ch3a-binary-numbers.ppt

1
Review on Number Systems
Decimal, Binary, and Hexadecimal

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
Decimal Number System
Base 10
Ten Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Example: 104510
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
Binary Number System
Base 2
Two Digits: 0, 1
Example: 10101102
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
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

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

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

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
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
Why choose 2’s complement?

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
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
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
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
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
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
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
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
Signed Numbers-8 bit example

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
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
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
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
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
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
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
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
“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
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
16’s Complement
FFF10
- B2CE
2
3
D
4

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
Why does 2’s complement
work?

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
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
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
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
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

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

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

ch3a-binary-numbers.ppt

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