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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
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
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
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
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
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
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
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
“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?
34
Sign Extension
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
40
Number Conversions
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
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
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
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
61C16
0
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

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ch3a-binary-numbers.ppt

  • 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 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. 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?
  • 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
  • 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
  • 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?
  • 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
  • 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 61C16 0
  • 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