The document discusses different number systems including binary, decimal, and hexadecimal. It explains that binary uses two digits (0,1), decimal uses ten digits (0-9), and hexadecimal uses sixteen digits (0-9 plus A-F). All of these systems are positional number systems where the value of each digit depends on its place value. The document then discusses binary addition and subtraction, two's complement representation for signed numbers, hexadecimal addition, and concepts like nibbles and bytes.

1. 1
Review on Number Systems
Decimal, Binary, and Hexadecimal

2. 2
Base-N Number System
Base N
N Digits: 0, 1, 2, 3, 4, 5, …, N-1
Example: 1045N
Positional Number System

• 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. 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).
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. 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).
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. 5
Definitions
nibble = 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

6. 6
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

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

7. 7
Binary Addition
•Single Bit Addition Table
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 10 Note “carry”

8. 8
Hex Addition
• 4-bit Addition
4 + 4 = 8
4 + 8 = C
8 + 7 = F
F + E = 1D Note “carry”

9. 9
Hex Digit Addition Table
+ 0 1 2 3 4 5 6 7 8 9 A B C D E F
0 0 1 2 3 4 5 6 7 8 9 A B C D E F
1 1 2 3 4 5 6 7 8 9 A B C D E F 10
2 2 3 4 5 6 7 8 9 A B C D E F 10 11
3 3 4 5 6 7 8 9 A B C D E F 10 11 12
4 4 5 6 7 8 9 A B C D E F 10 11 12 13
5 5 6 7 8 9 A B C D E F 10 11 12 13 14
6 6 7 8 9 A B C D E F 10 11 12 13 14 15
7 7 8 9 A B C D E F 10 11 12 13 14 15 16
8 8 9 A B C D E F 10 11 12 13 14 15 16 17
9 9 A B C D E F 10 11 12 13 14 15 16 17 18
A A B C D E F 10 11 12 13 14 15 16 17 18 19
B B C D E F 10 11 12 13 14 15 16 17 18 19 1A
C C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B
D D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C
E E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D
F F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E

10. 10
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

11. 11
Why choose 2’s complement?
if you try to subtract 4 from 6 (two positive numbers) you can 2's
complement 4 and add the two together 6 + (-4) = 6 - 4 = 2
This means that subtraction and addition of both positive and negative
numbers can all be done by the same circuit in the CPU..

12. 12
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

13. 13
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

14. 14
Two’s Complement Representation
for Signed Numbers
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
1
0
1
1
1
1
0
1
1
0
1
0
1
,
1















 d
d
d
d

15. 15
Negating in Two’s Complement
Theorem: To negate
a two’s complement
number, just complement it and add 1.
Proof (for the case of 3-bit numbers XYZ):
1
)
( 2
2 

 YZ
X
YZ
X
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

16. 16
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

17. 17
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

18. 18
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 Pos 0

19. 19
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

20. 20
Signed Numbers-8 bit example

21. 21
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.

22. 22
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

23. 23
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.

24. 24
Arithmetic Subtraction
Borrow Method
 This is the technique you learned in grade
school
 For binary numbers, we have

0 - 0 = 0
1 - 0 = 1
1 - 1 = 0
0 - 1 = 1 with a “borrow”
1

25. 25
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)

26. 26
Binary Subtraction - Example
Let n=4, A=01002 (410), and
B=00102 (210)
Let’s find A+B, A-B and B-A
0 1 0 0
+ 0 0 1 0
 (4)10
 (2)10
0 11 0 6
A+B

27. 27
Binary Subtraction - Example
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

28. 28
Binary Subtraction - Example
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

29. 29
“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 ?

30. 30
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

31. 31
16’s Complement
FFF10
- B2CE
2
3
D
4

32. 32
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.

33. 33
Why does 2’s complement
work?

34. 34
Sign Extension

35. 35
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?

36. 36
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

37. 37
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)
Sign bit is extended

38. 38
Sign Extension
 What about negative numbers?
 Let A=0101 and B=100
 Now B = 100  1100
Sign bit is extended
0101 (5)
+1100 (-4)
-------
10001 (1)
Throw away

39. 39
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
n

40. 40
Number Conversions

41. 41
Decimal to Binary Conversion
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

42. 42
Decimal to Binary Conversion
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

43. 43
Decimal to Binary Conversion
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

44. 44
Binary to Hex Conversion
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

45. 45
Hexadecimal to Binary Conversion
Example
1. Convert each hex digit to equivalent binary
(1 E 9 C)16
(0001 1110 1001 1100)2

46. 46
Decimal to Hex Conversion
Method II:
Use repeated division by radix.
16 | 1564
97 R = 12 = C
16|_____
6 R = 1
16|_____
0 R = 6

N = 61C 16