1. Number Systems & Operations
Part II

2. Arithmetic Operations with Signed Numbers
Because the 2’s complement form for representing signed numbers is the most widely used in computer systems. We’ll limit to 2’s complement arithmetic on:
Addition
Subtraction
Multiplication
Division

3. Addition 4 cases that can occur when 2 signed numbers are added:
Both numbers positive
Positive number with magnitude larger than negative number
Negative number with magnitude larger than positive number
Both numbers negative

4. Addition Both numbers positive:
ex:
The sum is positive and is therefore in true (uncomplemented) binary.

5. Addition Positive number with magnitude larger than negative number:
ex:
The final carry bit is discarded. The sum is positive and is therefore in true (uncomplemented) binary.
Discard carry

6. Addition Negative number with magnitude larger than positive number:
ex:
The sum is negative and therefore in 2’s complement form.

7. Addition Both numbers negative:
ex:
The final carry bit is discarded. The sum is negative and therefore in 2’s complement form.
Discard carry

8. Addition
Remark:
The negative numbers are stored in 2’s complement form so, as you can see, the addition process is very simple: Add the two numbers and discard any final carry bit.

9. Addition Overflow condition:
When two numbers are added and the number of bits required to represent the sum exceeds the number of bits in the two numbers, an overflow results as indicated by an incorrect sign bit.
An overflow can occur only when both numbers are + or -.
ex:
Magnitude incorrect
Sign incorrect

10. Addition Numbers are added two at a time:
Computer add strings of numbers two numbers at a time.
ex: add the signed numbers: , , , and
Add 1st two numbers
st sum
Add 3rd number
nd sum
Add 4th number
Final sum

11. Subtraction Subtraction is a special case of addition.
Subtraction is addition with the sign of the subtrahend changed.
The result of a subtraction is called the difference.
The sign of a positive or negative binary is changed by taking it’s 2’s complement.
_______________________________
Subtrahend = ตัวลบ, Minuend = ตัวตั้ง

12. Subtraction
Since subtraction is simply an addition with the sign of the subtrahend changed, the process is stated as follows:
To subtract two signed numbers, take the 2’s complement of the subtrahend and add. Discard any final carry.

13. Subtraction
ex: Perform each of the following subtraction of the signed numbers:
(a) – (b) –
(c) – (d)
(a) (b)
(c) (d)

14. Multiplication
The multiplication operation in most computer is accomplished using addition.
There are 2 basic methods:
Direct addition
Partial products

15. Multiplication Direct addition
Add the multiplicand a number of times equal to the multiplier.
For example, 8x3 = = 24
Disadvantage  lengthy operations
When two binary numbers are multiplied, both numbers MUST BE in true (uncomplemented) form.
Let’s try multiply the signed binary numbers: and using direct addition.
_______________________________
Multiplier = ตัวคูณ, multiplicand = ตัวตั้ง

16. Multiplication Partial products
The multiplicand is multiplied by each multiplier digit beginning with the LSD.
239 multiplicand
x 123 multiplier
717
478

17. Multiplication Partial products
The result of the multiplication of the multiplicand by a multiplier digit is called a partial product.
239 multiplicand
x 123 multiplier
717 1st part prod
478 2nd part prod
rd part prod

18. Multiplication Partial products
Each successive partial product is shifted one place to the left and when all the partial products have been produced, they are added to get the final product.
239 multiplicand
x 123 multiplier
717 1st part prod
478 2nd part prod
rd part prod
29397 Final Product

19. Multiplication
The sign of the product of a multiplication depends on the signs of the multiplicand and the multiplier according to the following two rules:
If the signs are the same, the product is positive.
If the signs are different, the product is negative.

20. Multiplication
Step 1: Determine if the signs of the multiplicand and multiplier are the same or different. This determines what the sign of the product will be.
(multiplicand)
(multiplier)
Multiplicand  +
Multiplier  -
Product  -

21. Multiplication
Step 2: Change any negative number to true form. Because most computers store negative numbers in 2’s comp, a 2’s comp operation is required to get the negative number into true form.
(multiplicand)
(multiplier)
(2’s comp)

22. Multiplication
(multiplicand)
(multiplier) x
Step 3: Starting with the LSB (multiplier), generate the partial products. Shift each successive partial product one bit to the left.
Step 4: add each successive partial product to the sum of the previous partial products to get the final product.

23. Multiplication
(multiplicand)
(multiplier)
(final)
(2’s comp)
According to step 1 
the sign bit of the product must be 1 (negative).
Hence:
Step 4: if the sign bit was determined in step 1 is negative, take the 2’s comp of the product. If positive, leave the product in true form. Attach the sign bit to the product.

24. Division
The number in a division are the dividend, the devisor, and the quotient.
dividend = quotient
devisor
The division op in computer is accomplished using subtraction. Since subtraction is done with an adder, division can also be accomplished with an adder.

25. Division
The quotient is the number of times that the divisor will go into the dividend. This means that the divisor can be subtracted from the dividend a number of times equal to the quotient. (let’s do 21/7) 21
- 7 14 7
The divisor was subtracted from the dividend 3 times before a remainder of zero was obtained. Therefore, the quotient is 3.

26. Division
The sign of the quotient depends on the signs of the dividend and the divisor according to the following two rules:
If the signs are the same, the quotient is positive.
If the signs are different, the quotient is negative.
When 2 binary numbers are divided, both numbers MUST BE in true form.

27. Division
Step 1: Determine if the signs of the dividend and divisor are the same or different. This determines what the sign bit of the quotient will be. Initialize the quotient to zero.
(dividend)
(divisor)
Dividend  +
Divisor  +
Quotient  +

28. Division
Step 2: Subtract the divisor from the dividend using 2’s complement addition to get the first partial remainder and add 1 to the quotient.
If this partial remainder is positive, go to step 3.
If the partial remainder is zero or negative, the division is complete.
(dividend)
(divisor)
(2’s comp)
Quotient = =
Note: The final carries are discarded.

29. Division
Step 3: Subtract the divisor from the partial remainder and add 1 to the quotient.
If this partial remainder is positive, repeat for the next partial remainder.
If the result is zero or negative, the division is complete.
(dividend)
(divisor)
(2’s comp)
Quotient = =
Quotient = =
Stop when the result is zero (or negative)
Quotient = =
Quotient = 4

30. Hexadecimal and Octal Numbers

31. Hexadecimal Numbers We will call it for short as “hex”.
It has 16 characters. Digits 0-9 and letters A-F.
It used primarily as a compact way of displaying or writing binary numbers since it is very easy to convert between bin and hex.
Hex is widely used in computer and microprocessor applications.

32. Hexadecimal Numbers Decimal Binary Hexadecimal 0000 1 0001 2 0010 3
0000 1
0001 2
0010 3
0011 4
0100 5
0101 6
0110 7
0111 8
1000 9
1001 10
1010 A 11
1011 B 12
1100 C 13
1101 D 14
1110 E 15
1111 F

33. Hexadecimal Numbers
If you see ‘h’ mixing in numbers (in the context of computer systems), please note that it’s most likely that the numbers are hexadecimal numbers. (Be careful. ‘h’ is not one of A-F using in hex).
For example
16h =
0Dh =

34. Hexadecimal Numbers Bin-to-Hex Conversion
Simply break the binary number into 4-bit groups, starting at the right-most bit and replace each 4-bit group with the equivalent hex symbol.
(a) (b)
C A F
= CA = 3F16916

35. Hexadecimal Numbers Hex-to-Bin Conversion
Reverse the process (of bin-to-hex) and replace each hex symbol with the appropriate four bits.
ex: Determine the binary numbers for the following hex numbers:
(a) 10A4h (b) CF8Eh (c) 9742h
A C F E

36. Hexadecimal Numbers Hex-to-Dec Conversion 2 methods:
Hex-to-Bin first and then Bin-to-Dec.
Multiply the decimal values of each hex digits by its weight and then take the sum of these products.

37. Hexadecimal Numbers Hex-to-Dec Conversion
Hex-to-Bin first and then Bin-to-Dec
ex: Convert the following hex numbers to decimal:
(a) 1Ch
1Ch = = = 2810
(b) A85h
A85h = = =

38. Hexadecimal Numbers Hex-to-Dec Conversion
Multiply the decimal values of each hex digits by its weight and then take the sum of these products.
ex: Convert the following hex numbers to decimal:
(a) E5h
E5h = (Ex16)+(5x1) = (14x16)+5 = = 22910
(b) B2F8h
B2F8h = (Bx4096)+(2x256)+(Fx16)+(8x1)
= (11x4096)+(2x256)+(15x16)+(8x1)
= 45, = 45,81610

39. Hexadecimal Numbers Dec-to-Hex conversion
Repeated division of a dec number by 16
ex: Convert the dec number 650 to hex
650/16 = x16 = 10 = A
40/16 = x = 8 = 8
2/16 = x16 = 2 = 2
LSD
Hence = 28Ah
MSD
Stop when whole number quotient is ZERO.

40. Octal Numbers
Like the hex, the “oct” provides a convenient way to express binary numbers and codes. (btw, it’s not as commonly used as hex).
8 digits: 0-7
0,1,2,3,4,5,6,7,10,11,12,13,14,15,16,17,20,…
Operations we learn about hex so far work the same way on oct just mark this:
Hex = 4 binary bits
Oct = 3 binary bits
Now, let’s crack the following exercises…

41. Octal Numbers Bin-to-Oct Conversion Oct-to-Bin Conversion
(a) (b)
(c) (d)
Oct-to-Bin Conversion
(a) 138 (b) 258 (c) 1408 (d) 75268
Oct-to-Dec Conversion
(a) 23748
Dec-to-Oct Conversion
(a) 35910