The document discusses different number systems including binary, decimal, and hexadecimal. It defines base-N number systems and provides examples of decimal, binary, and hexadecimal numbers. Key concepts covered include positional notation, bits and bytes, addition and subtraction in different bases, signed numbers represented using sign-magnitude and two's complement, and arithmetic operations like addition and subtraction on signed binary numbers. Sign extension is introduced as an important concept for performing arithmetic on numbers of different bit-widths in a signed system.
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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
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
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
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)
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
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
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