39. Addition and Subtraction (Examples)
• Additionn of Numbers in Twos Complement
(a) (+3) + (+4)
0011
+ 0100
0111 = 7
(b) (-7) + (+5)
1001
+ 0101
1110 =-2
(c) (-4) + (-1)
1100
+ 1111
11011 =-5
(d) (-4) + (+4)
1100
+0100
10000 =0
(e) (+5) + (+4)
0101
+0100
1001 =Overflow
(f) (-7) + (-6)
1001
+1010
10011 =Overflow

40. Addition and Subtraction (Examples)
• Subtraction of Numbers in Twos Complement
(a) (+2) – (+7)
0010
+ 1001
1011 = -5
(b) (+5) - (+2)
0101
+ 1110
1011 =3
(c) (-5) - (+2)
1011
+ 1110
11001 =-7
(d) (+5) - (-2)
0101
+0010
0111 =7
(e) (+7) - (-7)
0111
+0111
1100 =Overflow
(f) (-6) - (+4)
1010
+1100
10110 =Overflow

41. Hardware for Addition and Subtraction