number system

1. Number Systems
Decimal, Binary, Octal and
Hexadecimal
1

2. Base-N Number System
•Base N
•N Digits: 0, 1, 2, 3, 4, 5, …, N-1
•Example: 1045N
•Positional Number System
•
2
• Digit do is the least significant digit (LSD).
• Digit dn-1 is the most significant digit (MSD).
1 4 3 2 1 0
1 4 3 2 1 0
n
n
N N N N N N
d d d d d d



3. DECIMAL NUMBER SYSTEM
•Base 10
•Ten Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
•Example: 104510
•Positional Number System
•Digit d0 is the least significant digit (LSD).
•Digit dn-1 is the most significant digit (MSD).
3
1 4 3 2 1 0
1 4 3 2 1 0
10 10 10 10 1010
n
n
d d d d d d



4. BINARY NUMBER SYSTEM
•Base 2
•Two Digits: 0, 1
•Example: 10101102
•Positional Number System
•Binary Digits are called Bits
•Bit bo is the least significant bit (LSB).
•Bit bn-1 is the most significant bit (MSB).
4
1 4 3 2 1 0
1 4 3 2 1 0
2 2 2 2 2 2
n
n
b b b b b b



5. OCTAL NUMBER SYSTEM
•Base 8
•Two Digits: 0, 1,2,3,4,5,6,7
•Example: 2178
•Positional Number System
•Bit bo is the least significant bit (LSB).
•Bit bn-1 is the most significant bit (MSB).
5

6. DECIMAL NUMBER SYSTEM
•Base 10
•Ten Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
•Example: 104510
•Positional Number System
•Digit d0 is the least significant digit (LSD).
•Digit dn-1 is the most significant digit (MSD).
6
1 4 3 2 1 0
1 4 3 2 1 0
10 10 10 10 1010
n
n
d d d d d d



7. Hexadecimal Number System
•Base 16
•Sixteen Digits: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
•Example: EF5616
•Positional Number System
•
7
0000 0
0001 1
0010 2
0011 3
0100 4
0101 5
0110 6
0111 7
1000 8
1001 9
1010 A
1011 B
1100 C
1101 D
1110 E
1111 F
1 4 3 2 1 0
16 16 16 16 1616
n

8. 1’s Complements
• 1’s complement (or Ones’ Complement)
• To calculate the 1’s complement of a binary number just “flip” each bit of the
original binary number.
• E.g. 0  1 , 1  0
• 01010100100  10101011011
8

9. Why choose 2’s complement?
9

10. 2’s Complements
• 2’s complement
• To calculate the 2’s complement just calculate the 1’s complement, then add 1.
01010100100  10101011011 + 1=
10101011100
• Handy Trick: Leave all of the least significant 0’s and first 1 unchanged, and then
“flip” the bits for all other digits.
• Eg: 01010100100 -> 10101011100
10

11. Complements
• Note the 2’s complement of the 2’s complement is just the original
number N
• EX: let N = 01010100100
• (2’s comp of N) = M = 10101011100
• (2’s comp of M) = 01010100100 = N
11

12. Two’s Complement Representation
for Signed Numbers
1
0
1
1
1
1
0
1
1
0
1
0
1
,
1















 d
d
d
d
12
• Let’s introduce a notation for negative digits:
• For any digit d, define d = −d.
• Notice that in binary,
where d  {0,1}, we have:
• Two’s complement notation:
• To encode a negative number, we implicitly negate the leftmost (most significant) bit:
• E.g., 1000 = (−1)000
= −1·23 + 0·22 + 0·21 + 0·20 = −8

13. Negating in Two’s Complement
1
)
( 2
2 

 YZ
X
YZ
X
13
• Theorem: To negate
a two’s complement
number, just complement it and add 1.
• Proof (for the case of 3-bit numbers XYZ):
1
1
)
1
)(
1
(
1
11
100
)
1
(
)
(
2
2
2
2
2
2
2
2
2
















YZ
X
Z
Y
X
YZ
X
YZ
X
YZ
X
YZ
X
YZ
X
YZ
X

14. Signed Binary Numbers
• Two methods:
• First method: sign-magnitude
• Use one bit to represent the sign
• 0 = positive, 1 = negative
• Remaining bits are used to represent the magnitude
• Range - (2n-1 – 1) to 2n-1 - 1
where n=number of digits
• Example: Let n=4: Range is –7 to 7 or
• 1111 to 0111
14

15. Signed Binary Numbers
• Second method: Two’s-complement
• Use the 2’s complement of N to represent
-N
• Note: MSB is 0 if positive and 1 if negative
• Range - 2n-1 to 2n-1 -1
where n=number of digits
• Example: Let n=4: Range is –8 to 7
Or 1000 to 0111
15

16. Signed Numbers – 4-bit example
Decimal 2’s comp Sign-Mag
7 0111 0111
6 0110 0110
5 0101 0101
4 0100 0100
3 0011 0011
2 0010 0010
1 0001 0001
0 0000 0000
16
Pos 0

17. Signed Numbers-4 bit example
Decimal 2’s comp Sign-Mag
-8 1000 N/A
-7 1001 1111
-6 1010 1110
-5 1011 1101
-4 1100 1100
-3 1101 1011
-2 1110 1010
-1 1111 1001
-0 0000 (= +0) 1000
17

18. Signed Numbers-8 bit example
18

19. Notes:
•“Humans” normally use sign-magnitude
representation for signed numbers
• Eg: Positive numbers: +N or N
• Negative numbers: -N
•Computers generally use two’s-complement
representation for signed numbers
• First bit still indicates positive or negative.
• If the number is negative, take 2’s complement to
determine its magnitude
• Or, just add up the values of bits at their positions, remembering
that the first bit is implicitly negative.
19

20. Examples
•Let N=4: two’s-complement
•What is the decimal equivalent of
01012
Since MSB is 0, number is positive
01012 = 4+1 = +510
•What is the decimal equivalent of
11012 =
• Since MSB is one, number is negative
• Must calculate its 2’s complement
• 11012 = −(0010+1)= − 00112 or −310
20

21. Very Important!!! – Unless otherwise stated, assume two’s-
complement numbers for all problems, quizzes, HW’s, etc.
The first digit will not necessarily be
explicitly underlined.
21

22. Arithmetic Subtraction
• Borrow Method
• This is the technique you learned in grade school
• For binary numbers, we have
•
22
0 - 0 = 0
1 - 0 = 1
1 - 1 = 0
0 - 1 = 1 with a “borrow”
1

23. Binary Subtraction
• Note:
• A – (+B) = A + (-B)
• A – (-B) = A + (-(-B))= A + (+B)
• In other words, we can “subtract” B from A by “adding” –B to A.
• However, -B is just the 2’s complement of B, so to perform subtraction, we
• 1. Calculate the 2’s complement of B
• 2. Add A + (-B)
23

24. Binary Subtraction - Example
• Let n=4, A=01002 (410), and
B=00102 (210)
• Let’s find A+B, A-B and B-A
24
0 1 0 0
+ 0 0 1 0
 (4)10
 (2)10
0 11 0 6
A+B

25. Binary Subtraction - Example
25
0 1 0 0
- 0 0 1 0
 (4)10
 (2)10
10 0 1 0 2
A-B
0 1 0 0
+ 1 1 1 0
 (4)10
 (-2)10
A+ (-B)
“Throw this bit” away since n=4

26. Binary Subtraction - Example
26
0 0 1 0
- 0 1 0 0
 (2)10
 (4)10
1 1 1 0 -2
B-A
0 0 1 0
+ 1 1 0 0
 (2)10
 (-4)10
B + (-A)
1 1 1 02 = - 0 0 1 02 = -210

27. “16’s Complement” method
• The 16’s complement of a 16 bit Hexadecimal number is just:
=1000016 – N16
• Q: What is the decimal equivalent of B2CE16 ?
27

28. 16’s Complement
• Since sign bit is one, number is negative. Must calculate the 16’s
complement to find magnitude.
1000016 – B2CE16 = ?
• We have
10000
- B2CE
28

29. 16’s Complement
FFF10
- B2CE
29
2
3
D
4

30. 16’s Complement
• So,
1000016 – B2CE16 = 4D3216
= 4×4,096 + 13×256 + 3×16 + 2
= 19,76210
• Thus, B2CE16 (in signed-magnitude)
represents -19,76210.
30

31. Why does 2’s complement work?
31

32. Sign Extension
32

33. Sign Extension
• Assume a signed binary system
• Let A = 0101 (4 bits) and B = 010 (3 bits)
• What is A+B?
• To add these two values we need A and B to be of the same bit width.
• Do we truncate A to 3 bits or add an additional bit to B?
33

34. Sign Extension
• A = 0101 and B=010
• Can’t truncate A! Why?
• A: 0101 -> 101
• But 0101 <> 101 in a signed system
• 0101 = +5
• 101 = -3
34

35. Sign Extension
• Must “sign extend” B,
• so B becomes 010 -> 0010
• Note: Value of B remains the same
So 0101 (5)
+0010 (2)
--------
0111 (7)
35
Sign bit is extended

36. Sign Extension
• What about negative numbers?
• Let A=0101 and B=100
• Now B = 100  1100
36
Sign bit is extended
0101 (5)
+1100 (-4)
-------
10001 (1)
Throw away

37. Why does sign extension work?
•Note that:
(−1) = 1 = 11 = 111 = 1111 = 111…1
• Thus, any number of leading 1’s is equivalent, so long as the
leftmost one of them is implicitly negative.
•Proof:
111…1 = −(111…1) =
= −(100…0 − 11…1) = −(1)
•So, the combined value of any sequence of leading
ones is always just −1 times the position value of the
rightmost 1 in the sequence.
111…100…0 = (−1)·2n
37
n

38. Number Conversions
38

39. Decimal to Binary Conversion
39
Method I:
Use repeated subtraction.
Subtract largest power of 2, then next largest, etc.
Powers of 2: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2n
Exponent: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 , n
210 2n
29
28
20 27
21 22 23 26
24 25

40. Decimal to Binary Conversion
40
Suppose x = 156410
Subtract 1024: 1564-1024 (210) = 540  n=10 or 1 in the (210)’s position
Thus:
156410 = (1 1 0 0 0 0 1 1 1 0 0)2
Subtract 512: 540-512 (29) = 28  n=9 or 1 in the (29)’s position
Subtract 16: 28-16 (24) = 12  n=4 or 1 in (24)’s position
Subtract 8: 12 – 8 (23) = 4  n=3 or 1 in (23)’s position
Subtract 4: 4 – 4 (22) = 0  n=2 or 1 in (22)’s position
28=256, 27=128, 26=64, 25=32 > 28, so we have 0 in all of these positions

41. Decimal to Binary Conversion
41
Method II:
Use repeated division by radix.
2 | 1564
782 R = 0
2|_____
391 R = 0
2|_____
195 R = 1
2|_____
97 R = 1
2|_____
48 R = 1
2|_____
24 R = 0
2|__24_
12 R = 0
2|_____
6 R = 0
2|_____
3 R = 0
2|_____
1 R = 1
2|_____
0 R = 1

Collect remainders in reverse order
1 1 0 0 0 0 1 1 1 0 0

42. Binary to Hex Conversion
42
1. Divide binary number into 4-bit groups
2. Substitute hex digit for each group
1 1 0 0 0 0 1 1 1 0 0
0
Pad with 0’s
If unsigned number
61C16
Pad with sign bit
if signed number

43. Hexadecimal to Binary Conversion
Example
43
1. Convert each hex digit to equivalent binary
(1 E 9 C)16
(0001 1110 1001 1100)2

44. Decimal to Hex Conversion
44
Method II:
Use repeated division by radix.
16 | 1564
97 R = 12 = C
16|_____
6 R = 1
16|_____
0 R = 6

N = 61C 16

45. Definitions
•nybble = 4 bits
•byte = 8 bits
•(short) word = 2 bytes = 16 bits
•(double) word = 4 bytes = 32 bits
•(long) word = 8 bytes = 64 bits
•1K (kilo or “kibi”) = 1,024
•1M (mega or “mebi”) = (1K)*(1K) = 1,048,576
•1G (giga or “gibi”) = (1K)*(1M) = 1,073,741,824
45